Chemistry Part I Chapter 5 States Of Matter
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    NCERT Solution For Class 11 Chemistry Chemistry Part I

    States Of Matter Here is the CBSE Chemistry Chapter 5 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Chemistry States Of Matter Chapter 5 NCERT Solutions for Class 11 Chemistry States Of Matter Chapter 5 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Chemistry.

    Question 1
    CBSEENCH11005820

    What does the abbreviation STP stand for?

    Solution

    STP stands for standard temperature and pressure.

    Question 2
    CBSEENCH11005821

    What is atmospheric pressure?

    Solution

    The atmospheric pressure is defined as the weight exerted by the overhead atmosphere on a unit area of surface at normal temperature. 

    Question 3
    CBSEENCH11005822
    Question 4
    CBSEENCH11005823

    What is absolute temperature?

    Solution
    It is the temperature measured on the absolute or kelvin scale which begins with –273°C as the lowest temperature.
    Question 6
    CBSEENCH11005825

    What is the absolute zero temperature?

    Solution
    It is the temperature at which the volume of any gas is reduced to zero. It is taken as -273°C on the celsius scale.
    Question 7
    CBSEENCH11005826

    Can absolute zero temperature be attained for a gas?

    Solution
    No, the temperature of absolute zero (i.e. –273°C) cannot be attained for a gas since the gas changes to the liquid state much before this temperature is achieved.
    Question 8
    CBSEENCH11005827

    How pressure of a given sample of gas is related to absolute temperature at constant volume?

    Solution
    A pressure of a given sample of gas is directly proportional to absolute temperature (P ∝ N).
    Question 9
    CBSEENCH11005828

    How is the pressure of a gas related to the number of molecules of the gas at constant temperature and volume?

    Solution
    Pressure of a gas is directly proportional to the number of molecules of the gas (P ∝ N).
    Question 10
    CBSEENCH11005829

    What is standard (or normal) temperature and pressure (STP)?

    Solution

    Standard temperature is 273.15 K  (0°C). Standard pressure is one bar or 760 mm.

    Question 11
    CBSEENCH11005830

    What does SATP stand for? Define it.

    Solution

    SATP stands for standard ambident temperature and pressure. In SATP conditions, the standard pressure is taken as 1 bar or 105 Pa and standard temperature is taken as 298·15 K.

    Question 12
    CBSEENCH11005831

    What is the value of molar volume at STP?

    Solution
    The molar volume at SATP is 22·7L mo–1.
    Question 13
    CBSEENCH11005832

    What is standard molar volume?

    Solution
    The volume occupied by one mole of any gas under STP (1 bar pressure and 273K) is 22·4L. This volume is called standard molar volume.
    Question 15
    CBSEENCH11005834

    What is the value of gas constant in SI units?

    Solution
    R (gas constant) = 8·314 JK–1 mol–1.
    Question 16
    CBSEENCH11005835

    What would be the SI unit for the quantity fraction numerator PV squared straight T squared over denominator straight n end fraction ?

    Solution

    The unit of pressure is given as= Nm-2
    The unit of volume is given as= m3
    The unit of Temperature is given as= K2
    Unit of mol is givne as = mol-1.
    Thus SI unit of given quantity is given as,
    fraction numerator left parenthesis Nm to the power of negative 2 end exponent right parenthesis thin space left parenthesis straight m cubed right parenthesis squared space left parenthesis straight K right parenthesis squared over denominator mol end fraction space equals space Nm to the power of 4 straight K squared mol to the power of negative 1 end exponent

    Question 17
    CBSEENCH11005836

    What is meant by aqueous tension?

    Solution
    The partial pressure of water vapours at a particular temperature is known as aqueous tension.

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    Question 18
    CBSEENCH11005837

    What is the nature of gas consant R?

    Solution

    R represent work done per degree per mole.

    Question 19
    CBSEENCH11005838

    In terms of Charles’s law explain why –273°C is the lowest possible temperature.

    Solution

    The law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. It was found that for all gases (at any given pressure), the plots of volume vs. temperature is a straight line. If this line is extended to zero volume, then it intersects the temperature axis at -2730C.

    In other words, the volume of any gas at 273 0C is zero. This is because all gases get liquefied before reaching a temperature of 2730C.

    Hence, it can be concluded that -273 0C is the lowest possible temperature

    Question 20
    CBSEENCH11005839

    Can Dalton's law of partial pressure be applicable to a mixture of ammonia and hydrochloric acid gas?

    Solution
    No, it can not be applied since the two gases react to form ammonium chloride.
    Question 21
    CBSEENCH11005840
    Question 23
    CBSEENCH11005842

    How will you account for the fact that gases occupy the whole of the volume available to them?

    Solution

    Due to the absence of attractive forces between molecules, the molecules of gases can easily separate from one another.

    Question 25
    CBSEENCH11005844

    The tyre of an automobile is inflated to a slightly lesser pressure in summer than in winter. Why?

    Solution

    As the automobile moves, the temperature of the tyre increases due to friction against the road. Due to increase in temperature, the air inside it expands and pressure against the walls of the tyre increases and the tyre may burst.

    Question 26
    CBSEENCH11005845

    The size of the weather ballon becomes larger and larger as it ascends to higher altitudes. Explain.

    Solution
    As the weather balloon ascends, the pressure tends to decrease. Consequently, the volume of the gas inside the balloon is expected to increase. This increases the size of the balloon.
    Question 27
    CBSEENCH11005846

    Why a gas can be easily compressed?

    Solution
    A gas can be compressed easily due to the presence of intermolecular spaces called voids. When pressure is applied, the number of such like spaces decreases and the gas molecules come closer.
    Question 28
    CBSEENCH11005847

    What change is expected if the temperature is kept constant but pressure is reduced to one fourth of its original value?

    Solution

    According to the Boyle's law, at constant Temperature (T), a volume is inversely proportional to pressure. Such as, Vα 1 over straight P

    Therefore, the volume of gas will become four times its original value.

    Question 29
    CBSEENCH11005848

    How do you find the pressure of dry gas collected over water?

    Solution

    By deducting the aqueous tension at that temperature from the observed pressure of the gas. 
    space space space space straight P subscript dry space gas end subscript space equals space straight P subscript obs space minus space straight P subscript straight H subscript 2 straight O end subscript
where space straight P subscript straight H subscript 2 straight O end subscript space equals space aq. space tension. space

    Question 30
    CBSEENCH11005849

    How are kinetic energy and absolute temperature related?

    Solution
    Kinetic energy is directly proportional to temperature.
    KE ∝ T.
    Question 32
    CBSEENCH11005851

    Define Boltzman constant.

    Solution

    The Boltzmann constant k, is a physical constant relating energy at the individual particle level with temperature. It is  the gas constant R divided by the  Avogadro constant NA.

    open parentheses straight k space equals space straight R over straight N close parentheses.

    Question 33
    CBSEENCH11005852

    Give the numerical value of Boltzmann constant.

    Solution
    k = 1·38 × 10–23J deg–1 mol–1.
    Question 34
    CBSEENCH11005853

    When do gases deviate more from the ideal behaviour?

    Solution
    At low temperature and high pressure, gases deviate more from the ideal behaviour.
    Question 35
    CBSEENCH11005854

    What is Boyle's Temperature?

    Solution

    It is the temperature at which Non-ideal gas exhibits almost ideal behaviour for a considerable range of pressure. Boyle's temperature (TB) is related to Vander Waal’s constant a, b as follows:
    straight T subscript straight B space equals space straight a over Rb

    Question 36
    CBSEENCH11005855

    What is compressibility factor? What is its value for an ideal gas?

    Solution
    Compressibility space factor comma space straight Z space equals space PV over nRT
    For an ideal gas, compressibility factor is (Z)= 1.
    Question 37
    CBSEENCH11005856

    Write vander Waals equation for n moles of a gas.

    Solution
    open parentheses straight P plus an squared over straight V squared close parentheses space space left parenthesis straight V minus nb right parenthesis space equals space nRT
    is the Vander Waals equation for n moles of a gas.
    Question 39
    CBSEENCH11005858

    Critical temperature for carbon dioxide (CO2) and methane (CH4) are 31·1°C and –81 ·9°C respectively. Which of these has stronger intermolecular forces and why?

    Solution
    We know that higher the critical temperature, more easily the gas can be liquefied i.e. greater are intermolecular forces of attraction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Since CO2 has greater critical temperature than CH4, therefore CO2 has stronger intermolecular forces than CH4.

     

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    Question 40
    CBSEENCH11005859

    How is pressure caused by gas molecules?

    Solution
    The rapid motion and collisions of molecules with the walls of the container causes pressure. Pressure is proportional to the number of molecular collisions and the force of the collisions in a particular area. The more collisions of gas molecules with the walls, the higher the pressure.
    Question 41
    CBSEENCH11005860
    Question 42
    CBSEENCH11005861

    Why there is no change in temperature when an ideal gas expands?

    Solution

    In an ideal gas, there is no attraction between the molecules of a gas. Therefore, when it expands, it does no work against the intermolecular attraction. 


    Question 43
    CBSEENCH11005862

    Find the density of a gas with molecular mass 44.

    Solution

    A weight of 1L of gas at NTP is its density. 
    space space because space space Weight space of space 22.4 space straight L space of space gas space at space NTP space is space 44 space straight g
space space therefore space space space Weight space of space 1 straight L space of space gas space at space NTP space is space fraction numerator 44 over denominator 22.4 end fraction cross times 1 space equals space 2.858 space straight g

    Question 44
    CBSEENCH11005863

    Do all the molecules of the same gas have the same velocity? If not, give a reason for it ?

    Solution
    All the molecules are having different velocities. We measure only the average.
    Question 45
    CBSEENCH11005864

    Define critical temperature of a gas.

    Solution

    It is the highest temperature at which a gas can be liquefied by increasing pressure. It is denoted by Tc.

    Question 46
    CBSEENCH11005865

    Define critical pressure.

    Solution
    It is the minimum pressure required to liquefy a gas at its critical temperature. It is denoted by Pc.
    Question 47
    CBSEENCH11005866

    Define critical volume.

    Solution

    It is the volume per mole of a gas at its critical temperature and critical pressure. It is denoted by Vc

    Question 48
    CBSEENCH11005867

    Define evaporation.

    Solution

    Evaporation is the process of the escape of liquid molecules from its surface into the gas phase. It depends on upon the nature of liquid, temperature and surface area of the liquid.

    Question 49
    CBSEENCH11005868

    Define vapour pressure.

    Solution

    It is the pressure exerted by the vapours of the liquid in equilibrium with the liquid at particular temperature. A liquid having weaker intermolecular interaction has lower vapour pressure.

    Question 50
    CBSEENCH11005869

    Define boiling point of liquid.

    Solution
    It is the temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure.
    Question 51
    CBSEENCH11005870

    The heat of vaporisation of water at 100°C is 40·6 kJ per mole and that of benzene at 80°C is 31kJ per mole. Which out of the two has stronger attractive forces ?

    Solution

    Attractive forces between water molecules are stronger than between benzene molecules. Since the energy required to vaporisation of  the water is more than that of benzene.

    Question 52
    CBSEENCH11005871

    Why vegetables are cooked with difficulty at a hill station?

    Solution
    The atmospheric pressure is less and so the boiling point is lowest.
    Question 53
    CBSEENCH11005872

    Arrange water, benzene and acetone in order of increasing boiling point. 

    Solution
    Order of increasing boiling point is,
    Acetone < Benzene < Water.
    Question 54
    CBSEENCH11005873

    Vapour pressure of ether and alcohol is higher than water. Why?

    Solution
    This is due to the fact that there are weaker attractive forces in ether and alcohol as compared to water. 
    Question 55
    CBSEENCH11005874

    Arrange ether, water and ethyl alcohol in order of decreasing intermolecular attractive forces.

    Solution
    Order of decreasing intermolecular attractive forces.
    Water > Ethyl alcohol > Ether.
    Question 56
    CBSEENCH11005875

    What is the difference between a normal boiling point and standard boiling point ?

    Solution
    When the external pressure is equal to one atmospheric pressure, the boiling point is called normal boiling point. When the external pressure is 1 bar, it is called standard boiling point.   
    Question 57
    CBSEENCH11005876

    Name two compounds which lower the surface tension of water.

    Solution

    Alcohol and camphor lower the surface tension of water.

    Question 58
    CBSEENCH11005877

    Insects can walk on the surface of water. Why?

    Solution

    This is due to the existence of surface tension. The surface tension makes water surface to behave like an elastic membrane and prevents the insects from drowning.

    Question 59
    CBSEENCH11005878

    Why can a glass be filled above its rim with water?

    Solution
    This is due to the existence of surface tension. The water piles up, trying to assume a spherical shape. However, if too much water is added to the glass, the force of gravity dominates over surface tension and the ‘skin’ breaks and water overflows.
    Question 60
    CBSEENCH11005879

    What is the effect of temperature on: (i) surface tension (ii) viscosity.

    Solution

    Surface tension and Viscosity both decrease with increase in temperature.

    Question 61
    CBSEENCH11005880

    What are SI units of surface tension and coefficient of viscosity?

    Solution
    SI unit of surface tension is Nm–1.
    SI unit of coefficient of viscosity is kg m-1 s-1.
    Question 62
    CBSEENCH11005881

    Viscosity of liquids decreases as the temperature rises. Why?

    Solution

    This is because, at high temperature, molecules have high kinetic energy and can overcome the intermolecular forces to slip fast one another between the layers.

    Question 63
    CBSEENCH11005882

    What are intermolecular forces?

    Solution

    These are the weak attractive forces existing in gases, liquids, and solids. For examples,
    (i) Dispersion forces or London forces. 
    (ii) Dipole- dipole interactions .
    (iii) Dipole-induced dipole interactions. 
    (iv) Hydrogen bond -stronger intermolecular interaction.

    Question 64
    CBSEENCH11005883

    Name two solids which directly change into vapours on heating.

    Solution

    Iodine, ammonium chloride.

    Question 65
    CBSEENCH11005884

    Does the volume of a solid change on slight heating?

    Solution

    No, there is no change in the volume of a solid on slight heating.

    Question 66
    CBSEENCH11005885

    What are characteristic properties of gases?

    Solution

    Properties of gases:
    The gaseous state is the simplest of the three physical states and shows the greatest regularity in behaviour. The chief characteristic properties of gases are:
    1. A gas neither possesses a definite volume nor a definite shape.
    2. All the parts of a gas or a gaseous mixture have similar composition throughout.

    3. Gases can expand indefinitely to fill up all the available space.

    4. The molecules or atoms of a gas are in a state of continuous zig-zag motion in all directions.

    5. Two or more gases mix swiftly and easily to give a homogeneous mixture.

    6. On cooling and applying pressure gases can be liquefied.

    7. The gases exert pressure on the walls of the container as a result of collisions of the particles against the walls of the container.

    Question 67
    CBSEENCH11005886

    How are the following properties of a gas measured:
    (i) Mass    (ii) Volume?

    Solution

    (i) Mass: The mass of a gas can be determined by weighing the container containing the gas and then emptying the container by taking out the gas and weighing the empty container gain. The difference between the two weights gives the mass of the gas. The mass of the gas can be converted into moles by using the following relation:
    Number space of space moles space space equals space fraction numerator Mass space in space grams over denominator Molar space mass end fraction
    The moles can be converted into number of molecules by using the relation
    1 Mole = 6·023 × 1023 molecules.

    (ii) Volume: Since a gas occupies the entire space of the container available to it, the volume of the gas is the same as the volume of the container.
    In S.I. units volume is measured in cubic metre (m3). 1m3 = 103 dm3 = 106 cm3

    Question 68
    CBSEENCH11005887

    How pressure of a gas is determined experimentally?

    Solution

    The pressure of a gas is measured with the help of an instrument called manometer. Two types of manometers are used:
    (i) Open end manometer
    (ii) Closed end manometer.

    (i) Open end manometer. It consists of a V-shaped tube with one limb shorter than the other. It is partially filled with mercury. The mercury in the longer limb is subjected to the pressure of the gas in the vessel. Thus, the difference in the levels of mercury in the two limbs represents the difference between atmospheric pressure and gas pressure.
        straight P subscript difference space in space levels space of space Hg end subscript space equals space straight P subscript atm space minus space straight P subscript gas

therefore space space straight P subscript gas space equals space straight P subscript atm space minus space straight P subscript difference space in space columns space of space Hg end subscript
    Thus, by knowing the atmospheric pressure, gas pressure can be determined.

    (ii) Closed end manometer: It consists of a V-shaped tube with one limb shorter than the other. It is partially filled with mercury. The shorter limb is connected to the vessel containing,

    the gas whereas the longer limb is closed. In this manometer, the downward pressure exerted by the excess height of mercury column in the closed limb is balanced by the pressure of the gas in the vessel. Since there is a nearly complete vacuum in the closed longer limb, the difference in the levels of mercury in the two limbs directly gives the pressure of the gas.

    Question 69
    CBSEENCH11005888

    What are the units of pressure?

    Solution

    Pressure represents force per unit area. The pressure expressed in terms of height of mercury column can be converted into the units of force per unit area.
    Let a mercury column h cm high and A cm2 in ara of cross-section exert a downward force equal to the weight of mercury column. Therefore,  per unit area of the surface is P.

    straight P space equals space fraction numerator Force space over denominator Area end fraction space equals space fraction numerator Mass space cross times space Acceleration over denominator Area end fraction space equals space mg over straight A
    where m = mass of mercury in the column
              g = acceleration due to gravity
     But mass = density x volume
                   therefore space space space straight P space equals space ρVg over straight A

    Where ρ and V represent density and volume of mercury respectively.
    But space volume space equals space space area space cross times space height space equals space straight A space cross times space straight h
therefore space space space space space straight P space equals space ρAhg over straight A space equals space ρhg


    Standard atmospheric pressure. A standard pressure of one atmosphere (1 atm) is defined as the pressure exerted by a column of mercury 76 cm high at 273K (density of mercury is 13·595 g cm–3) and at standard gravity (i.e. 9·81 cm–2). That is,
    1 atm = 76·0 cm of Hg = 760 mm of Hg = 760 torr.
    The S.I. unit of pressure is pascal (Pa). It is defined as the pressure exerted when a force of 1 newton (1N) acts on a 1 m2 area. Pascal is related to atmosphere or bar as
    1 bar =1 atm = 1 ·0133 × 105 Nm–2
    = 1 ·0133 × 105 Pa = 101·33 k Pa
    However, for approximate work,
    1 bar 1 atm = 102 k Pa = 105 Pa

    Question 71
    CBSEENCH11005890

    Why mercury is generally used as a liquid in barometer?

    Solution

    It is because of the following reasons:
    (i) Mercury is non-volatile at ordinary temperature, therefore, the pressure exerted by the vapours above the mercury column is very small and can be neglected.
    (ii) The height of the column in a barometer is inversely proportional to the density of the liquid. Since mercury is very dense, it is best for barometer as it supports column of convenient height.

    Question 72
    CBSEENCH11005891

    State Boyle's law. How this law is expressed mathematically?

    Solution
    It states: “At constant temperature, the volume of a given mass of a gas is inversely proportional to its pressure. ” Mathematical representation: It may be expressed mathematically as
    straight V proportional to 1 over straight P space left square bracket at space constant space temp. space for space straight a space given space mass space of space gas right square bracket
or space space space straight V space equals space straight k over straight p space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket where space straight k space is space straight a space constant right square bracket
or space PV space equals space straight k left parenthesis constant right parenthesis
    where V is the volume of a given mass of a gas and P is the pressure.
    For a gas having pressure P1 and V1
    P1V1 = constant               ...(1)
    If the pressure changes from P1 to P2 and volume changes from V1 to V2, then
    P2V2 = constant ...(2)
    From (1) and (2),
    P1 V1 = P2V2 (Temperature constant) Hence, Boyle’s law may also be stated as:
    “At a constant temperature, the product of the pressure and volume of a given mass of gas is constant.”
    Question 73
    CBSEENCH11005892

    How will you represent the Boyle's law graphically in different ways?

    Solution

    This law can be represented in three ways:
    (i) Graph between P (pressure) and V (volume):

    If a graph is plotted between P and V of gas at a particular temperature, a curve of the form of the regular hyperbola is obtained. This shows that volume is inversely proportional to pressure. 
    (ii) Graph between P (pressure) and 1 over straight V : (1/volume):

    If a graph is plotted between P and 1 over straight V comma space straight a
    a straight line is obtained which shows that pressure and volume are inversely related to each other. 
    (iii) Graph between P(pressure) and PV (Product of pressure and volume):

    If a graph is plotted between P and PV, a horizontal straight line is obtained. This illustrates the relationship
    PV = constant.
    These curves plotted at particular temperature are known as isotherms.


            
     

    Question 74
    CBSEENCH11005893

    Why does the volume of a gas decrease on increasing pressure?

    Solution

    Gas molecules have space in between them. When the pressure is increased, the molecules of the gas are forced  to come closer and as a result, the volume occupied by the gas decreases.

    Question 75
    CBSEENCH11005894

    A weather balloon has a volume of 175 L when filled with hydrogen at a pressure of 1.00 bars. Calculate the volume of the balloon when it rises to a height of 2000m, where the atmospheric pressure is 0.80 bar. Assume that the temperature is constant ?

    Solution
    From the given data,
    V1 = 175 L ;
    V2 = ?
    P1 = 1·00 bar
    P2 = 0·80 bar
    According to Boyle’s law,
    P1V1 = P2V

    Substituting the values,
    we have,
    1.0 space cross times space 175 space equals space 0.80 space cross times space straight V subscript 2
or space space space space straight V subscript 2 space equals space fraction numerator 1.0 space cross times space 175 over denominator 0.80 end fraction space equals space 218. space 75 space straight L
    Question 76
    CBSEENCH11005895

    A balloon is filled with hydrogen at room temperature. It will burst if the pressure exceeds 0.2 bar. If at 1 bar pressure, the gas occupies 2.27 L volume, up to what volume can the balloon be expanded?

    Solution

    From the given data,
    space space space space straight V subscript 1 space equals space 2.27 space straight L space space space space space space space space space space space
space space straight V subscript 2 space equals space ?
space space straight P subscript 1 space equals 1 space bar space space space space space space space space space space space space space space
space space straight P subscript 2 space equals space 0.2 space bar
According space to space Boyle apostrophe straight s space law comma

space space space space space space space space space space space space space space space space space straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 straight V subscript 2

therefore space space space space space space straight V subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight P subscript 2 end fraction
    Substituting the values, we have
                 space space space straight V subscript 2 space equals fraction numerator 1 space bar space cross times space 2.27 space straight L over denominator 0.2 space bar end fraction space equals space 11.35 space straight L
    Since balloon bursts at 0·2 bar pressure, the volume of the balloon should be less than 11·35L.

    Question 77
    CBSEENCH11005896

    What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 at 30C?

    Solution

    From the given data,
    straight V subscript 1 space equals space 500 space dm cubed comma space space space straight P subscript 1 space equals space 1 space bar
straight V subscript 2 space equals space 200 space dm cubed comma space space straight P subscript 2 space equals space ?
    According to Boyle's law,
                       straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 straight V subscript 2
    or space space space straight P subscript 2 over straight P subscript 1 space equals space straight V subscript 1 over straight V subscript 2
    Substituting the values, we have,
       fraction numerator straight P subscript 2 over denominator 1 space bar end fraction space equals space fraction numerator 500 space dm cubed over denominator 200 space dm cubed end fraction
therefore space space space fraction numerator straight P subscript 2 over denominator 1 space bar end fraction space equals space 2.5
or space space space space space straight P subscript 2 space equals space 2.5 space bar

    Question 78
    CBSEENCH11005897

    A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1·2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure ?

    Solution

    Here,
    straight V subscript 1 space equals space 12 space ml semicolon space space space space space straight P subscript 1 space equals space 1.2 space space bar
straight V subscript 2 space equals space 180 space mL space semicolon space space straight P subscript 2 space equals ?
    According to Boyle's law,
                          straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 straight P subscript 2
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#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>
    Substituting the values, we have,
                      fraction numerator straight P subscript 2 over denominator 1.2 space bar end fraction space equals space fraction numerator 120 space mL over denominator 180 space mL end fraction
or space space space space straight P subscript 2 space equals space 1.2 space bar space cross times 2 over 3 space equals space 0.8 space bar

    Question 79
    CBSEENCH11005898

    A gas occupies a volume of 300 mL at 740 mm Hg at 20°C. What additional pressure is required to reduce the gas volume to 250 ml at 20°C?

    Solution

    From the given data.
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#2 [internal function]: _hx_lambda->execute('Http Error #404')
#3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(532): call_user_func_array(Array, Array)
#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
#9 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(458): haxe_Http->customRequest(true, Object(haxe_io_BytesOutput), NULL, NULL)
#10 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(40): haxe_Http->request(true)
#11 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(80): com_wiris_plugin_impl_HttpImpl->request(true)
#12 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array)
#13 {main}</pre>
    Substituting the values in the above equation,
        740 space cross times space 300 space equals space straight P subscript 2 space cross times space 250
    therefore space space space space straight P subscript 2 space equals space fraction numerator 740 space cross times space 300 over denominator 250 end fraction space equals space 888 space space mm space space Hg
Additional space pressure space required space equals 888 space minus space 740 equals 148.0 space mm
      

    Sponsor Area

    Question 80
    CBSEENCH11005899

    What pressure must be applied to a given sample of a gas in order to compress it to three-fourth of its original volume ?

    Solution
    According to Boyle’s law
    P1V1 = P2V
    From the given data,
             straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 space cross times space 3 over 4 straight V subscript 1
therefore space space space space space space space straight P subscript 2 space equals space 4 over 3 straight P subscript 1
    Thus, the pressure applied must be 4 over 3 times the original pressure.
    Question 81
    CBSEENCH11005900

    State and explain Charle’s law.

    Solution

    Charle's law states:  Pressure remaining constant the volume of a given mass of gas increases or decreases by 1 over 273 space of space its volume at 0 degree straight C for each one-degree rise or fall in temperature. 
    Mathematically,
     If V0 is the volume at 0°C, then the volume at various other temperatures can be written as:
         Volume space at space 10 degree straight C space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times 100 over denominator 273 end fraction
Volume space at space 40 degree straight C space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times 40 over denominator 273 end fraction
In space general comma
Volume space at space straight t degree straight C comma space space straight V subscript straight t space equals space straight V subscript 0 space plus space fraction numerator straight V subscript 0 cross times space straight t over denominator 273 end fraction
equals space straight V subscript 0 space open parentheses 1 plus straight t over 273 close parentheses
    But if the gas is cooled to –273°C, then its volume becomes zero.
    Volume space at space minus 273 degree straight C space equals space straight V subscript 0 open parentheses 1 minus 273 over 273 close parentheses space equals space 0
    This implies that a gas at -273°C will have zero or no volume i.e. it will cease to exist. In actual practice, all gases liquefy before this temperature is reached. Also –273°C should be the lowest possible temperature because any further cooling would lead to a volume of less than zero or negative volume which is meaningless. Therefore, this temperature (–273° C) was termed as absolute zero of temperature.
    Another statement of Charle’s law:
    According to Charle’s law, volume of a given mass of a gas at different temperatures is related to its volume (V0) at 0°C as follows:
    Volume at Volume space at space straight t subscript 1 superscript degree straight C comma
    straight V subscript straight t subscript 1 end subscript space equals space straight V subscript 0 space open square brackets 1 plus straight t subscript 1 over 273 close square brackets space equals space straight V subscript 0 space open square brackets fraction numerator 273 space plus space straight t subscript 1 over denominator 273 end fraction close square brackets
straight V subscript straight t subscript 1 end subscript space equals space straight V subscript 0 straight T subscript 1 over 273 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
    Volume space at space straight t subscript 2 superscript degree straight C comma space space space space straight V subscript straight t subscript 2 end subscript space equals space straight V subscript 0 space open square brackets 1 plus straight t subscript 2 over 273 close square brackets
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals straight V subscript 0 open square brackets fraction numerator 273 plus straight t subscript 2 over denominator 273 end fraction close square brackets
straight V subscript straight t subscript 2 end subscript space equals space straight V subscript 0 space straight T subscript 2 over 273 space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    Dividing (1) by (2), we get,
         
     straight V subscript straight t subscript 1 end subscript over straight V subscript straight t subscript 2 end subscript space equals space fraction numerator straight V subscript 0 straight T subscript 1 over denominator 273 end fraction cross times fraction numerator 273 over denominator straight V subscript 0 straight T subscript 2 end fraction space equals space straight T subscript 1 over straight T subscript 2
    or  <pre>uncaught exception: <b>Http Error #404</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php line 61<br />#0 [internal function]: com_wiris_plugin_impl_HttpImpl_0(Object(com_wiris_plugin_impl_HttpImpl), NULL, 'http://www.wiri...', 'Http Error #404')
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#2 [internal function]: _hx_lambda->execute('Http Error #404')
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#4 [internal function]: haxe_Http_5(true, Object(com_wiris_plugin_impl_HttpImpl), Object(com_wiris_plugin_impl_HttpImpl), Array, Object(haxe_io_BytesOutput), true, 'Http Error #404')
#5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/php/Boot.class.php(769): call_user_func_array('haxe_Http_5', Array)
#6 [internal function]: _hx_lambda->execute('Http Error #404')
#7 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/HttpImpl.class.php(27): call_user_func_array(Array, Array)
#8 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/haxe/Http.class.php(444): com_wiris_plugin_impl_HttpImpl->onError('Http Error #404')
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#13 {main}</pre>
    [Pressure and mass of the gas constant]
    Hence Charle’s law may also be stated as Pressure remaining constant, the volume of a given mass of a gas is directly proportional to absolute temperature i.e. straight V over straight T space equals space constant. space
    Thus, if V1 is the initial volume of the gas at temperature T1 (in degree kelvin) and V2 is the final volume of the gas at temperature T2 (in degree kelvin), keeping pressure constant, then
    straight V subscript 1 over straight T subscript 1 space equals space straight V subscript 2 over straight T subscript 2

    Question 82
    CBSEENCH11005901

    How will you verify Charle's law graphically? What is the practical importance of Charle's law?

    Solution
    Graphical verification: Charle’s law can be verified by the volume-temperature graph. At constant pressure, when a graph is plotted between volume (along Y-axis) and temperature on kelvin scale (along X-axis), a straight line passing through the origin is obtained. The plot obeys the linear equation V = m T (c.f., y = mx, equation of a line passing through the origin). Thus, we conclude that at constant pressure, the volume of a given mass of a gas is directly proportional to absolute temperature.

    Practical importance: The use of hot air balloons in sports and for meteorological observations is an interesting application of Charle’s law. According to Charle’s law, gases expand on heating and hot air is less dense than cool air. This causes a hot air balloon to rise by displacing the cooler air of the atmosphere.
    Question 83
    CBSEENCH11005902

    Is it possible to cool a gas below absolute zero?

    Solution
    No, the gas cannot be cooled below absolute zero. At this temperature, the kinetic energy of the molecules is zero. The molecules come to rest and the pressure also becomes zero. So it is impossible to cool a gas below absolute zero since there is no heat left to be removed from the gas.
    Question 84
    CBSEENCH11005903

    What is absolute or kelvin scale of temperature?

    Solution

    With a temperature of absolute zero  (i.e. –273°C) as the lowest temperature, a new scale for expressing the temperature has been developed which is called absolute scale or kelvin scale and the temperature expressed on this scale is called absolute or kelvin temperature.
    In the kelvin scale, the size of the degree is kept the same as on the celsius scale but the zero is shifted to –273·15 below zero on the Celsius scale.
    Thus,  –273.15° C = 0 K
    or         0° C   = 273.15 K
    or         t° C = (273.15 + t) K


    Hence temperature on absolute or kelvin scale may be obtained by adding 273.15 to the temperature on celsius scale. 
    TK = 273.15 +  t°C
    In practice, the above relation is generally taken as
    TK = 273 + t° C where T is the temperature on kelvin scale and t is the temperature on celsius scale.

    Question 85
    CBSEENCH11005904

    Why is kelvin scale of temperature regarded better than centigrade scale?

    Solution
    The kelvin scale of temperature regarded better than centigrade because it has an only positive value of temperature. We do not have any temperature below zero degrees. But on the centigrade scale of temperature, there can be positive as well as negative temperatures. Further, many equations and relationship (such as Charle’s law) can be written in a much simpler form in terms of kelvin scale (also called Thermodynamic scale of temperature) than in centigrade (or celsius) scale.
    Question 86
    CBSEENCH11005905

    A sample of helium gas has a volume of 500 cm3 at 373K. Calculate the temperature at which the volume becomes 260 cm3. Assume that the pressure is kept constant.

    Solution

    From the given data,
    straight V subscript 1 space space equals space 500 space cm cubed comma space space space straight T subscript 1 space equals space 373 space straight K
straight V subscript 2 space equals space 260 space cm cubed comma space space straight T subscript 2 space equals space ?
    According to Charles' law
                         
        straight V subscript 1 over straight V subscript 2 space equals space straight T subscript 1 over straight T subscript 2
    Putting the values in the above equation,
                                 
       500 over 260 space equals space 373 over straight T subscript 2
    therefore space space straight T subscript 2 space equals space fraction numerator 373 space cross times space 260 over denominator 500 end fraction space equals 193.96 space straight K

    Question 87
    CBSEENCH11005906

    On a ship sailing in Pacific ocean where the temperature is 23.4 °C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches Indian ocean, where the temperature is 26.1°C ? 

    Solution

    From the given data,
    straight V subscript 1 space equals space 2 straight L semicolon space space space straight T subscript 1 space equals space 273 plus 23.4 space equals space 296.4 space straight K
straight V subscript 2 space equals space ? space space space space straight T subscript 2 space equals space 273 plus 26.1 space equals space 299.1 space straight K
    According to Charle's law,
                          straight V subscript 1 over straight V subscript 2 space equals space straight T subscript 1 over straight T subscript 2
    or                straight V subscript 2 space equals space fraction numerator straight V subscript 1 straight T subscript 2 over denominator straight T subscript 1 end fraction
    Substituting the values,  we have
              straight V subscript 2 space equals space fraction numerator 2 straight L space cross times space 299.1 space straight K over denominator 296. space 4 straight K end fraction space equals space 2 straight L cross times 1.009 space equals space 2.018 space straight L
                        

    Question 88
    CBSEENCH11005907

    A cylinder contains 450 mL of a gas at 27°C at constant pressure. The cylinder is placed in a refrigerator at –10°C. Calculate the resultant volume of the gas.

    Solution

    From the given data,
    straight V subscript 1 space equals space 450 space mL semicolon space space space space straight T subscript 1 space equals space 273 plus space 27 space equals space 300 space straight K
straight V subscript 2 space equals space ? space space space space space space space space space space space space space space space straight T subscript 2 space equals space 273 minus 10 space equals space 263 space straight K
    According to Charle's law
           
     straight V subscript 1 over straight V subscript 2 space equals space straight T subscript 1 over straight T subscript 2
    Substituting the values,
                   
       fraction numerator 450 space mL over denominator straight V subscript 2 end fraction space equals space fraction numerator 300 space straight K over denominator 263 space straight K end fraction
    therefore space space space space space space straight V subscript 2 space equals space fraction numerator 450 space cross times space 263 over denominator 300 end fraction mL space equals space 394.5 space mL

    Question 89
    CBSEENCH11005908

    What volume of air will be expelled from a vessel containing 380 cm3 at 10°C when it is heated to 30°C at the same pressure ?

    Solution
    From the given data,
    straight V subscript 1 space equals space 380 space cm cubed semicolon space space space straight T subscript 1 space equals space 273 plus 10 space equals space 283 space straight K
straight V subscript 2 space equals space ? space space space space space space space space space space space space space space straight T subscript 2 space equals space 273 plus 30 space equals space 303 space straight K
    According to Charle's law,
                             straight V subscript 1 over straight V subscript 2 space equals space straight T subscript 1 over straight T subscript 2
    Substituting the values,  
    we have,
                               380 over straight V subscript 2 space equals space 283 over 303
    therefore space space space straight V subscript 2 space equals space fraction numerator 380 space cross times space 303 over denominator 283 end fraction space equals space 406.85 space cm cubed
therefore space space space Air space that space will space be space expelled space equals space 406.85 space minus space 380 space equals space 26.85 space cm cubed
    Question 91
    CBSEENCH11005910

    State pressure-temperature relationship law (Gay-Lussac's law).

    Solution
    At constant volume, the pressure of a given mass of a gas is directly proportional to its absolute temperature.
    Mathematically,
                             
    straight P space proportional to space space straight T space left parenthesis at space constant space volume right parenthesis
    or space space space space space straight P over straight T space equals constant.
    In general, if P1 and P2 are the pressures at the absolute temperatures T1 and T2respectively, then       straight P subscript 1 over straight T subscript 1 space equals space straight P subscript 2 over straight T subscript 2
     
     or space space space straight P subscript 1 over straight P subscript 2 space equals space straight T subscript 1 over straight T subscript 2
    Question 92
    CBSEENCH11005911

    A gas cylinder containing cooking gas can withstand a pressure of 14.9 bar. The pressure gauge of the cylinder indicates 12 bar at 27°C. Due to sudden fire in the building, the temperature starts rising. At what temperature will the cylinder explode?

    Solution

    Since the gas is confirmed in a cylinder, its volume will remain constant. 
    From the given data,
     straight P subscript 1 space equals space 12 space bar semicolon space space space space space straight T subscript 1 space equals space 27 plus 273 space equals space 300 space straight K
straight P subscript 2 space equals space 14.9 space bar semicolon space space space straight T subscript 2 space equals space ?
    Applying pressure temperature law,
     straight P subscript 1 over straight P subscript 2 space equals space straight T subscript 1 over straight T subscript 2
     or space space space space straight T subscript 2 space equals space fraction numerator straight T subscript 1 space cross times space straight P subscript 2 over denominator straight P subscript 1 end fraction
    Substituting the values in the above equation,
     straight T subscript 2 space equals space fraction numerator 300 space cross times space 14.9 over denominator 12 end fraction space equals space 372.5 space straight K
    Temperature space in space degree straight C space equals space 372.5 space minus space 273
space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 99.5 space degree straight C
space
space space space space
    Hence the cylinder will at 99.5 space degree straight C.

    Question 93
    CBSEENCH11005912

    State Avogadro's law and prove that volume of gases at constant temperature and pressure is directly proportional to their number of moles.

    Solution

    Avogadro’s law states: Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
    For example, 22·4 litres of any gas at STP (273K and 760mm Hg) contain Avogadro number, N0 (6·02×1023) of molecules irrespective of its nature.
    It follows that volume of gas V is directly proportional to N, the number of molecules at constant temperature and pressure.
    V ∝ N (at constant temperature and pressure)
    Also number of moles of a gas (n) ∝ number of molecules (N)
    No. space of space moles space left parenthesis straight n right parenthesis space equals space fraction numerator No. space of space molecules over denominator Avogadro apostrophe straight s space no. space left parenthesis straight N subscript 0 right parenthesis end fraction
space space space space space space space space space space space space space space space space space space space straight n space equals space straight N over straight N subscript 0
therefore space space space space space space space space space straight n space proportional to space space straight N
or space space space space space space space space space straight V space proportional to space space straight n
or space space space space space space space space space straight V space space equals space constant space straight x space straight n
    Hence the volume of gases at constant temperature and pressure is directly proportional to their number of moles.

    Question 94
    CBSEENCH11005913

    Two flasks A and B have equal to volumes. Flask A contains H2 and is maintained at 300K while the flask B contains an equal mass of CH4 gas and is maintained at 600K. Which flask contains a greater number of molecules ? How many times?

    Solution

    The number of molecules of a gas is proportional to the number of moles.
      Let the mass of each gas in the two flasks be =  mg
       Molecular mass of H2 = 2
        Molecular mass of CH4 =16
    space space therefore space space space Number space of space moles space of space straight H subscript 2 space equals space straight m over 2
space space Number space of space moles space of space CH subscript 4 space equals space straight m over 16
Ratio space between space the space number space of space moles space of space straight H subscript 2 space and space CH subscript 4
space equals space straight m subscript 2 over 2 colon straight m over 16 space space or space space 16 colon 2 space or space space 8 colon 1.
    Thus, the number of molecules of H2 in the flask. A will be eight times the number of molecules of CH4 in flask B.

    Question 95
    CBSEENCH11005914

    With the help of Boyle's and Charle's laws, derive an expression for the ideal gas equation.

    Solution

    Derivation of the ideal gas equation: Let the volume of a given mass of a gas change from V1 to V2 when the pressure is changed from Pi to P2 and temperature is changed from T1 to T2. Suppose this change from initial state (P1 V1T1) to final state (P2V2T2) occurs in two steps:
    First step: Suppose the volume of the given mass of the gas changes from V1 to Vx when pressure is changed from P1 to P2, keeping temperature T1 constant.
    therefore space space space space According space to space Boyle apostrophe straight s space law comma
straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 straight V subscript 2 space left parenthesis Temperature space and space mass space of space gas space constant right parenthesis
or space space space space straight V subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight P subscript 2 end fraction space space space space space space space space space space... left parenthesis 1 right parenthesis
    Second step: Now let the volume changed from Vx to V2 when the temperature is changed from T1 to T2, keeping pressure P2 constant.
    therefore     According to Charle's law,
                          straight V subscript 1 over straight T subscript 1 space equals space straight V subscript 2 over straight T subscript 2
    or space space space space space space space space straight V subscript 1 space space equals space fraction numerator straight V subscript 2 straight T subscript 1 over denominator straight T subscript 2 end fraction space space space space space space space space space space... left parenthesis 2 right parenthesis
    From (1) and (2), we have
                  fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight P subscript 2 end fraction space equals space fraction numerator straight V subscript 2 straight T subscript 1 over denominator straight T subscript 2 end fraction
    or space space space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis
or space space space space space PV over straight T space equals space straight k space space space space where space straight K space is space constant space space... left parenthesis 4 right parenthesis
    The numerical value of the constant K is independent of the nature of the gas but depends on upon the amount of the gas. But at constant temperature and pressure the volume of a gas is proportional to the number of moles (n), this means that K is directly proportional to the number of moles (n).
    or space space space straight K space proportional to space space space straight n
straight i. straight e. space space space space straight K space equals space nR space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 5 right parenthesis
    where R is a constant of proportionality which is independent of nature as well as the amount of the gas and is known as a universal gas constant. From (4) and (5)
      space space space space space space space PV over straight T space equals space nR
or space space space space space space space space space PV space equals space nRT space space space space space space space space space space... left parenthesis 6 right parenthesis
    Equation (6) is known as an ideal gas equation. Alternative derivation of ideal gas equation
    : According to Boyle’s law,
                    space straight V proportional to space 1 over straight P space space space space space space space space space space space space space open square brackets at space constant space straight T space and space straight n close square brackets space space space space... left parenthesis 1 right parenthesis
    According to Charle's law,
     
     straight V proportional to space space straight T space space space space space left square bracket at space constant space straight P space and space straight n right square bracket space space space space... left parenthesis 2 right parenthesis
    According to Avogadro's law,
     
     straight V space proportional to space space straight n space space space space space space open square brackets at space constant space straight P space and space straight T close square brackets space space space... left parenthesis 3 right parenthesis
    Combining (1), (2) and (3), we have,
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#13 {main}</pre>
    .
    or space space space space straight V space equals space straight R nT over straight P
or space space space space space PV space equals space nRT                                  

     

    Question 96
    CBSEENCH11005915

    Derive the dimensions of the gas constant. 
    Or
    What do you mean by the nature of universal consant R?

    Solution

    The ideal gas equation is
                  PV = nRT
         or space space straight R space equals space PV over nT
space space space space space space equals space fraction numerator Pressure space cross times space Volume over denominator No. space of space moles space cross times space Temperature left parenthesis straight K right parenthesis end fraction

space space space space space space equals space Force over Area cross times space fraction numerator Area space cross times space Length over denominator No. space of space moles space cross times space Temperature space left parenthesis straight K right parenthesis end fraction

space space space space space space space equals space fraction numerator Force space cross times space Length over denominator No. space of space moles space cross times space Temperature space left parenthesis straight K right parenthesis end fraction

space space space space space space space space equals space fraction numerator Energy space left parenthesis or space Work right parenthesis over denominator No. space of space moles space cross times space Temperature space left parenthesis straight K right parenthesis end fraction
    therefore R represents energy or work done per degree per mole. 

    Question 98
    CBSEENCH11005917

    How much time would it take to distribute one Avogadro number of wheat grains, if 1010grains are distributed each second?

    Solution
    One Avogadro number = 6·022 × 1023 Now 1010 grains are distributed in 1 second
    therefore space space 6.022 space cross times space 10 to the power of 23 space grains space would space be space distributed
space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 6.022 space cross times space 10 to the power of 23 over denominator 10 to the power of 10 end fraction seconds
space space space space space space space space space space space space space space space space space space space space space space space equals space 6.022 space cross times space 10 to the power of 13 space seconds
space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 6.022 space cross times space 10 to the power of 13 over denominator 60 space cross times space 60 space cross times 24 space cross times 365 end fraction years

space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.9089 space cross times space 10 to the power of 16 space years space approx.
    Question 99
    CBSEENCH11005918

    Write the units of the gas constant R in different scales of measurement ?

    Solution

    The units of the gas constant (or numerical value) R in different scales of measurement may be calculated as follows:

    1. In SI units:
    One mole of a gas under standard conditions of temperature and pressure (STP or NTP) occupies a volume of 22·7L.
    therefore space space space Pressure space left parenthesis straight P right parenthesis space equals space 1 space bar space equals space 10 to the power of 5 Pa
Temperature space left parenthesis straight T right parenthesis space equals space 273.15 space straight K
Volume space left parenthesis straight V right parenthesis space equals space 22.7 space straight L space space equals space 22.7 space cross times space 10 to the power of negative 3 end exponent straight m cubed
According space to space ideal space gas space equation comma
space space space space space space space space space PV space equals space nRT
or space space space space space space space straight R space equals space PV over nT
or space space space space space space space space straight R space equals space fraction numerator 10 to the power of 5 Pa space cross times space 22.7 space cross times space 10 to the power of negative 3 end exponent straight m cubed over denominator 1 space mol space cross times space 273.15 space straight K end fraction
space space space space space
               space equals space fraction numerator 10 to the power of 5 Nm to the power of negative 2 end exponent space cross times space 22.7 space cross times space 10 cubed straight m cubed over denominator 1 space mol space cross times space 273.15 space straight K end fraction
space equals space 8.314 space Nm space mol to the power of negative 1 end exponent straight K to the power of negative 1 end exponent
space equals space 8.314 space straight J space mol to the power of negative 1 end exponent straight K to the power of negative 1 end exponent left parenthesis 1 space NM space equals space 1 straight J right parenthesis

    2. Under NTP condition (for one mole of the gas):One mole of any gas at 0°C (i.e. 273·15K.) and under 1 bar pressure occupies 22.7 L
        therefore space space space space Pressure space left parenthesis straight P right parenthesis space equals space 1.0 space bar
space Temperature space left parenthesis straight T right parenthesis space equals space space 273.15 space straight K
Volume space left parenthesis straight V right parenthesis space equals space 22.7 space dm cubed
space space space space space space space space space space space space straight n space equals space 1 space mol
    According to ideal gas equation,
                        PV = nRT
    or space space space space space space space space space straight R space equals space PV over nT
    Substituting the values, we have,
                   straight R space equals space fraction numerator left parenthesis 1.0 space bar right parenthesis space cross times space left parenthesis 22.7 space dm cubed right parenthesis over denominator left parenthesis 1 space mol right parenthesis space cross times space left parenthesis 273.15 space straight K right parenthesis end fraction
space space space space equals 0.083 space bar space dm cubed space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent

    Question 100
    CBSEENCH11005919

    Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.

    Solution

    The equation of state is given by,
    pV = nRT …. (i)

    Where, p → Pressure of gas

    V → Volume of gas

    n→ Number of moles of gas

    R → Gas constant

    T → Temperature of gas

    From equation (i) we have,

    p = n RT/V

    Where n= Mass of gas(m)/ Molar mass of gas(M)

    Putting value of n in the equation, we have

     p = m RT/ MV ------------(ii)

    Now density(ρ) = m /V ---------(iii)

    Putting (iii) in (ii) we get

    P = ρ RT / M

    OR

    ρ = PM / RT

    Hence, at a given temperature, the density (ρ) of gas is proportional to its pressure (P).

    Question 101
    CBSEENCH11005920

    250 mL of a gas is confined in a vessel at at 15°C and 750 mm pressure. How much volume will it occupy at S.T.P.?

    Solution

    Original conditions Final conditions (S.T.P.)
    straight V subscript 1 space equals space 250 space mL space space space space space space space space space space space space space space space space space space space space straight V subscript 1 space equals space ?
straight P subscript 1 space equals space 750 space mm space space space space space space space space space space space space space space space space space space space straight P subscript 2 space equals space 760 space mm
straight T subscript 1 space equals space 273 plus 15 space equals space 288 straight K semicolon space space space space space space space straight T subscript 2 space equals space 273 thin space straight K
    Applying the gas equation
                  fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction
    Substituting the values, we get,
     fraction numerator space 750 space cross times space 250 over denominator 288 end fraction space equals space fraction numerator 760 space cross times space straight V subscript 2 over denominator 273 end fraction
therefore space space space space space space space space space space straight V subscript 2 space equals space fraction numerator 750 space cross times space 250 space cross times space 273 over denominator 288 space cross times space 760 end fraction space equals space 233.8 space mL

    Question 102
    CBSEENCH11005921

    A sample of nitrogen gas occupies a volume of 1.00L at a pressure of 0.50 bar and at 40°C. Calculate the pressure if the gas is compressed to 0·225 cm3 at –6°C.

    Solution

    Original conditions:
    straight V subscript 1 space equals space 1 space straight L space equals space 1000 space mL
straight P subscript 1 space equals space 0.50 space bar
straight T subscript 1 space equals space 273 space plus space 40 space equals space 313 space straight K
    Final conditions:
      straight V subscript 2 space equals space 0.225 space mL
straight P subscript 2 space equals space ?
straight T subscript 2 space equals space 273 minus 6 space equals space 267 space straight K
    Applying gas equation fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction.
    Substituting the values in the above equation, we .
    have,
    fraction numerator 0.50 space cross times space 1000 over denominator 313 end fraction space equals space fraction numerator straight P subscript 2 space cross times space 0.225 over denominator 267 end fraction
    therefore space space space space space space space straight P subscript 2 space equals space fraction numerator 0.50 space cross times space 1000 space cross times space 267 over denominator 313 space cross times space 0.225 end fraction
space space space space space space space space space space space space space space space equals space 1895.6 space bar

     

     

     

     

    Question 103
    CBSEENCH11005922

    At 25°C and 760 mm of Hg pressure, a gas occupies 600 mL volume. What will be its pressure at a height where the temperature is 10°C and volume of the gas are 640 mL ?

    Solution
    Original conditions Final conditions.
    straight V subscript 1 space equals space 600 space mL space space space space space space space space space space space space space space space space space space space space straight V subscript 2 space equals space 640 space mL
straight P subscript 1 space equals space 760 space mm space Hg space space space space space space space space space space space space space space space space straight P subscript 2 space equals space ?
straight T subscript 1 space equals space 273 plus 25 space equals space 298 space straight K space space space space space space space straight T subscript 2 space equals space 273 plus 10 space equals space 283 space straight K
    Applying the gas equation,
                           fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction
     or space space space space space space space space straight P subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 straight T subscript 2 over denominator straight V subscript 2 straight T subscript 1 end fraction
    Substituting the values, we have
         straight P subscript 2 space equals space fraction numerator 760 space mm space cross times space 600 space mL space cross times space 283 space straight K over denominator 640 space mL space cross times space 298 space straight K space end fraction
space space space space space equals space 676.6 space mm space Hg
    Question 104
    CBSEENCH11005923

    To what temperature must a neon gas sample be heated to double its pressure, if the initial volume of the gas at 75°C is decreased by 15%?

    Solution
    Initial conditions:
    straight V subscript 1
straight P subscript 1
straight T subscript 1 space equals space 273 space plus space 75 straight K space equals space 384
Final space conditions colon
space space space space space space space straight V subscript 2 space equals space straight V subscript 1 minus space 15 over 100 comma space space space straight V subscript 1 space equals space fraction numerator 85 space straight V subscript 1 over denominator 100 end fraction
space space space space space space space space space straight P subscript 2 space equals space 2 straight P subscript 1
space space space space space space space space space straight T subscript 2 space equals space ?
    Applying the gas equation
             fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction
    Substituting the values, we have,
      fraction numerator straight P subscript 1 space cross times space straight V subscript 1 over denominator 348 end fraction space equals space fraction numerator 2 straight P subscript 1 space cross times space 85 space straight V subscript 1 over denominator 100 space cross times space straight T subscript 2 end fraction
therefore space space space space space space space straight T subscript 2 space equals space fraction numerator 2 space cross times space 85 space cross times space 348 space over denominator 100 end fraction space equals space 591.6 space straight K
Temperature space in space degree straight C space equals space 591.6 space minus space 273 space equals space 318.6 space degree straight C
    Question 105
    CBSEENCH11005924

    The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0·15g of aluminium reacts?

    Solution

    Al reacts with caustic soda to produce dihydrogen.
    stack 2 Al with 2 cross times 27 straight g below space plus 2 NaOH plus 2 straight H subscript 2 straight O rightwards arrow space 2 NaAlO subscript 2 space plus space stack 3 straight H subscript 2 with 3 cross times 22400 space mL space at space STP below
     
    54g Al on reacting with caustic soda produces
                                                    equals space 3 space cross times space 22400 space mL space straight H subscript 2
    therefore space space 0.15 space straight g space Al space on space reacting space with space caustic space soda space would space produce space
space space equals space fraction numerator 3 space cross times space 22400 over denominator 54 end fraction cross times 0.15 space mL space straight H subscript 2 space equals space 186.7 space mL
    Applying the gas equation
            left parenthesis STP space condition right parenthesis space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction space left parenthesis Reqd. space condition right parenthesis
space space space space space fraction numerator 1 space atm space cross times space 186.7 space mL over denominator 273 straight K end fraction space equals space fraction numerator 0.987 space atm space cross times space straight V subscript 2 over denominator 293 space straight K end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space 1 space bar space equals space 0.987 space atm close square brackets
space space space therefore space space space straight V subscript 2 space equals space fraction numerator 1 space cross times space 186.7 over denominator 273 end fraction cross times fraction numerator 293 over denominator 0.987 end fraction space equals space 203.0 space mL

    Question 106
    CBSEENCH11005925

    Calculate the volume occupied by 7.0 g of nitrogen gas at 27°C and 750 mm Hg pressure.

    Solution

    Molecular mass of nitrogen  = 28
    Number of moles of nitrogen (n)
                              equals space fraction numerator Amount space of space nitrogen over denominator Molecular space mass end fraction space equals space 7 over 28
    Now space space space space space space space straight P space equals space 750 space mm space equals space 750 over 760 space bar
space space space space space space space space space space space space space straight T space equals space 273 space plus 27 space equals space 300 space straight K
    Applying ideal gas equation PV = nRT, we have,
    straight V space equals space nRT over straight P space equals space fraction numerator 7 space cross times space 0.0831 space cross times space 300 over denominator 28 space cross times space begin display style 750 over 760 end style end fraction space equals space 6.316 space straight L

    Question 107
    CBSEENCH11005926

    Calculate the volume occupied by 8.8 g of CO2 at 31·1°C and 1 bar pressure. R = 0.083 bar LK–1 mol–1).

    Solution

    Here mass of CO2 = 44 g mol-1

    Number space of space moles space left parenthesis straight n right parenthesis space equals space fraction numerator Amount space of space CO subscript 2 over denominator Molar space mass end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 8.8 over denominator 44 end fraction space equals space 0.2
straight P space equals space 1 space bar comma space space space straight T space equals space 273 plus 31.1 space space equals space 304.1 space straight K
Applying space ideal space gas space equation comma
space space space space space space PV space equals space nRT
or space space space space space straight V space equals space nRT over straight P
    Substituting values, we have,
     
      straight V space equals space fraction numerator 0.2 space mol space cross times space 0.083 space bar space LK to the power of negative 1 end exponent space cross times space 304.1 straight K over denominator 1 space bar end fraction
space space equals 5.05 space straight L

    Question 108
    CBSEENCH11005927

    Calculate the temperature of 4·0 moles of a gas occupying 5 dm3 at 3·32 bar (R = 0·083 bar dm3 K–1 mol–1).

    Solution

    Here,
    n = 4.0 mol
    P = 3.32 bar
    V = 5 dm3,  
    R = 0.083 bar dm3 K-1 mol-1

    Applying ideal gas equation PV = nRT,
    or space space space space straight T space equals space PV over nR
    Substituting the values, we have,
     straight T space equals space fraction numerator 3.32 space bar space cross times space 5 space dm cubed over denominator 4 space mol space cross times space 0.083 space bar space dm cubed space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent end fraction
space space space space equals space 50 space straight K         

    Question 109
    CBSEENCH11005928

    What will be the pressure exerted (in pascal) by a mixture of 32g of methane and 4·4g of carbon dioxide contained in a 9 dmflask at 27°C?

    Solution

    Number of moles of methane (n1)
                             equals space fraction numerator Mass space of space methane over denominator Molecular space mass end fraction
equals space fraction numerator 3.2 space straight g over denominator 16.0 space straight g space mol to the power of negative 1 end exponent end fraction
equals space 0.2 space mol
    Number of moles of carbon dioxide (n2)
              space equals space fraction numerator Mass space of space carbon space dioxide over denominator Molecular space mass end fraction
equals space fraction numerator 4.4 space straight g over denominator 44 straight g space mol to the power of negative 1 end exponent end fraction
equals space 0.1 space mol

    therefore space space Total space number space of space moles space equals space 0.2 space mol space plus space 0.1 space mol space equals space 0.3 space mol
Temperature space left parenthesis straight T right parenthesis space equals space 27 plus 273 space equals space 300 space straight K
Using space ideal space gas space equation space PV space equals space nRT
pr space space straight P space equals space nRT over straight V space space... left parenthesis 1 right parenthesis
Here space straight n space equals space 0.3 space mol comma space space straight R space equals space 0.083 space bar space dm cubed space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent comma
straight T space equals space 300 space straight K comma space space space straight V space equals space 9 space dm cubed
space space Substituting space the space values space in space eq. space left parenthesis 1 right parenthesis comma space we space have
    straight P space equals space fraction numerator 0.3 space mol space cross times space 0.083 space bar space dm cubed space straight K to the power of negative 1 end exponent space mol to the power of minus to the power of 1 space cross times space 300 straight K over denominator 9 space dm cubed end fraction
space space space equals fraction numerator 0.3 space cross times space 0.083 space cross times space 300 over denominator 9 end fraction bar
space space space equals 0.83 space bar
space space equals space 0.83 space cross times 10 to the power of 5 Pa space equals space 8.3 space cross times space 10 to the power of 4 Pa

    Question 110
    CBSEENCH11005929

    At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide? 

    Solution

    We know,
    straight d space equals space MP over RT
    At the same temperature and for same density 
               straight M subscript 1 straight P subscript 1 space equals space straight M subscript 2 straight P subscript 2 space space space left square bracket because space space straight R space is space constant right square bracket space space... left parenthesis 1 right parenthesis
    (Gaseous oxide) (N2)
     Here P1 = 2 bar;    
    P2 = 5 bar;
    M2 = molecular mass of N2 = 28
    Substituting the values in equation(1), we have
          M1 x 2 = 28 x 5
    therefore space space space space straight M subscript 1 space equals space 70

    Question 111
    CBSEENCH11005930

    Density of a gas is found to be 5.46 g/dm3 at 7°C at 2 bar pressure. What will be its density at STP? 

    Solution
    We know,
    straight d space equals space MP over RT
    For the same gas at different temperature and pressure,
                  straight d subscript 1 over straight d subscript 2 space equals space straight P subscript 1 over straight T subscript 1 cross times straight T subscript 2 over straight P subscript 2                     ...(1)
    Initial conditions,                  STP conditions
    straight d subscript 1 space equals space 5.46 space straight g space dm to the power of negative 3 end exponent space space space space space space space space space space space space space space space space space space space space space space space space straight d subscript 2 space equals space ?
straight T subscript 1 space equals space 273 plus 27 space equals space 300 straight K space space space space space space space space space space space space space space space space space straight T subscript 2 space equals space 273 space straight K
straight P subscript 1 space equals space 2 space bar space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight P subscript 2 space equals space 1 space bar space space space space space space space space space space space space space space space space space space space space space space space space space space space
    Substituting the values, we have 
    fraction numerator 5.46 space straight g space dm to the power of negative 3 end exponent over denominator straight d subscript 2 end fraction space equals space fraction numerator 2 space bar space cross times space 273 space straight K over denominator 300 straight K space cross times 1 space bar end fraction space space space or space space straight d subscript 2 space equals space 3 straight g space dm to the power of negative 3 end exponent
    Question 112
    CBSEENCH11005931

    If the density of a gas at the sea level at 0° is 1·29 kg m–3, what is the molar mass? (Assume that pressure is equal to 1 bar).

    Solution
    Using ideal gas equation,
                  PVn space equals space RT space space space space space or space space space space straight P. straight M over straight d space equals space RT
                or space space space space space straight M equals space dRT over straight P space space space space space.... left parenthesis 1 right parenthesis
         
     Here space space space space space space space straight P space equals space 1 space bar space equals space 1.0 space cross times space 10 to the power of 5 Nm to the power of negative 2 end exponent
space space space space space space space space space space space space space straight T space equals space 273.15 space straight K
space space space space space space space space space space space space space straight R space equals space 8.314 space Nm space straight K to the power of negative 1 end exponent mol to the power of negative 1 end exponent
space space space space space space space space space space space space space straight d space equals space 1.29 space kg space straight m to the power of negative 3 end exponent
    Substituting the values in eq. (1), we have,
       straight M space equals space fraction numerator left parenthesis 1.29 space kg space straight m to the power of negative 3 end exponent right parenthesis space cross times space left parenthesis 8.314 space straight N space mk to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis space cross times space left parenthesis 273.15 space straight K right parenthesis over denominator 1.0 space cross times space 10 to the power of 5 Nm to the power of negative 2 end exponent end fraction
space space space space equals space fraction numerator 1.29 space cross times space 8.314 space cross times space 273.15 space kg space mol to the power of negative 1 end exponent over denominator 1 space cross times space 10 to the power of 5 end fraction
space space space space equals space 0.0293 space kg space mol to the power of negative 1 end exponent
space space space space equals space 29.3 space straight g space mol to the power of negative 1 end exponent
therefore space space space Molar space mass space is space 29.3 space straight g space mol to the power of negative 1 end exponent
    Question 113
    CBSEENCH11005932
    Question 114
    CBSEENCH11005933

    2.9 g of a gas at 95°C occupied the same volume as 0·184g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas? 

    Solution

    For gas:
                    PV space equals space nRT               ...(1)
    Here comma space space straight n space equals space fraction numerator Mass space of space gas over denominator Molecular space mass space of space gas end fraction space equals fraction numerator 2.9 space straight g over denominator straight M subscript gas end fraction
space space space space space space space space space straight T space equals space 95 plus 273 space equals space 368 space straight K
    Substituting the values  in eq. (1), we have,
                   PV space equals space fraction numerator 2.9 straight g over denominator straight M subscript gas end fraction cross times straight R cross times 368 straight K space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    For hydrogen gas:
      straight n space equals space fraction numerator Mass space of space hydrogen over denominator Molecular space mass space of space hydrogen end fraction
space space space space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction
straight T space equals space 17 plus 273 space equals space 290 space straight K
    therefore Substituting the values in eq. (1) we have,
       
     PV space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction cross times space straight R space cross times space 290 straight K   ...(3)
    .
    From eq. (2) and (3), we have
    fraction numerator 2.9 straight g over denominator straight M subscript gas end fraction cross times space straight R cross times 368 space straight K space equals space fraction numerator 0.184 space straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction space cross times space straight R space cross times space 290 space straight K
therefore space space space space straight M subscript gas space equals space fraction numerator 2.9 space cross times space 368 space cross times space 2 over denominator 0.184 space cross times space 280 end fraction straight g space mol to the power of negative 1 end exponent
space space space space space space space space space space space space space space equals space fraction numerator 2134.4 space over denominator 53.36 end fraction straight g space mol to the power of negative 1 end exponent space equals space 40 space straight g space mol to the power of negative 1 end exponent

    Question 115
    CBSEENCH11005934

     6·02 × 1022 oxygen molecules are present at –13°C in a 1500mL flask. What would be the pressure of the gas?

    Solution

     Here comma space space straight V space equals space 1500 over 1000 straight L space equals 1.5 space straight L semicolon
space space space space space space space space space straight T space equals space 273 space minus space 13 space equals space 260 space straight K
space space space space space space space space space straight n space equals space fraction numerator 6.02 space cross times space 10 to the power of 22 over denominator 6.02 space cross times space 10 to the power of 23 end fraction space equals space 0.1
    Applying ideal gas equation,    PV = nRT
    Substituting the values, in the above equation, we have
      straight P space cross times space 1.5 space equals space 0.1 space cross times space 0.0821 space cross times space 260
therefore space space space space space space straight P space equals space 1.423 space bar

    Question 116
    CBSEENCH11005935

    Calculate the mass of methane in a 9L cylinder at 16 bar and 27°C. (R = 0·08L bar K–1 mol–1).

    Solution
    Let the mass of methane by weight.
    space space Molecular space mass space of space methane space left parenthesis CH subscript 4 right parenthesis space equals space 16
space space straight V space equals space 9 straight L comma space space space straight T space equals space 273 space plus space 27 space equals space 300 space straight K
space space space straight n space equals space straight w over 16 comma space space space straight P space equals space 16 space bar
    Applying ideal gas equation PV = nRT
    Substituting the values in the above equation, we have,
               space space 16 space cross times space 9 space equals space straight w over 16 cross times 0.08 space cross times space 300
    space space therefore space space space space straight w space equals space fraction numerator 16 space cross times space 16 space cross times space 9 over denominator 0.08 space cross times space 300 end fraction space equals space 96
    Thus, mass of methane = 96g
        
    Question 117
    CBSEENCH11005936

    Two flasks A and B have equal volumes. Flask A contains H2 and is maintained at 300K while the flask B contains an equal mass of CH4 gas and is maintained at 600K. In which flask is the pressure greater? How many times ?

    Solution
    Let the mass of each gas in the two flasks = mg
    Molecular mass of H2 = 2
    Molecular mass of CH4 = 16
    therefore space space Number space of space moles space of space straight H subscript 2 left parenthesis nH subscript 2 right parenthesis equals space straight m over 2
Number space of space moles space of space CH subscript 4 left parenthesis nCH subscript 4 right parenthesis space equals space straight m over 16
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#13 {main}</pre>
    For space CH subscript 4 space gas colon
straight P subscript CH subscript 4 end subscript space equals space fraction numerator CH presuperscript straight n subscript 4 RT over denominator straight V end fraction space equals space fraction numerator CH presuperscript straight n subscript 4 space cross times space straight R space cross times space 600 over denominator straight V end fraction space space space... left parenthesis 2 right parenthesis
    Dividing (1) and (2), we get,
      
    straight P subscript straight H subscript 2 end subscript over straight P subscript CH subscript 4 end subscript space equals space fraction numerator straight H presuperscript straight n subscript 2 cross times straight R cross times 300 over denominator straight V end fraction cross times space fraction numerator straight V over denominator CH presuperscript straight n subscript 4 space cross times space straight R space cross times space 600 end fraction
space space space space space space space space space space space equals fraction numerator straight H presuperscript straight n subscript 2 over denominator CH presuperscript straight n subscript 4 end fraction cross times 300 over 600
space space space space space space space space space space space equals space straight m over 2 cross times 16 over straight m cross times 300 over 600 equals space space fraction numerator 8 cross times 300 over denominator 600 end fraction space equals space 4
therefore space space space space space straight H presuperscript straight P subscript 2 space equals space 4 space cross times space straight P subscript straight H subscript 2 end subscript
    Thus, the pressure of H2 in flask A will be four times the pressure of CH4 in flask B.
    Question 118
    CBSEENCH11005937

    Pressure of 1g of an ideal gas A at 27° C is found to be 2 bar. When 2g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.

    Solution

    Let the molecular masses of A and B be MA and MB.
    therefore space space Number space of space moles space of space straight A left parenthesis straight A presuperscript straight n right parenthesis space equals space 1 over straight M subscript straight A
    Number space of space moles space of space straight B left parenthesis straight n subscript straight B right parenthesis space equals 2 over straight M subscript straight B
Now comma space space space space straight P subscript straight A space equals 2 space bar comma space space space space straight P subscript straight A plus straight P subscript straight B space equals space 3 space space bar
straight i. straight e. space space space space space space straight P subscript straight B space equals space 3 space minus space straight P subscript straight A
space space space space space space space space space space space space space space space space equals space 3 minus 2 space equals space 1 space bar
    Applying ideal gas equation
                PV = nRT.
    For ideal gas A:
            PAV = nART
    For ideal gas B:
          PBV = nBRT
    therefore space space space space straight P subscript straight A over straight P subscript straight B space equals space straight n subscript straight A over straight n subscript straight B space equals space fraction numerator 1 divided by straight M subscript straight A over denominator 2 divided by straight M subscript straight B end fraction equals space fraction numerator straight M subscript straight B over denominator 2 straight M subscript straight A end fraction
or space space space space space fraction numerator 2 space bar over denominator 1 space bar end fraction space equals space fraction numerator straight M subscript straight B over denominator 2 straight M subscript straight A end fraction
or space space space space space space straight M subscript straight B space equals space 4 straight M subscript straight A

    Sponsor Area

    Question 120
    CBSEENCH11005939

    State and explain Dalton's law of partial pressure ?

    Solution

    It states,“If two or more gases, which do not react chemically, are present together in an enclosed space, the total pressure exerted by the gaseous mixture is equal to the sum of partial pressures which each gas would exert if present alone in that space at the same temperature.”
    Mathematically, Ptotal = p1 + p2 + p3 + .... where Ptotal = total pressure exerted by the mixture in a given volume (V) at a given temperature (T)
    p1,p2,p3 .... = partial pressure exerted by each
    gas. 
    But space space space straight P space equals space nRT over straight V space space left square bracket from space gas space equation right square bracket
space therefore space space space space space space space straight P subscript total space equals space fraction numerator straight n subscript 1 RT over denominator straight V end fraction plus fraction numerator straight n subscript 2 RT over denominator straight V end fraction plus fraction numerator straight n subscript 3 RT over denominator straight V end fraction plus....
    
    space space space equals space RT over straight V left parenthesis straight n subscript 1 plus straight n subscript 2 plus straight n subscript 3 plus..... right parenthesis
space space equals space fraction numerator total presuperscript straight n space RT over denominator straight V end fraction space space space space space open square brackets because space space space PT over straight V equals straight P subscript total over straight N subscript total close square brackets
therefore space space space Partial space pressure comma space
space space space space space space space space space space space space space space space straight p subscript 1 space equals space straight n subscript 1 over straight n subscript total space cross times space straight P subscript total space equals space straight X subscript 1 space cross times space straight P subscript total
space space space space space space space space space space space space space space space straight p subscript 2 space equals space straight n subscript 2 over straight n subscript total cross times straight P subscript total space equals space straight X subscript 2 space cross times space straight P subscript total
and space so space on....
    Here X1, X..... are mole fractions of respective gases.
    The partial pressure of a gas in a mixture is equal to the mole fraction of the gas multiplied by the total pressure exerted by the gaseous mixture.

    Question 121
    CBSEENCH11005940

    How is Dalton's law of partial pressures useful in calculating the pressure of a dry gas?

    Solution
    Gases like hydrogen, nitrogen, oxygen etc. which do not dissolve in water are collected over it. The pressure of the gas recorded is the pressure of the moist gas. The pressure of the dry gas can be calculated with the help of Dalton’s law of partial pressures. Mathematically:
    Pmoist gas = Pdry gas + Pwater vapours

    But pressure exerted by the water vapours is called aqueous tension. Thus,
    Pdry gas = Pmoist gas - Aqueous tension
    Question 122
    CBSEENCH11005941

    300 mL of a gas A at a pressure of 600 mm is mixed with 200mL of another gas B at a pressure of 700 mm in a vessel of 2-litre capacity. What will be the total pressure of the resulting mixture, if the temperature is kept constant?

    Solution

    Let the partial pressure of the gases A and B be represented by PA and PBrespectively.
    (i) Partial pressure of gas A in 2 litre flask:
       V1 = 300 mL;      P2 = 600 mm
       V2 = 2 L = 2 x 1000 = 2000 mL;  PA = ?
    According to Boyle's law,  P1V1 = P2V2
    i.e.  600 x 300 = PA x 2000
    therefore space space space space space space straight P subscript straight A space equals space fraction numerator 600 space cross times space 300 over denominator 2000 end fraction space equals space 90 space nm
    (ii) Partial pressure of gas B in 2 litre flask:
           V1 = 200 mL,   P1 = 700 mL
           V2 = 2 litre = 2 x 1000 = 2000 mL
           PB = ?
    According to Boyle's law,
                  P1V1 = P2V2
    i.e.     700 x 200 = PB X 2000
    therefore space space space space straight P subscript straight B space equals space fraction numerator 700 space cross times space 200 over denominator 2000 end fraction space equals space 70 space mm
    therefore  Total pressure exerted by the mixture
          straight P space equals space straight P subscript straight A plus straight P subscript straight B
space space space space space space space space left square bracket By space Dalton apostrophe straight s space law space of space partial space pressure right square bracket
space space space space space space space space space space space space space space equals space 90 plus 70 space equals space 160 space mm
     

    Question 123
    CBSEENCH11005942

    What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C? 

    Solution
    Calculation of partial pressure of H2.
    straight V subscript 1 space equals space 0.5 space straight L semicolon space space space space space space space space straight V subscript 2 space equals space 1.0 space straight L
straight P subscript 1 space equals space 0.8 space bar space space space space space space space space straight P subscript 2 space equals space ?
    According to Boyle's law
                          straight P subscript 1 straight V subscript 1 space equals space straight P subscript 2 straight V subscript 2 space space space or space space space straight P subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight V subscript 2 end fraction
    Substituting the values, we have
             
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#13 {main}</pre>
    Calculation of partial pressure of O2.
    V1 = 2.0 L                     V2 = 1 L
    P1 = 0.7 bar                    P2 = ?
    According to Boyle's law
    P1V1 = P2V2    or   straight P subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight V subscript 2 end fraction
    Substituting the values, we have
    straight P subscript straight O subscript 2 end subscript space equals space fraction numerator 0.7 space bar space cross times space 2.0 space straight L over denominator 1 straight L end fraction space equals 1.4 space bar
therefore space space space Total space pressure space space equals space PH subscript 2 plus space PO subscript 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.4 space bar space plus space 1.4 space bar space equals space 1.8 space bar

    Question 124
    CBSEENCH11005943

    A gaseous mixture containing 12g of oxygen and 52 g of nitrogen were enclosed in a vessel of 10L capacity at 27°C. Calculate:
    (i) Partial pressure of each gas
    (ii) Total pressure of the gaseous mixture.

    Solution

    Weight of oxygen  = 12 g
    Molar mass of oxygen = 32g mol-1
    space space therefore space space Number space of space moles space of space oxygen space equals space 12 over 32 space equals space 0.375 space
Weight space of space nitrogen space space equals space 52
Molar space mass space of space nitrogen space equals space 28 space straight g space mol to the power of negative 1 end exponent
therefore space space space Number space of space moles space equals space 52 over 28 space equals space 1.857
    (a) Partial pressure of oxygen gas:
    According to general gas equation,
                          PV = nRT
    H e r e space straight P space equals space straight P subscript straight O subscript 2 end subscript space space colon space space space space straight V space equals space 10 space straight L
space space space space space space straight T space equals space 273 plus 27 space equals space 300 space straight K
    Substituting the value.
     straight P subscript straight O subscript 2 end subscript space cross times space 10 space equals space 0.375 space cross times space 0.0821 space cross times space 300
space space space space space therefore space space space straight P subscript straight O subscript 2 end subscript space equals space fraction numerator 0.375 space cross times space 0.0821 space cross times space 300 over denominator 10 end fraction space equals space 0.92 space bar
    b) The partial pressure of nitrogen gas: According to the general gas equation.
                       PV = nRT
      straight i. straight e. space space space straight P subscript straight N subscript 2 end subscript space cross times space 10 space equals space 1.857 space cross times space 0.0821 space cross times space 300
therefore space space space straight P subscript straight N subscript 2 end subscript space equals space fraction numerator 1.857 space cross times space 0.0821 space cross times space 300 over denominator 10 end fraction space equals space 4.57 space bar
    (c) Total pressure of the gaseous mixture : According to Dalton’s law of partial pressure,
    straight P space equals space straight P subscript straight O subscript 2 end subscript space plus space straight P subscript straight N subscript 2 end subscript
space space space equals space 0.92 plus 4.57 space equals space 5.49 space bar

     

    Question 125
    CBSEENCH11005944

    A neon-dioxygen mixture contains 70.6 g dioxygen and 167·5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar, what is the partial pressure of dioxygen and neon in the mixture? 

    Solution

    Weight of dioxygen (O2) = 70.6 g
    Molar mass of dioxygen  = 32 g mol-1

    therefore space space space Number space of space moles space of space dioxygen
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 70.6 over denominator 32 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 2.21 space mol
Weight space of space neon space equals space 167.5 space straight g
Molar space mass space of space neon space equals space 20 space straight g space mol to the power of negative 1 end exponent
    Number space of space moles space of space neon space equals space fraction numerator 167.5 space straight g over denominator 20 space straight g space mol to the power of negative 1 end exponent end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8.375 space mol
Mole space fraction space of space dioxygen space equals space fraction numerator 2.21 over denominator 2.21 plus 8.375 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 2.21 over denominator 10.585 end fraction space equals 0.21 space space space space space space space space space space space space space space space
    Mole fraction of neon
                  space equals space fraction numerator 8.375 over denominator 2.21 plus 8.375 end fraction space equals space fraction numerator 8.375 over denominator 10.585 end fraction space equals space 0.79
    We know,
       Partial pressure of a gas = Mole fraction x Total pressure
          therefore   Partial pressure of  a gas  = Mole fraction x total pressure
                            = 0.21 x 25 bar = 5.25 bar
    Partial pressure of neon = 0.79 x 25 bar = 19.75 bar

    Question 127
    CBSEENCH11005946

    A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen. 

    Solution

    Since the mixture H2 and O2 contains 20% by weight of hydrogen, therefore if H2 = 20g, then O2 = 100 - 20 = 80 g
    therefore   No. of moles of dihydrogen (H2)
           space equals space 20 over 2 space equals space 10 space moles   
    No. of moles of dioxygen (O2) = equals space 80 over 32 space equals space 2.5 space moles
    Partial pressure of a gas = mole fraction of the gas X total pressure,
    therefore space space space Partial space pressure space of space dihydrogen space left parenthesis straight p subscript straight H subscript 2 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator straight n subscript straight H subscript 2 end subscript over denominator straight n subscript straight H subscript 2 end subscript plus space straight n subscript straight O subscript 2 end subscript end fraction space cross times space straight P subscript total
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 10 over denominator 10 plus 2.5 end fraction cross times 1 space bar space equals space 0.8 space bar                                 

    Question 128
    CBSEENCH11005947

    Calculate the total pressure in a mixture of 8g of dioxygen and 4g of dihydrogen confined in a vessel of 1 dm3 at 27°C. R = 0·083 bar dm3 k–1 mol–1

    Solution
    Number of moles of dioxygen (n1,)
                      equals space fraction numerator Mass space of space dioxygen over denominator Molecular space mass end fraction space equals space fraction numerator 8 straight g over denominator 32 space straight g space mol to the power of negative 1 end exponent end fraction space equals space 0.25 space mol
    Number of moles of dihydrogen (n2)
                      equals space fraction numerator Mass space of space dihydrogen over denominator Molecular space mass end fraction space equals space fraction numerator 4 straight g over denominator 2 straight g space mol to the power of negative 1 end exponent end fraction space equals space 2 space mol
    Temperature (T) = 27+ 273 = 300 K
    Now let us calculate the partial pressure of dioxygen and dihydrogen by using ideal gas equation,
    space space space space straight P subscript straight O subscript 2 end subscript space equals space fraction numerator straight n subscript 1 RT over denominator straight V end fraction
space equals space fraction numerator left parenthesis 0.24 space mol right parenthesis space cross times left parenthesis 0.083 space bar space dm cubed space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis space cross times space 300 space straight K over denominator 1 space dm cubed end fraction
space space space equals 6.225 space bar
straight P subscript straight H subscript 2 end subscript space equals space fraction numerator straight n subscript 2 RT over denominator straight V end fraction
equals space fraction numerator left parenthesis 2 space mol right parenthesis space cross times space left parenthesis 0.083 space bar space dm cubed thin space straight K to the power of negative 1 end exponent space mol to the power of negative 1 end exponent right parenthesis space cross times space 300 straight K over denominator 1 space dm cubed end fraction
space space space space space space equals space 49.8 space bar
therefore space space space Total space pressure space of space gaseous space mixtu
equals space 6.225 space space bar space plus space 49.8 space bar space equals space 56.025 space bar
space space space space space space space space space space space space space space space space space
    Question 129
    CBSEENCH11005948

    What do you mean by kinetic molecular theory of gases? What is its purpose?

    Solution

    This model is used to describe the behaviour of gases. The model itself along with the assumptions regarding elastic collisions, momentum, kinetic energy, random motion, temperature, volume and pressure of gases is known as the kinetic molecular theory of gases. The main postulates of kinetic theory of gases are rapidly moving molecules. The molecules collide with one another and with the walls of the containing vessel.
    Purpose: The main purpose of the kinetic theory of gases is to develop a physical picture of the nature of the gaseous state capable of explaining various laws of gaseous behaviour. This theory is applicable only to a perfect or an ideal gas.

    Question 130
    CBSEENCH11005949

    Outline the basic assumptions/postulates of kinetic theory of gases.

    Solution

    The important assumptions/postulates of kinetic theory of gases are:
    (i) A gas consists of a large number of identical molecules of mass m. The dimensions of these molecules are very-very small as compared to the space between them. Hence the molecules are treated as point masses.
    (ii) The molecules of a gas are always in a state of zig-zag motion. They collide with one another and with the walls of the containing vessel. The direction of motion and speed of the molecules change continuously.
    (iii) The collisions between the gas molecules are perfectly elastic i.e. there is no loss of energy during these collisions.
    (iv) There are practically no attractive forces between the molecules. The molecules, therefore, move independently.
    (v) The pressure of the gas is the result of collisions of molecules with the walls of the container.
    (vi) The average kinetic energy of the colliding molecules is directly proportional to its absolute temperature.
    In view of those assumptions, the above model of a gas is called Kinetic Gas Model or Dynamic Particle Model.

    Question 131
    CBSEENCH11005950

    Write a short note on distribution of velocities of gas molecules.

    Solution

    We know that molecules in a gas go on colliding with each other and hence their velocities go on changing though the average velocity remains unchanged. After the collision, the faster molecule slows down and slower molecule speeds up. The collisions are very rapid (many collisions per second) and hence, it is not possible to find the velocity of any single molecule. However, there must be a fraction of the total number of molecules that has a particular velocity at any time. The distribution of velocities amongst the gaseous molecules i.e. at a given instant how many molecules are moving with a particular speed was calculated by Maxwell and Boltzmann using the laws of probability. The fraction of molecules (∆N/N) possessing a particular velocity is taken on the Y-axis and the velocity on the X-axis. The curve obtained is shown in the figure. It is called ‘normal distribution curve of molecular velocities’.

    The graph supplies the following information:
    (i) The area under the curve gives the total number of molecules.
    (ii) A very small fraction of molecules has very low (close to zero) velocities.
    (iii) The fraction of molecules having higher and higher velocities keep on increasing with the increase in velocity, reaches a peak value and then starts decreasing for very high velocity. Once again, a very small fraction of molecules has very high velocities.
    (iv) The maximum fraction of molecules possesses a velocity corresponding to the peak of the curve. This velocity is termed as most probable velocity (a). Thus, most probable velocity may be defined as the velocity possessed by the maximum fraction of molecules at a given temperature.

    Question 132
    CBSEENCH11005951

    What is the effect of temperature on the distribution of molecular velocities?

    Solution

    On increasing the temperature, the motion of the gas molecules becomes rapid and hence the value of most probable velocity also increases. As a result, the entire distribution curve becomes flatter and peak shifts to regions of higher velocities as shown in the figure.

    α1 = Most probable velocity at temp. T1

    α2 = Most probable velocity at temp. T2

    α2 > α1

    Question 133
    CBSEENCH11005952

    Even though carbon dioxide is heavier than air, it does not form the lower layer, Explain. 

    Solution
    The diffusion of gases is independent of the gravitational pull of the earth. As a result, carbon dioxide diffuses throughout the atmosphere and does not form a lower layer of the atmosphere.
    Question 134
    CBSEENCH11005953

    What are ideal and real gases? Give the points of difference between them.

    Solution
    Ideal gases: A gas which obeys the general gas equation and other gas laws under all conditions of temperature and pressure is known as an ideal gas or perfect gas.
    Characteristics of an ideal gas: The molecules of an ideal gas
    (i) occupy a negligible volume
    (ii) have no intermolecular attractive forces.

    Real gases: A gas which does not obey general gas equation and all other gas laws strictly but tends towards ideality at low pressure and high temperature is known as real gas or non-ideal gas.
    Points of differences:

    Ideal gas Real gas
    1. It obeys gas laws (PV = RT) under all conditions of temp, and pressure. 1. It obeys gas laws only at high temp. and low pressure. 
    2. It does not exist in actual practice. The concept of an ideal gas is only hypothetical. 2. All gases are real gases.
    3. The volume occupied by molecules is negligible as compared to the total volume of gas. 3. The volume occupied by molecules is not negligible.
    4. There are no intermolecular forces of attraction or repulsion among the molecules. 4. There are intermolecular forces of attraction due to which the pressure exerted is less than that calculated from gas laws.
    Question 135
    CBSEENCH11005954

    What is compressibility factor? What is its value for ideal gas and real gas?

    Solution
    Compressibility factor: The deviation of real gases from ideal behaviour can be studied by plotting a graph between PV over nRT space and space straight P. The quantity PV over nRT is called compressibility factor and is denoted by Z.
                          straight Z space equals space PV over nRT

    For an ideal gas,   Z = 1 under all conditions of temperature and pressure. 
    therefore space space space space space space space PV over nRT space equals space 1
or space space space PV space equals space nRT space left parenthesis for space ideal space gas right parenthesis
For space real space gases comma space straight Z less than 1 space space space or space space straight Z greater than 1
Thus comma space PV space not equal to space nRT space for space real space gases
The space graph space PV over nRT versus space straight P space for space some space common
    gases such as H2, N2, CH4 and CO2 supplies the following information:
    (i) Z = 1, for an ideal gas.
    (ii) Z < 1, it is called negative deviation. It means that the gas is more compressible than expected from ideal behaviour.
    (iii) If Z > 1, it is called positive deviation. It means that the gas is less compressible than expected from ideal behaviour.
    It may be noted that:
    (a) For N2, CH4 and CO2, Z < 1 at low pressure and Z > 1 at high pressures. This means that these gases are more compressible at low pressures and less compressible at high pressures than expected from ideal behaviour.
    (b) For H2 and He, Z > 1 at low and high pressures. This means that these gases are less compressible than expected from ideal behaviour at all pressures.
    (c) At high-pressure Z shows a large deviation from ideal behaviour. However, at very low-pressure Z is nearly equal to 1.
    (d) It may also be noted that the deviations decrease with the increase in temperature.
    Thus, the real gases behave in a nearly ideal manner only at low pressures and high temperatures.
    Question 136
    CBSEENCH11005955

    What are the faulty assumptions in the kinetic theory of gases which are responsible for deviations from ideal behaviour of gases?

    Solution

    Vander Waal pointed out that following two assumptions in kinetic theory are faulty:
    (i) The volume occupied by the gas molecule is negligible in comparison to the total volume of the gas. This assumption is nearly valid if the pressure is low.
    (ii) The molecules do not exert any force of attraction upon one another. This assumption is nearly valid when the pressure is low and the temperature is high.

    Question 137
    CBSEENCH11005956

    Why do real gases show deviation from ideal behaviour?

    Solution

    Real gases deviate from ideal behaviour. The disobedience of the ideal gas equation, PV = n RT by real gases may be due to the following two faulty assumptions of kinetic theory of gases:
    (i) The actual volume of molecules is negligible in comparison to the total volume of the gas. This assumption is true only at low pressure and high temperature. But at high pressures or at low temperatures, the total volume of the gas decreases while the volume of the molecules being incompressible remains the same. Thus, under these conditions, the volume of the molecules can not be neglected.
     
    (ii) The forces of attraction or repulsion between the gas molecules are negligible. No doubt, this assumption is true only at low pressure and high temperature since the molecules lie far apart from one another. But at high pressure and low temperature, the volume of the gas is quite small. Gas molecules come closer to each other. The attractive forces between the gas molecules under these conditions are quite appreciable and cannot be neglected.

    Question 138
    CBSEENCH11005957

    What modifications were applied by Vander Waal to overcome the deviations of the gases from ideal gas behaviour?

    Solution
    Vander Waal applied corrections for both molecular volume and molecular attraction and has given a new equation in place of the general gas equation, PV= n RT. This equation is called Vander Waal’s equation of state.
    open parentheses straight P plus an squared over straight V squared close parentheses space space left parenthesis straight V minus nb right parenthesis space equals space nRT
    [For n moles of the gas]
    where a and b are constants called Vander Waal's constant and n represents the number of moles of the gas present in volume V litres of the gas. 
    For one mole of a gas, the equation reduces to
    open parentheses straight P plus straight a over straight V squared close parentheses space left parenthesis straight V minus straight b right parenthesis space equals space RT.
    Question 139
    CBSEENCH11005958

    Derive an expression for the Vander Waal's equation of state.

    Solution

    Vander Waal in 1873, modified the general gas equation (PV = nRT) by introducing two correction terms:
    (i) Volume correction. He pointed out that for real gases the volume of the gas molecules is not to be neglected in comparison to the total volume of the gas. This means that the molecules are not free to move in the whole of volume V but the free volume is less than the observed volume. In other words, the ideal volume of the gas is less than the observed volume. Vander Waal suggested that
    Videal = (V – b) for one mole
    where b is the correction term known as co-volume or excluded volume. It is related (not equal) to the actual volume of gas molecules. It has been found that b is about four times the actual volume of the gas molecules.
    For n moles of a gas,
    Videal = (V – nb)
    (ii) Pressure correction. There is an attractive force of attraction between molecules of a gas.
    A molecule (M) in the interior of the container is surrounded uniformly on all sides by other molecules, so there is no net force of attraction on it.
    But when a molecule (N) is about to strike the wall of the container, the molecules are present only on one side of it will attract the striking molecule and pull it back. Thus, the molecule strikes the wall with a lesser force than it would have done if there were no attractive forces. Therefore, the observed pressure is less than the ideal pressure. Consequently, some correction factor must, therefore, be added to the observed pressure in order to calculate ideal pressure.

    therefore space space space space space space space straight P subscript ideal space equals space straight P plus space straight beta

    where P is the observed pressure and β is the correction term. The necessary correction term (β) will depend on upon the number of molecules in the surface layer and also the number of molecules in the body of the gas. Both these are directly proportional to the pressure and inversely proportional to the volume.
    For 1 mole of a gas, Vander Waal suggested the factor straight a over straight V squared

     therefore space space space space space straight P subscript ideal space equals space straight P space plus space straight a over straight V squared space for space one space mole
and space space space space space straight P subscript ideal space equals space straight P space plus space an squared over straight V squared space for space straight n space moles
    Substituting the corrected values in PV = nRT, we have,
    open parentheses straight P plus an squared over straight V squared close parentheses space left parenthesis straight V minus nb right parenthesis space equals space nRT space for space straight n space moles
open parentheses straight P plus straight a over straight V squared close parentheses space left parenthesis straight V minus straight b right parenthesis space equals space RT space for space one space mole.
    The constants a and b are called Vander Waal's constants.
    Question 140
    CBSEENCH11005959

    What is the significance of the Vander Waal's constants? Write its units for Vander Waal's constants.

    Solution

    Vander Waal's equation of state for n moles of the gas is
    open parentheses straight P plus an squared over straight V squared close parentheses space left parenthesis straight V minus nb right parenthesis space equals space nRT
    The significance of constant 'a'. The value of constant a gives the idea of the magnitude of attractive forces between the molecules of the gas. Larger the value of a the larger will be the intermolecular forces between the molecules. Units of a. Pressure correction (β) is given by
                         straight beta space equals space an squared over straight V squared
    therefore space space space straight a space equals space space fraction numerator straight beta space cross times space straight V squared over denominator straight n squared end fraction
space space space space space space space space space equals space fraction numerator Pressure space cross times space left parenthesis volume right parenthesis squared over denominator left parenthesis mole right parenthesis squared end fraction
space space space space space space space space space equals space fraction numerator atm space cross times space left parenthesis litre right parenthesis squared over denominator left parenthesis mole right parenthesis squared end fraction space equals space atm space straight L squared space mol to the power of negative 2 end exponent
    The significance of constant 'b'.  The constant b is called co-volume or excluded volume per mole of a gas. It is a measure of the effective size of gas molecules.
    Units of b. Units of b must be units of volume i.e. L mol–1 or dm3 mol–1.

    Question 141
    CBSEENCH11005960

    What are the limitations of Vander Waal's equation of state?

    Solution

    The values of 'a' and 'b' are not constant over the entire ranges of temperatures and pressures, so van der Waal’s equation is also valid for specific ranges of temperatures and pressures. Real gases deviate from van der Waal’s equation especially at too low temperatures or too high pressures.

    Question 142
    CBSEENCH11005961

    Write a short note on liquefaction of gases.

    Solution

    For the liquefaction of gas, the condition should be so arranged that intermolecular attraction forces are increased and kinetic energy is decreased. If the pressure is sufficiently high and the temperature is lowered enough, the intermolecular attractive forces overcome the tendency of molecules to move apart and the gas molecules condense to the liquid state. Gases cannot be liquefied by pressure alone, they must be at the same time below their critical temperatures for this purposes. The following requirements are necessary for the liquefaction of gases:
    (i) The gases should be pure and free from water vapours.
    (ii) The temperature of the gas must be brought below its critical temperature.
    (iii) The pressure applied must be great enough to cause the liquefaction.

    Question 143
    CBSEENCH11005962

    Explain what is meant by 'critical phenomenon'? What are 'critical constants' of a gas?

    Solution

    There are three critical constants for each gas namely critical temperature (Tc), the critical pressure (Pc) and the critical volume (Vc). If we go on cooling a gas to a sufficiently low temperature and increase the pressure on it, the gas can be liquefied. But for every gas there is a characteristic temperature above which it cannot be converted to the liquid state, however, we may increase the pressure on it. This limiting value of temperature is not the same for all gases but is different for different gases. This limiting value of temperature is called the critical temperature and is denoted by Tc. Hence critical temperature is defined as the temperature below which the continuous increase of pressure on a gas ultimately leads to its liquefication and above which no liquefication is possible, however, great the pressure be applied.

    At the critical temperature, a certain minimum pressure has to be applied to the gas to liquefy it. It is called critical pressure and denoted by Pc. This critical pressure may be defined as ‘the minimum pressure which must be applied to a gas at its critical temperature to liquefy it.’

    When the gas is at both critical temperature (Tc) and critical pressure (Pc), it is said to be under a critical state or at a critical point. At this stage, both gas and its corresponding liquid would occupy the same volume and therefore their densities are equal. In other words, we can say that at this stage, it is not possible to distinguish between the liquid and gaseous states of the substance as the two forms are existing in equilibrium. The phenomenon of a smooth merging of a gas into its liquid state under a critical state or critical point is referred to as ‘critical phenomenon’. The density of the gas at the critical point is called the critical density.

    The volume occupied by one mole of a gas at its critical temperature and critical pressure is called the critical volume and is denoted by Vc.
    Tc, Pc and Vc are known as critical constants of the gas.

    Question 144
    CBSEENCH11005963

    Describe briefly the Isotherm of Carbon dioxide.
    Or
    Briefly describe the Isotherm of Carbon dioxide as studied by Andrews.

    Solution
    Carbon dioxide has been studied in detail by Andrews. At different temperatures, he studied the behaviour of the gas. Keeping temperature constant, the various pressures were plotted against volumes, as obtained from his experiments.These P as V plots at constant temperature are called isotherms. Andrews obtained a number of typical isotherms from his experiments on carbon dioxide as shown below. These isotherms can be roughly classified as follows:

    A) Isotherms at a temperature much above the critical temperature (31·1°C): These are very much like the ideal gas isotherms (Isotherm I).

    (B) Isotherms at a temperature just above the critical temperature : They have a point of inflexion which is not found in the ideal gas isotherms (Isotherm II).

    (C) Isotherms at a temperature just above the critical temperature : These show a straight horizontal part along which the gas gets converted to the liquid state (Isotherms III and IV).

    This is the case of discontinuous liquefaction as the properties of the substance in the gaseous and liquid forms are markedly different from each other.

    In each of these isotherms, like isotherm (IV) at 13°C, it is clear that with the increase of pressure, the volume of the gas decreases along AB according to Boyle’s law and at B, the liquefaction starts. On decreasing the volume further, there is no change of pressure (along BC) until point C is reached where the whole of the gas has been liquefied.

    Curve CD is almost vertical showing that liquid obtained is slightly compressible.

    At higher temperature, say 21°C, curve III is a similar isotherm except that horizontal position is shorter. If the temperature is further raised, the horizontal part of the isotherm becomes shorter and shorter until it is reduced to a point F as in curve II at the critical temperature.

    Liquefaction of a gas does not occur at all along the isotherm above the critical temperature. It is only along the critical temperature isotherm that a gas undistinguishably continues into the liquid state.

    All gases behave similarly except that the length of horizontal portions and peak of the dotted parabola (F) are different depending on the nature of the gas.

    Question 145
    CBSEENCH11005964

    What are the main postulates of Kinetic energy Molecular Theory of liquids?

    Solution

    A liquid state is regarded as a condensed gas or a molten solid. The following are the postulates of kinetic molecular theory of liquids:
    1. A liquid is made up of small particles called molecules and these molecules have restricted random motion.
    2. The molecules are quite close to each other.
    3. There are appreciable attractive forces between the molecules.
    4. The average kinetic energy of the molecules in a liquid is proportional to its absolute temperature.

    Question 146
    CBSEENCH11005965

    On the basis of kinetic molecular theory of liquids, how can you explain the following properties of liquids:
    (i) Volume                             (ii) Density
    (iii) Compressibility                (iv) Diffusion?

    Solution

    1. Volume. Liquids have a definite volume. This is because the intermolecular forces of attraction in a liquid are quite strong. As a result, the molecules are not completely free to occupy all the space available (as gases do).

    2. Density. Liquids have much higher density than gases. This is because the molecules in the liquid state are closely packed. Hence, the volume for the same amount of the substance in the liquid state is less as compared to the volume in the gaseous state. Consequently, the density (m v) is greater in a liquid state.

    3. Compressibility. Liquids are much less compressible than gases. This is due to the fact that in comparison to gases, there is very little empty space between the molecules in the case of liquids.

    4. Diffusion. Liquids diffuse slowly in comparison to gases. This is because the molecules in the liquid state are quite close to each other. Therefore, a molecule of the liquid has to undergo a number of collisions in the neighbouring molecules. As a result of innumerable collisions, there is a more obstruction for the movement of the molecules of a liquid. Thus diffusion takes place slowly.

    Question 147
    CBSEENCH11005966

    What is evaporation? 

    Solution

    The spontaneous change in which a liquid change into vapours (or gaseous) at the surface of the liquid is called evaporation. 
    Evaporation may be explained in terms of the motion of molecules. At any given temperature, all the molecules do not possess the same kinetic energy (average kinetic energy is. however, same). Some molecules move slowly, some at intermediate rates and some move very fast. Rapidly moving molecules near the surface of the liquid, possessing extremely high energies are sufficient to overcome the intermolecular forces of attraction. These molecules tend to escape into space above the liquid surface and form vapours.

    Question 148
    CBSEENCH11005967

    Explain briefly the factors on which evaporation depends.

    Solution

    Evaporation depends on upon the following factors:
    (i) Nature of the liquid. The evaporation depends on the strength of intermolecular forces. The liquids having low intermolecular forces evaporate faster in comparison to the liquids having high intermolecular forces. For example, ether evaporates more quickly than alcohol, and alcohol evaporates more quickly than water. This is because their intermolecular forces of attraction are in the order
    Ether < Alcohol < Water
    (ii) Surface area. Evaporation is a surface phenomenon. Larger the surface area, greater the opportunity of the molecules to escape. Thus, the rate of evaporation increases with an increase of surface area.
    A rate of evaporation ∝ Surface area
    (iii) Temperature. The rate of evaporation increases with
    the increase of temperature as the kinetic energy of the molecules increases with the rise in temperature.
    The rate of evaporation ∝ Temperature
    (iv) The flow of air current over the surface. The flow of air helps the molecules to go away from the surface of the liquid and therefore, increases the evaporation.

     
    Question 149
    CBSEENCH11005968

    What do you mean by heat of vaporisation? On what factors does it depend?

    Solution

    The quantity of heat required to evaporate a given liquid at constant temperature is known as heat of evaporation or heat of vaporisation. The heat of vaporisation depends on the strength of intermolecular forces within the liquid.
    The amount of heat required to evaporate 1 mole of a given liquid at constant temperature is known as molar heat of vaporisation or molar heat of evaporation. For example, the heat of vaporisation of water at 100°C is 40·6 kJ per mole while that of benzene at 80°C is 31 kJ per mole. The relatively high molar heat of vaporisation of water suggests strong attractive forces in liquid water.

    Question 150
    CBSEENCH11005969

    Explain briefly the term "vapour pressure" . What are the factors on which vapour pressure of a liquid depends?

    Solution
    Vapour pressure: If a liquid is allowed to evaporate in a closed vessel, then molecules escaping from the liquid surface strike the walls of the container, rebound and may strike the surface of the liquid, where they are trapped. The return of the molecules from the vapour state to the liquid state is known as condensation. As evaporation proceeds, the number of molecules in the vapour state increases and, in turn, the rate of condensation increases. The rate of condensation soon becomes equal to the rate of evaporation i.e. the vapour in the closed container is in equilibrium with the liquid.
    straight H subscript 2 straight O space left parenthesis straight l right parenthesis rightwards harpoon over leftwards harpoon from Condensation to Evaporation of space straight H subscript 2 straight O left parenthesis straight g right parenthesis
    At equilibrium the concentration of molecules in the vapour phase and the pressure that it exerts remains unchanged. The pressure exerted by the vapour in equilibrium with liquid, at a given temperature, is called the vapour pressure of the liquid at that temperature.
    Factors on which it depends:
    (i) Nature of liquid. Liquids having weak intermolecular forces of attraction have a tendency to evaporate and therefore, have a high vapour pressure. On the other hand, liquids having strong intermolecular attractive forces have a small tendency to evaporate and therefore, have low vapour pressure.

    (ii) Temperature. Vapour pressure of a liquid is directly proportional to its temperature. With the increase in temperature, the fraction of molecules having larger kinetic energy increases. Therefore, more molecules escape from the surface of the liquid. As a result, the concentration of molecules in the vapour phase will increase. Hence, the vapour pressure of the liquid increases.
    Question 151
    CBSEENCH11005970

    What do you mean by state of dynamic equilibrium?

    Solution
    During the evaporation of a liquid in a closed vessel, a stage is reached when the rate of evaporation becomes equal to the rate of condensation. In such a state, the system is not at rest though there is no change in the system because the amount of liquid level in the vessel does not change. This does not mean that evaporation has stopped. In fact, the number of molecules which escape from the liquid to the vapour phase (due to evaporation) becomes equal to the number of molecules which return to the liquid from the vapour phase (due to condensation). This state is known as dynamic equilibrium. At dynamic equilibrium state, the two opposing processes (evaporation and condensation) occur at an equal rate.

    Question 152
    CBSEENCH11005971

    Separate portions of chloroform and water at the same temperature are poured on your hands. The chloroform feels colder. Account for this in terms of attractive forces.

    Solution
    The intermolecular forces of attraction in chloroform are weaker than those in water (which also involves hydrogen bonding). As a result, the boiling point of chloroform is lesser. This means that chloroform evaporates faster than water. Consequently, for evaporation, it absorbs more of heat from the hand which gets the feeling of coldness.
    Question 153
    CBSEENCH11005972

    Why is cooling caused in evaporation?

    Solution
    During evaporation, the liquid molecules with higher kinetic energy leave the liquid surface. As a result, the average kinetic energy of the remaining liquid molecules continues to fall. Consequently, the temperature of liquid falls (since KE ∝ T) or cooling is caused by evaporation.
    Question 154
    CBSEENCH11005973

    Explain why alcohol produces a greater cooling effect on skin that water does.

    Solution
    The boiling point temperature of alcohol (78°C) is less than that of water (100°C). This means that alcohol changes into vapour state faster than water. Consequently, alcohol will absorb the greater heat of vaporisation than done by water from the skin. In other words, alcohol will cause a greater cooling effect on the skin.

    Question 155
    CBSEENCH11005974

    Why does the temperature of the boiling liquid remain constant?

    Solution

    When a liquid starts boiling, each of the escaping molecules takes away more than the average kinetic energy of the liquid. As such any heat being supplied increases the energy of the remaining molecules to such an extent that it exceeds escaping energy of molecules. Hence, liquid molecules continue to escape as vapours. As a result, the temperature of the boiling liquid, therefore, can not increase till the whole of it has vapourised.

    Question 156
    CBSEENCH11005975

    Explain briefly the boiling point of a liquid. What is the effect of change in external pressure on the boiling point?

    Solution

    When a liquid is heated in an open container, its vapour pressure keeps on increasing with the rise in temperature till it becomes equal to the atmospheric pressure. At this state, molecules of the liquid are readily converted to the vapour phase and the liquid is said to boil. The temperature at which the vapour pressure of the liquid becomes equal to the atmospheric pressure (i.e. external pressure) is called the boiling point of the liquid. When the external pressure is equal to the standard atmospheric pressure (i.e. 760 mm of mercury), the boiling point is called normal boiling point. For example, normal boiling points of water and benzene are 100°C and 80°C respectively.

    Effect of change in external pressure. It should be possible to change the boiling point of a liquid by changing the external pressure. Clearly, if the external pressure is higher, more heat will be required to make the vapour pressure equal to the external pressure and hence higher will be the boiling point. Similarly, if the external pressure is decreased, the boiling point is lowered. Boiling at a lower temperature under reduced pressure is used in the purification of such liquids which decompose at or below their normal boiling points.

    Question 157
    CBSEENCH11005976

    Will water boil at a higher temperature at sea level or on the top of a mountain ?

    Solution
    Water will boil at a lower temperature at the top of a mountain since the atmospheric pressure is lesser as compared to sea level. Thus, lesser energy will be required to change water into vapours at the top of a mountain.
    Question 158
    CBSEENCH11005977

    How will you differentiate between evaporation and boiling?

    Solution

    Evaporation

    Boiling

    1. It is slow process

    1. It is a fast process

    2. Evaporation occurs spontaneously at all temperatures.

    2. Boiling occurs at the boiling point at which the vapour pressure of the liquid is equal to the atmospheric pressure. 

    3. Evaporation takes place only at the surface of the liquid

    3. Boiling involves the formation of bubbles of vapour below the surface of the liquid.

    4. The rate of evaporation increases by increasing the surface area

    4. Surface area has no effect on the rate of boiling.

    Question 159
    CBSEENCH11005978

    What do you understand by vapour pressure of solids?

    Solution
    Solids also evaporate and thereby exhibit a vapour pressure. The molecules of a crystalline solid operate with respect to their equilibrium position and are in a position to collide constantly with the nearest neighbours which lead to a distribution of kinetic energies. A small fraction of molecules at the surface of the solid possesses enough kinetic energy to break away from the surface. As a result, the free molecules enter the vapour state. The process in which solid is directly converted to its vapour state is called sublimation. This accounts for the vapour pressure of solids.
    Question 160
    CBSEENCH11005979

    Explain the term surface tension.

    Solution
    To understand it, consider a molecule ‘M’ somewhere within the bulk of a liquid. It is attracted equally in all directions by the surrounding molecules. Hence, the net force of attraction acting on the molecule M is zero. However, a molecule ‘N’ on the surface of a liquid is attracted only by the molecules below and beside it and experiences a net downward attractive force. Thus, all molecules lying on the surface of liquid experience a net inward pull. This force which exerts an inward pull on the surface of a liquid and therefore, tends to keep the surface to a minimum is called surface tension. The liquid

    thus behaves as if it were in a state of tension or strain. Thus, surface tension can be defined as the force acting at right angles to the surface along a unit length of the surface. As surface tension tends to keep the surface area of a liquid to the minimum due to inward pull, therefore, if the surface area of a liquid is to be increased, some work has to be done against this inward pull. Hence, surface tension may also be defined as a measure of work that must be done to expand the surface of a liquid by unit area. Surface tension is generally represented by γ and is expressed is Jm–2 (Joules/m2) or Nm–1 (Newton/m).
    Question 161
    CBSEENCH11005980

    Explain the following:
    (i) Drops of liquid assume a spherical shape.
    (ii) The level of mercury in a capillary tube is lower than the level outside when a capillary tube is inserted in mercury.

    Solution

    (i) The spherical shape of drops: Surface tension accounts for the spherical shape of liquid drops. The effect of the surface tension is to decrease the surface area of the liquid to a minimum. Hence, liquid drops become spherical because a sphere has a minimum surface area for a given volume.
    (ii) Fall of the mercury level in a capillary tube. This is because the force of attraction between the mercury molecules (cohesive forces) is greater than the adhesive forces between mercury and glass. This induces mercury to detach from the surface of the glass and level of the mercury goes down in the capillary tube.

    Question 162
    CBSEENCH11005981

    Liquid that wets glass rises in a capillary tube or oil rises in the wick of an oil Jamp. Explain.

    Solution

    This rise is clearly due to the inward pull of surface tension acting on the surface which pushes the liquid into the capillary tube. It is because of the same reason that oil rises into the wick of an oil lamp.

    Question 163
    CBSEENCH11005982

    Explain the following:
    (i) The boiling point of water (373 K) is abnormally high when compared to that of H2S (211·2K).
    (ii) Liquids like ether and acetone are kept in cool places. 
    (iii) Tea or coffee is sipped from a saucer when it is quite hot.
         

    Solution
    (i) The hydrogen bonding in water molecules increases the intermolecular forces of attraction in water. Due to this, greater energy is needed to separate the water molecule so as to convert it into water vapours and hence boiling point of water is relatively higher.

    (ii) Liquids like ether and acetone are quite volatile. Since the rate of evaporation decreaes with the decrease in temperature, so these liquids are kept in a cool place so as to avoid their evaporation.
    (iii) It is because evaporation causes cooling and the rate of evaporation increases with the increase in surface area. Saucer has a large surface area, therefore, tea or coffee in it cools quickly. Hence, tea or coffee is sipped from the saucer when it is quite hot.
    Question 164
    CBSEENCH11005983

    Explain briefly the term viscosity. Define coefficient of viscosity. What are its units?

    Solution

    Viscosity. The viscosity of the liquid is its property by virtue of which it tends to oppose the relative motion between two adjacent layers. In other words, all liquids exhibit resistance to flow. It is clear that the resistance to flow is governed by intermolecular forces. If the intermolecular forces are strong, there is greater resistance to flow i.e. viscosity has a high value. For example, glycerol, honey and concentrated H2SO4 are highly viscous due to stronger intermolecular attractive forces.
    The coefficient of viscosity: It has been found that the force of friction (F) per unit area is directly proportional to the difference in velocity v of the two adjacent layers and inversely proportional to the distance x, between them
    straight F space proportional to space straight v over straight x
straight F space equals space straight eta straight v over straight x
where space straight eta space is space the space coefficient space of space viscosity.
    The coefficient of viscosity may be defined as the force in newton per square metre required to maintain a difference of velocity of one metre per second, between two parallel layers of the liquid at a distance of one metre from each other. Unit. It is measured in kg m–1 s–1.

    Question 166
    CBSEENCH11005985

    What is the effect of temperature on:
    (i) density            
    (ii) surface tension
    (iii) viscosity            
    (iv) the vapour pressure of a liquid?

    Solution

    (i) Density:  The density of a liquid generally decreases with the increase in temperature because liquids expand on heating i.e. volume increases and therefore, 
    density space open parentheses straight d space equals space straight m over straight v close parentheses space decreases.
    (ii) Surface tension. The surface tension of a liquid decreases with the increase in temperature because with the rise in temperature, the kinetic energy of the molecules increases and therefore, intermolecular attraction decreases.
    (iii) Viscosity: The viscosity of a liquid decreases with the increase in temperature because with the rise in temperature, the average kinetic energy of the molecules increases which can overcome the intermolecular forces.
    (iv) Vapour pressure: Vapour pressure of a liquid increases with the increase in temperature because with the rise in the temperature, the average kinetic energy of the liquid molecules increases. Thus, the number of energetic molecules capable of escaping into the vapour state also increases. As a result, the number of molecules present in the vapour state at equilibrium will also be more. Consequently, the vapour pressure will naturally be greater.

    Question 167
    CBSEENCH11005986

    What is the effect of pressure on:
    (i) volume
    (ii) boiling point
    (iii) viscosity of a liquid?

     

    Solution

    (i) Volume: The volume of a liquid decreases very little with increasing pressure because the liquids are almost incompressible.
    (ii) Boiling point: The boiling point of a liquid increases with the increase of external pressure because more heating will be required to make the vapour pressure equal to the external pressure.
    (iii) Viscosity: The viscosity of a liquid increases with the increase in pressure because with the increase in pressure, intermolecular forces in liquids become a bit stronger.

    Question 168
    CBSEENCH11005987

    Explain briefly:
    (i) Dispersion or London forces
    (ii) Dipole-dipole forces.

    Solution

    (i) Dispersion or London forces: Such types of forces operate between the molecules in all substances. On an average, the electrons or the negative charge of the electron are distributed symmetrically about an atom. For brief periods, there may be some asymmetry in the distribution of electrons around the nucleus resulting into an instantaneous electric dipole which causes an electric field. Due to this field, the electron distribution in the neighbouring atom or molecule is distorted and neighbouring atom or molecule itself acquires a dipole moment. The two dipoles will attract and form the basis of dispersion or London forces. These forces are attractive and interaction energy is directly proportional to 1/r6. Therefore, such forces are important only at short distances (≃ 500 pm). London forces also depend on the polarizability of the molecule.
    (ii) Dipole-dipole forces: Dipole-dipole forces occur between molecules having a permanent electric dipole. A bond between two dissimilar atoms is polar in nature. For example, in H – Cl molecule, the shared electron pair between hydrogen and chlorine lies closer to the chlorine atom because of its higher electronegativity as compared to a hydrogen atom. Thus, an electric dipole results. This dipole interacts with another neighbouring dipole (hydrogen chloride polar molecule) and produces a net attractive interaction. The interaction energy is directly proportional to 1/r6 where r is the distance between the polar molecules.



    Question 169
    CBSEENCH11005988

    Describe briefly dipole-induced dipole forces.

    Solution
    Attractive forces are also operative between a molecule having a dipole moment and a molecule without any dipole moment like CH4. As the size of the atom increases, the influence of the electric dipole on it also increases. The electron cloud of the molecule is distorted in the electric field of the permanent dipole. This causes a shift in the centre of gravity of the negative charge relative to the nuclear charge and leads to the formation of an induced dipole moment. The interaction energy is proportional to 1/r6and also depends on upon the dipole moment of the permanent dipole and the polarizability of the second molecule. The interaction between two dipolar molecules also includes additive dipole-dipole interactions.


    Question 170
    CBSEENCH11005989

    What is hydrogen bond? How is it formed?

    Solution

    A hydrogen bond is defined as the weak electronic force of attraction which exists between the covalently bonded hydrogen of one molecule and highly electronegative atom of another molecule. Hydrogen bond has a strength of the order of 10 – 100 kJ mol–1as compared to another covalent bond which has a strength of the order of 200 – 400 kJ mol–1. A hydrogen bond is denoted by dotted lines (.....). The hydrogen bonding in hydrogen fluoride is represented as:

    Cause for the formation of hydrogen bond: Whenever a hydrogen atom is attached to a highly electronegative atom (for example, N, O, F), the shared pair of electrons between the two atoms is attracted towards the more electronegative atom. Consequently, the highly electronegative atom acquires a partial negative charge (δ) while the hydrogen atom acquires a partial positive charge (δ+). The partial negatively charged atom of one molecule tends to attract partial positive hydrogen atom of the other molecule. This weak electrostatic attraction constitutes hydrogen bond. For example.

    Question 171
    CBSEENCH11005990

    What are the conditions for hydrogen bonding?

    Solution

    These are:
    (i) Hydrogen atom should be bonded to highly electronegative atom (N, O, F). The strength of the hydrogen bond increases with an increase in the electronegativity of the other atom. For example, the electronegativity of N, O, F increases as N < O < F.
    Therefore, strength of hydrogen bond also increases as,
    N – H ... N < O – H ... O < F – H ... F
    (ii) The size of the electronegative atom should be small. Smaller the size of the atom, more the electrostatic force of attraction, stronger will be the hydrogen bond.

    Question 172
    CBSEENCH11005991

    Water is a liquid while hydrogen sulphide is a gas. Explain.

    Solution

    The intermolecular forces, namely hydrogen-bonding, in H2O are stronger than in H2S. This is because in H2O the bonds are more ionic and there is a larger partial positive charge on H and partial negative on O, so neighboring H2O molecules can bind to each other and cause a stronger network of bonded molecules, thus requiring more energy (higher temperature) to break these bonds and form a gas (where intermolecular forces no longer act).

    Question 173
    CBSEENCH11005992

    Out of NH3 and PH3, which has higher boiling point and why?

    Solution
    The boiling point of NH3 is higher than that of PH3. The nitrogen atom is more electronegative (3·0) than phosphorus (2·1). As a result, NHexhibits intermolecular hydrogen bonding while PHdoes not show any hydrogen bonding. Thus, extra energy is needed to break these hydrogen bonds in NH3. Thus, NH3 has higher boiling point than PH3.
    Question 174
    CBSEENCH11005993

    Nitrogen and chlorine have the same electronegativity but only former shows intermolecular hydrogen bonding. Discuss.

    Solution

    Both N and Cl have the same electronegativity (3.0). But the atomic size of nitrogen is much smaller than that of chlorine. Thus, nitrogen atom polarises the covalent bond N–H much more than done by Cl atom in H–Cl bond. As a result, the only nitrogen atom is able to exhibit intermolecular hydrogen bonding.

    Question 175
    CBSEENCH11005994

    Why does ice float over water?

    Solution
    We know that in ice, each oxygen atom

    is surrounded by four hydrogen atoms in such a way that the two hydrogen atoms are linked to an oxygen atom by covalent bonds whereas the other two hydrogen atoms are linked by hydrogen bonds. In ice (solid state), a water molecule is associated with four other water molecules through hydrogen bonding in a tetrahedral manner. This gives rise to open cage-like structure which prevents the close packing of molecules (lower density). When ice absorbs heat and melts to form water, the hydrogen bonds break and close packing of water molecules takes place. Due to this close packing, the density of water is higher than that of ice and hence ice floats over water.
    Question 176
    CBSEENCH11005995

    Give reasons for the following:
    (i) The density of water is maximum at 277 K (4°C).

    (ii) Why glycerol (glycerine) is more viscous than ethyl alcohol?

    Solution

    (i) Ice has a cage-like structure with large empty space between the molecules. As water is warmed from 273K (melting point of ice) onward, more and more hydrogen bonds are broken. Naturally, water molecules come closer and acquire lesser volume. Hence, density increases. But this decrease in volume is opposed by the normal tendency for expansion on heating.
    Up to 277 K: Contraction factor is more than the expansion effect, so there is a net decrease in volume or increase in density.
    Above 277K: Expansion factor (due to increase in thermal energy) becomes greater than the contraction effect (caused by breaking of hydrogen bonds), naturally, volume increases and the density decreases. Thus, the density of water is maximum at 277 K.
    (ii) This is because of the presence of three -OH groups in glycerol as compared to only one -OH group in ethyl alcohol. Hence, there are three hydrogen bonding sites in glycerol while there is only one in the case of ethyl alcohol molecule. Hence, there are greater intermolecular forces of attraction in glycerol as compared to ethyl alcohol, therefore, it is more viscous than ethyl alcohol.

    Question 179
    CBSEENCH11008049

    Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is:

    • 2 straight p subscript straight i open parentheses fraction numerator straight T subscript 1 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
    • 2 straight p subscript straight i open parentheses fraction numerator straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
    • 2 straight p subscript straight i space open parentheses fraction numerator straight T subscript 1 straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses
    • straight p subscript straight i open parentheses fraction numerator straight T subscript 1 straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses

    Solution

    B.

    2 straight p subscript straight i open parentheses fraction numerator straight T subscript 2 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses

    Initially,
    Number of moles of gases in each container = fraction numerator straight p subscript straight i straight V over denominator RT subscript straight i end fraction
    Total number of moles of gases in both containers
    equals space 2 fraction numerator straight p subscript straight i straight V over denominator RT subscript straight i end fraction
    After mixing, number of moles in left chamber
    fraction numerator straight p subscript straight f straight V over denominator RT subscript straight i end fraction
    Number of moles in right chamber = fraction numerator straight p subscript straight f straight V over denominator RT subscript 2 end fraction
    Total number of moles
    =   fraction numerator straight p subscript straight f straight V over denominator RT subscript 1 end fraction space plus space fraction numerator straight p subscript straight f straight V over denominator RT subscript 2 end fraction
equals space fraction numerator straight p subscript straight f straight V over denominator straight R end fraction open parentheses 1 over straight T subscript 1 plus 1 over straight T subscript 2 close parentheses
    As the total number of moles remains constant. 
    Hence,

    fraction numerator 2 straight p subscript straight i straight V over denominator RT subscript straight i end fraction space equals space fraction numerator straight p subscript straight f straight V over denominator RT subscript straight i end fraction space plus space fraction numerator straight p subscript straight f straight V over denominator RT subscript 2 end fraction space rightwards double arrow space straight p subscript straight f space equals space 2 straight p subscript straight i open parentheses fraction numerator straight T subscript 1 over denominator straight T subscript 1 plus straight T subscript 2 end fraction close parentheses

    Question 180
    CBSEENCH11008052

    The intermolecular interaction that is dependent on the inverse cube of the distance between the molecule is:

    • ion-ion interaction

    • ion-dipole interaction

    • London force

    • hydrogen bond

    Solution

    B.

    ion-dipole interaction

     ion-ion interaction is dependent on the square of the distance,
    i.e, ion-ion interaction ∝ 1/r2
    Similarly, 
    ion-dipole interaction ∝ 1/r3
    London forces ∝ 1/r6
    and dipole-dipole interactions ∝ 1/r3
    superficially it seems as both ions interaction and hydrogen bonding vary with the inverse cube of the distance between the molecules but when we look at the exact expressions of field (force) created in two situations it comes as,
    In the case of ion-dipole interaction:
    vertical line straight E vertical line space or space vertical line straight F vertical line space equals space fraction numerator 2 vertical line straight P vertical line over denominator 4 πε subscript straight o straight r cubed end fraction
    and, In the case of dipole-dipole interaction
    straight F space equals space fraction numerator 2 straight q squared straight r minus 4 straight q squared straight a over denominator 4 πε subscript straight o straight r cubed end fraction
    From the above it is very clear, the ion-dipole interaction is the better answer as compared dipole-dipole interaction i.e. hydrogen bonding.

    Question 181
    CBSEENCH11008060

    If Z is a compressibility factor, Vander Waal's equation at low pressure can be written as 

    • straight Z space equals 1 plus RT over pb
    • straight Z space equals space 1 minus straight a over VRT
    • straight Z space equals space 1 minus pb over RT
    • straight Z space equals space 1 space plus space pb over RT

    Solution

    B.

    straight Z space equals space 1 minus straight a over VRT

    According to Vander Waal's equation
    open parentheses straight p plus straight a over straight V squared close parentheses space left parenthesis straight V minus straight b right parenthesis space equals space RT
At space low space pressure comma space open parentheses straight p plus straight a over straight V squared close parentheses straight V space equals space RT
pV space plus straight a over straight V space equals space RT
pV space equals space RT minus straight a over straight V
Divide space both space side space by space RT comma
pV over RT space equals space 1 minus straight a over RTV

    Question 183
    CBSEENCH11008083

    The compressibility factor for a real gas at high pressure is

    • 1 plus RT over pb
    • 1
    • 1 plus pb over RT
    • 1 minus pb over RT

    Solution

    C.

    1 plus pb over RT

    Vander Waal's equation for one mole of real gas is
    open parentheses straight p plus straight a over straight V squared close parentheses space left parenthesis straight v minus straight b right parenthesis space equals space RT
When space pressure space is space high space straight p space greater than greater than straight a over straight V squared
such space that space open parentheses straight p space plus straight a over straight V squared close parentheses space equals space straight p
thus comma space straight p left parenthesis straight v minus straight b right parenthesis space equals space RT
pV minus pb space equals space RT
pv space equals space RT space plus pb
therefore space COmpressibillity space factor comma
straight z space equals space pV over RT space equals space open parentheses 1 plus pb over RT close parentheses

    Question 184
    CBSEENCH11008094

    'a' and 'b' are van der Waals constants for gases. Chlorine is more easily liquefied than ethane because :

    • a and b for Cl2 > a and b for C2H6

    • a and b for Cl2 < a and b for C2H6

    • a and Cl2 < a for C2H6 but b for Cl2 > b for C2H6

    • a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

    Solution

    D.

    a for Cl2 > a for C2H6 but b for Cl2 < b for C2H6

    Vander Waals, constant is due to force of attraction and b due to the infinite size of molecules. Thus, greater the value a and smaller the value b, larger the liquefaction.

      a b
    Cl2  6.579 L2 bar mol-2 0.05622 L bar mol-2
    C2H5 5.562 L2 bar mol-2 0.06380 L mol-1
    Question 185
    CBSEENCH11008098

    If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in the vapour phase when equilibrium is established?(Given: Vapour pressure of H2O at 300 K is 3170 Pa; R = 8.314 J K–1 mol–1)

    • 5.56 x 10-3 mol

    • 1.53 x 10-2 mol

    • 4.46 x 10-2 mol

    • 1.27 x 10-3 mol

    Solution

    D.

    1.27 x 10-3 mol

    The volume occupied by water molecules in vapour phase is (1-10-4) dm3, that is approximately (1 x 10-3)m3
    pvapV = nH2O (mol)
    3170 (pa) x 1 x 10-3 (m3) - nH2O(mol) x 8.314 (JK-1 mol-1) x 300K

    Question 188
    CBSEENCH11008199

    The formation of the oxide ion O2-(g) requires first an exothermic and then an endothermic step as shown below
    straight O space left parenthesis straight g right parenthesis space space plus straight e to the power of minus space straight O to the power of minus left parenthesis straight g right parenthesis space increment straight H to the power of straight o space equals space minus space 142 space kJ space mol to the power of negative 1 end exponent
straight O to the power of minus space left parenthesis straight g right parenthesis space space plus straight e to the power of minus straight O to the power of 2 minus end exponent space left parenthesis straight g right parenthesis increment straight H to the power of straight o space equals space 844 space kJ space mol to the power of minus space

    • Oxygen is more electronegative

    • O- ion has comparatively larger size than oxygen atom

    • O- ion will tend to resist the addition of another electron

    • Oxygen has high electron affinity

    Solution

    C.

    O- ion will tend to resist the addition of another electron

    straight O to the power of minus space left parenthesis straight g right parenthesis space plus space straight e to the power of minus space rightwards arrow with space space on top space straight O to the power of 2 minus end exponent space left parenthesis straight g right parenthesis space increment straight H to the power of straight o space minus space 844 space kJ space mol to the power of negative 1 end exponent
    This process is unfavourable in the gas phase because the resulting increase in electron-electron repulsion overweights the stability gained by achieving the noble gas configuration.
    Question 190
    CBSEENCH11008205

    In Vander Waals equation of state of the gas law, the constant ‘b’ is a measure of 

    • intermolecular repulsions

    • intermolecular collisions per unit volume

    • Volume occupied by the molecules

    • intermolecular attraction

    Solution

    C.

    Volume occupied by the molecules

    Vander Waals' equation for one mole of a gas is 
    open square brackets straight P space plus straight a over straight V squared close square brackets left square bracket straight V minus straight b right square bracket space equals space RT
    where, b is the volume correction i.e., the volume occupied by the molecules.

    Question 191
    CBSEENCH11008207

    An ideal gas expands in volume from 1×10-3 m3 to 1×10-2 m3 at 300 K against a constant pressure of 1×105 Nm-2. The work done is

    • -900 J

    • 900 J

    • 2780 J

    • -2780 J

    Solution

    A.

    -900 J

    Work done due to change in volume against constant pressure is 
    W = - p(V2-V1)
     = - 1 x 105 Nm-2 (1 x 10-2- 1 x 10-3)m3

    Question 192
    CBSEENCH11008257

    Dipole-induced dipole interactions are present in which of the following pair?

    • H2O and alcohol

    • Cl2 and CCl4

    • HCl and He atoms

    • SiF4 and He atoms

    Solution

    C.

    HCl and He atoms

    Dipole-induced dipole interaction is present in the pair in which the first species is polar and the other is non-polar.
    H2O and alcohol both are non-polar so there exist dipole-dipole interactions in between them.
    Cl2 and CCl4 both are non-polar so there exists induced dipole -induced dipole interactions in between them. Similarly is true for SiCl4 and He atoms pair.

    HCl is a polar molecule, whereas He atoms are non-polar, so in between them dipole-induced dipole interactions exist.
    Question 194
    CBSEENCH11008285

    pA and pB are the vapour pressure of pure liquid components, A and B, respectively of and ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be

    • pA + xA(pB-pA)

    • pA + xA(pA-pB)

    • pB + xA(pB-pA)

    • pB + xA(pA-pB)

    Solution

    D.

    pB + xA(pA-pB)

    Total pressure, 
    pT =p'A +p'B
    We know that,
    p'A =pAxA
    p'B = pBxB
    substituting the values of p'A and p'B in eq. (i)
    pT = pAxA +pBxB
    [xA+xB =1
    xA = 1-xB or xB = 1-xA]
    =pAxA +pB (1-xA)
    =pAxA +pB -pBxA)
    pT = pB +xA (pA-pB)

    Question 195
    CBSEENCH11008289
    Question 196
    CBSEENCH11008291

    For real gases van der Waal's equation is written as,

    open parentheses straight p space plus an squared over straight V squared close parentheses open parentheses straight V minus nb close parentheses space equals space nRT
    where, a and b are van der Waal's constants.
    Two sets of gases are

    I. O2, CO2, H2 and He
    II. CH4, O2, and H2

    The gases given in set- I increasing order of 'b' and gases given in set -II in decreasing order of 'a' are arranged below. Select the correct order from the following.

    • (I) He < H2 < CO< O2
      (II)  CH4>  H2> O2

    • (I) O2 < He < H2 < CO2
      (II) H2> O2 > CH4

    • (I) O2 < He < O2 < CO2
      (II) CH4 > O2 > H2  

    • (I) H2 > O2 <He < CO2
      (II) O2 > CH4 > H2  

    Solution

    C.

    (I) O2 < He < O2 < CO2
    (II) CH4 > O2 > H2  

    From experimental values of a and b, sequence of Van der Waals constant (b) is

    Gas

    H2

    He

    O2

    CO2

    Value of b

    0.0273

    0.0318

    0.0273

    0.0427

    Hence correct order is

    H2 < He < O2 < CO2

    And sequence of van der Waals constant (a) is

    Gas

    CH4

    O2

    H2

    Value of a

    2.25

    1.36

    0.244

    Hence, correct order is

    (II) CH4 >O2> H2

    Question 202
    CBSEENCH11008322

    A gas such as carbon monoxide would be most likely to obey the ideal gas law at

    • high temperatures and low pressures

    • low temperatures and high pressures

    • high temperatures and high pressures

    • low temperatures and low pressures

    Solution

    A.

    high temperatures and low pressures

    Real gases show ideal gas behaviour at high temperatures and low pressures.
     
    PV = nRT

    Question 204
    CBSEENCH11008368

    The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129o C is (Atomic masses C = 12.01, H = 1.01 and R = 8.314 JK-1 mol-1)

    • 215216 Pa

    • 13409 Pa

    • 41648 Pa

    • 31684 Pa

    Solution

    C.

    41648 Pa

    Given,
    Volume , V = 0.03 m3
    Temperature, T = 129 + 273 = 402 K
    mass of methane, w = 6.0 g
    mol. Mass of methane, M = 12.01 + 4 x 1.01 = 16.05
    From, ideal gas equation,
    pV = nRT
    straight p space equals fraction numerator 6 over denominator 16.05 end fraction space straight x space fraction numerator 8.314 space straight x space 402 over denominator 0.03 end fraction space space equals space 41648 space Pa

    Question 205
    CBSEENCH11008394

    What  is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?

    • Hydrogen bonding

    • Dipole-dipole interaction

    • Covalent bonds

    • London dispersion force

    Solution

    A.

    Hydrogen bonding

    When H atom is directly linked with N or O, or F, inter or intramolecular H - bonding is formed.
    In between CH3OH molecules, intermolecular H-bonding exists. 


    Hence, it is the intermolecular H - bonding that must be overcome in converting liquid CH3OH to gas.

    Question 206
    CBSEENCH11008402

    If a gas expands at constant temperature, it indicates that

    • the kinetic energy of molecules decrease

    • the pressure of the gas increases

    • the kinetic energy of molecules remains the same

    • the number of the molecules of as increases

    Solution

    C.

    the kinetic energy of molecules remains the same

    KE space equals space 3 over 2 space RT
KE proportional to space straight T
    here, the temperature of the gas remains constant, hence kinetic energy of molecules remains the same.
    Question 207
    CBSEENCH11008409

    What volume of oxygen gas (O2) measured at 0oC and 1 atm, is needed to burn completely 1 L of propane gas (C3H8)  measured under the same conditions?

    • 7 L

    • 6 L 

    • 5 L 

    • 10 L

    Solution

    C.

    5 L 

    Volume of a gas at STP = 22.4 L 
    C38 +    5O2 → 3CO2 + 4H2O
    22.4 L       5 x 22.4 L
    Therefore,
    to burn 22.4 L C3H8 the oxygen required is 
     = 5 x 22.4 L 
    To burn 1 L C3H8  the oxygen required will be  = 5 x 22.4 / 22.4  = 5 L 

    Question 208
    CBSEENCH11008461

    20-litre container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be

    Given that: SrCO3 ⇌ SrO(s) + CO2(g) kp = 1.6 atm)

    • 5 litre

    • 10 litre

    • 8 litre

    • 3  litre

    Solution

    A.

    5 litre

    Max. the pressure  of CO2= Pressure of CO2 at equilibrium
    For reaction, SrCO3 ⇌ SrO(s) + CO2(g)
    Kp= PCO2 = 1.6 atm = maximum pressure of CO2
    Volume Container at this stage,
    V= nRT/P .... (i)

    Since container is sealed and reaction was not  earlier at equilibrium
    therefore, n = constant
    n= Pv/RT  .... (ii)
    from equation (i) and (ii) we get
    V = (0.4 x 20)/RT = 5 Litre

    Question 209
    CBSEENCH11008479

    The compressibility of a gas is less than unity at STP, therefore,

    • Vm > 22.4 L

    • Vm < 22.4 L

    • Vm = 22.4

    • Vm = 44.8 L

    Solution

    B.

    Vm < 22.4 L

    compressibility (Z) = PVnRTZ less than unity, thereforePVnRT <1or PV<nRT [At STP, P =1 atm, T = 273K, R =0.0821 L atm K-1]i.e or  1 atm x V < 1 x 0.0821 x 273or V <22.4 litres.

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