System of Particles and Rotational Motion

  • Question 1
    CBSEENPH110024932

    Determine the volume of 1 mole of any gas at S.T.P., assuming it behaves like an ideal Gas.

    Solution

    1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere).

    These figures are actually only true for an ideal gas, and we'll have a look at where they come from.

    We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure.

    First, we have to get the units right.

    0°C is 273 K. T = 273 K

    1 atmosphere = 101325 Pa. p = 101325 Pa

    We know that n = 1, because we are trying to calculate the volume of 1 mole of gas.

    And, finally, R = 8.31441 J K-1 mol-1.

    Slotting all of this into the ideal gas equation and then rearranging it gives:

    And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres.

    The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

    And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.

    Question 2
    CBSEENPH110024936

    1. how does carnot cycle operates
    2. why does absolute zero not correspond to zero energy

    Solution
    1. The Carnot cycle consists of the following four processes: A reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs qin amount heat from a heat source at a high temperature Th, expands and does work on surroundings.
    2. As you know , absolute zero means the temperature in Kelvin scale is 0K or in Celcius scale is -273.15°C . actually, in this temperature, Gaseous molecule be rest , there is no motion of molecule in their position. due to this reason we say that energy of Gaseous molecule at this is zero. means absolute zero correspond to zero energy .

      well, it's hard to achieve Absolute zero or 0K or -273.15°C temperature. it is just theoretical. molecular can't achieve Absolute zero.

    Question 3
    CBSEENPH11020356

    A thin uniform rod of length

    • 1 third fraction numerator calligraphic l squared straight omega squared over denominator straight g end fraction
    • 1 over 6 fraction numerator calligraphic l straight omega over denominator straight g end fraction
    • 1 half fraction numerator calligraphic l squared straight omega squared over denominator straight g end fraction
    • 1 over 6 fraction numerator calligraphic l squared straight omega squared over denominator straight g end fraction

    Solution

    D.

    1 over 6 fraction numerator calligraphic l squared straight omega squared over denominator straight g end fraction 1 half open parentheses fraction numerator straight m space calligraphic l squared over denominator 3 end fraction close parentheses straight omega squared space equals space mgh
rightwards double arrow space straight h space equals space fraction numerator straight omega squared calligraphic l squared over denominator 6 straight g end fraction
    Question 4
    CBSEENPH11020357

    Let P(r) = Qr/πR4 be the charge density distribution for a solid sphere of radius R and total charge Q. for a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is

    • fraction numerator straight Q over denominator 4 πε subscript straight o straight r subscript 1 superscript 2 end fraction
    • 0

    • fraction numerator Qr subscript 1 superscript 2 over denominator 4 πε subscript straight o straight R to the power of 4 end fraction
    • fraction numerator Qr subscript 1 superscript 2 over denominator 3 πε subscript straight o straight R to the power of 4 end fraction

    Solution

    C.

    fraction numerator Qr subscript 1 superscript 2 over denominator 4 πε subscript straight o straight R to the power of 4 end fraction fraction numerator straight Q over denominator 4 πε subscript 0 straight r subscript 1 superscript 2 end fraction space integral subscript straight r space equals 0 end subscript superscript straight r subscript 1 end superscript space fraction numerator 4 space πr cubed dr over denominator πR to the power of 4 end fraction
space equals space fraction numerator straight Q over denominator 4 πε subscript 0 end fraction fraction numerator straight r subscript 1 superscript 2 over denominator straight R subscript 4 end fraction

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