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States of Matter

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Question
CBSEENCH11008461
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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

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Question
CBSEENCH11008272
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50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is 36, the molecular mass of gas A will be

  • 96

  • 128

  • 20.2

  • 64

Solution

C.

20.2

Given,
VA =VB = 50 mL
TA = 150 s
TB = 200 s
MB = 36
MA = ?
From Graham's law of effusion
fraction numerator straight r subscript straight B over denominator straight r subscript straight A space end fraction space equals space square root of M subscript A over M subscript B end root space equals space fraction numerator V subscript B T subscript A over denominator T subscript B. V subscript A end subscript end fraction rightwards double arrow space square root of M subscript A over 36 end root space equals space fraction numerator V subscript A x 150 over denominator 200 space x space V subscript A end fraction O r space square root of M subscript A over 36 end root space equals space 15 over 20 space equals space 3 over 4 M subscript A over 36 space equals space 9 over 16 M subscript A space equals space fraction numerator 9 space x space 36 over denominator 16 end fraction space equals space fraction numerator 9 space x space 9 space over denominator 4 end fraction space equals space 81 over 4 space equals space 20.2 space

Question
CBSEENCH11008289
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A  certain gas takes three times as long to effuse out as helium. Its molecular mass will be 

  • 27 u

  • 36 u 

  • 64 u

  • 9 u

Solution

B.

36 u 

From Graham's diffusion law,
straight r subscript 1 over straight r subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root O r V subscript 1 over t subscript 1 space x space t subscript 2 over V subscript 2 space equals space square root of M subscript 2 over M subscript 1 end root because space V o l u m e space i s space s a m e therefore space 3 over 1 equals space square root of straight M subscript 2 over 4 end root space or space 9 equals straight M subscript 2 over 4 Or space straight M subscript 2 space equals space 9 space straight x space 4 space equals space 36 space straight u

Question
CBSEENCH11008342
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A bubble of air is underwater at temperature 15o C  and the pressure 1.5 bar. If the bubbles rises to the surface where the temperature is 25o C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

  • Volume will become greater by a factor of 1.6 

  • Volume will become  greater by a factor of 1.1

  • Volume will become  smaller by a factor of 0.70

  • Volume will become greater by a factor of 2.9

Solution

A.

Volume will become greater by a factor of 1.6 

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 left parenthesis ideal space gas space equation right parenthesis or fraction numerator 1.5 space straight x space straight V subscript 1 over denominator 288 end fraction space equals space fraction numerator 1 space straight x space straight V subscript 2 over denominator 298 end fraction therefore space space straight V subscript 2 space equals space 1.55 space straight V subscript 1 space
i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.

Question
CBSEENCH11008322
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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

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