Chemical Thermodynamics Question 322

Question: Work for the following process ABCD on a monoatomic gas is

Options:

A) $ W=-2,P_0V_0ln,2 $

B) $ W=-2,P_0V_0(1+ln,2) $

C) $ W=-,P_0V_0(1+ln,2) $

D) $ W=-,P_0V_0ln,2 $

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Answer:

Correct Answer: A

Solution:

  • At A and D the temperatures of the gas will be equal, so $ \Delta E=0 $ , $ W_{AB}=-P_0(2V_0-V_0)=-P_0V_0 $

$ W_{BC}=-P_0\times 2V_0 $ ln $ ( \frac{4V_0}{2V_0} )=-2P_0V_0 $ ln 2

$ W_{CD}=\frac{P_0}{2}(4V_0-2V_0)=P_0V_0 $

Now $ W=W_{AB}+W_{Bc}+W_{CD}=-P_0V_0-P_0V_0 $ ln $ 2+P_0V_0 $

$ =-2P_0V_0 $ ln 2



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