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 $
Show Answer
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
