Question

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to a final state $Z$. The final state $Z$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct?
[take $\Delta S$ as change in entropy and $w$ as work done]

• A. $\Delta S_{x\to z}=\Delta S_{x\to y}+\Delta S_{y\to z}$
• B. $w_{x\to z}=w_{x\to y}+w_{y\to z}$
• C. $w_{x\to y\to z}=w_{x\to y}$
• D. $\Delta S_{x\to y\to z}=\Delta S_{x\to y}$

Answer 1

Answer: (A), (C)

$\Delta S_{x\to z}=\Delta S_{x\to y}+\Delta S_{y\to z}$
$w_{x\to y\to z}=w_{x\to y}$

Change in entropy from $X$ to $Z$ is the sum of Entropy change from $X$ to $Y$ and $Y$ to $Z$.

So, $\Delta S_{x\to z}=\Delta S_{x\to y}+\Delta S_{y\to z}$.

Work done does not depend upon the path of change. So,

$w_{x\to y\to z}=w_{x\to y}$.

So, correct answers are option A and C.