For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the final state $Z$

Question

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the 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]\

Professor PC Thomas Classes · Answer

(a, c) : As $\Delta S$ does not depend on path and only depends on initial and final stages i.e., it is a state function thus\ $\Delta S_{X ightarrow Z}=\Delta S_{X ightarrow Y}+\Delta S_{Y ightarrow Z}$\ and $\Delta S_{Y ightarrow Z}$ is not zero thus\ $\Delta S_{X ightarrow Y ightarrow Z} 
eq \Delta S_{X ightarrow Y}$\ As we know that work is not a state function and depends on path,\ Thus, $w_{X ightarrow Z} 
eq w_{X ightarrow Y}+w_{Y ightarrow Z}$\ $w_{X ightarrow Y}=P d V \quad$ ( $P$ is constant.)\ $w_{Y ightarrow Z}=0$\ ( $V$ is constant.)\ $w_{X ightarrow Z}=2.303 n R T \log \frac{V_{2}}{V_{1}}$\ $w_{X ightarrow Y ightarrow Z}=w_{X ightarrow Y}+w_{Y ightarrow Z}$\ As $w_{Y ightarrow Z}=0$, hence $w_{X ightarrow Y ightarrow Z}=w_{X ightarrow Y}$\