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 choices is correct? [Take $\Delta S$ as change in entropy and $w$ as work done.]

Professor PC Thomas Classes · Answer

(a) : 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}
$
As we know that work is not a state function and depends on path,
Thus, $w_{X ightarrow Z} 
eq w_{X ightarrow Y}+w_{Y ightarrow Z}$
$
\begin{array}{rlr}
w_{X ightarrow Y} & =P d V & (P 	ext { is constant }) \
w_{Y ightarrow Z} & =0 & (V 	ext { is constant }) \
w_{X ightarrow Z} & =2.303 n R T \log \frac{V_2}{V_1}
\end{array}
$
As $w_{Y ightarrow Z}=0$, hence $w_{X ightarrow Y ightarrow Z}=w_{X ightarrow Y}$