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 a change in entropy and $\omega$ as work done]

• A. $\Delta S_{X\to Z}=\Delta S_{X\to Y}+\Delta S_{Y\to Z}$
• B. ${\omega }_{X\to Z}={\omega }_{X\to Z}+{\omega }_{Y\to Z}$
• C. ${\omega }_{X\to Y\to Z}\ne {\omega }_{X\to Y}$
• D. $\Delta S_{X\to Y\to Z}=\Delta S_{X\to Y}$

Answer 1

The correct option is A $\Delta S_{X\to Z}=\Delta S_{X\to Y}+\Delta S_{Y\to Z}$

Solution:-

Since entropy is a state function, thus we can simply add the entropies.

$\therefore {\Delta S}_{X\to Z}={\Delta S}_{X\to Y}+{\Delta S}_{Y\to Z}$

As the process $Y\to Z$ is an isochoric process.

$\therefore W_{Y\to Z}=0$

Hence, $W_{X\to Y\to Z}=W_{X\to Y}$