Question

For an ideal gas, consider only P-V work in going from an initial state X to 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)

The correct options are
A ${\Delta S}_{x\to z}={\Delta S}_{x\to y}+{\Delta S}_{y\to z}$
C $w_{x\to y\to z}=w_{x\to y}$

$X\to Y\Rightarrow P-constant,V-increases,\Delta S>0$

$Y\to Z\Rightarrow P-decreases,V-constant,\Delta S>0$

$X\to Z\Rightarrow P-decreases,V-increases,\Delta S>0$
Entropy is a state function. Its value is independent of the path followed.

​Hence, option A is correct and option D is incorrect.
Work is a path function. Its value depends on the path followed.

​This is because $△W_{Y\to Z}=0$ as the volume change from Y to Z is zero.
Thus, option C is correct and option B is incorrect.