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 DS as change in entropy and w as work done].
(a) $\Delta S_{x \rightarrow z} = \Delta S_{x \rightarrow y} + \Delta S_{y \rightarrow z}$
(b) $w_{x \rightarrow z} = w_{x \rightarrow y} + w_{y \rightarrow z}$
(c) $w_{x \rightarrow y \rightarrow z} = w_{x \rightarrow y}$
(d) $\Delta S_{x \rightarrow y \rightarrow z} = \Delta S_{x \rightarrow y}$

Answer 1

This is your required answer.

(A,C)

Entropy is state function

so  ΔSx→z=ΔSx→y+ΔSy→zΔSx→z=ΔSx→y+ΔSy→z

and Wx→y→z=Wx→yWx→y→z=Wx→y

because process y→z is constant volume so work done is zero.