For an ideal gas, consider only P-V work in g | vedclass

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(A) Entropy $(S)$ is a state function, meaning its change depends only on the initial and final states, not the path taken. Therefore, $\Delta S_{X \r

Vedclass · Accepted Answer

$(A, C)$

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(A) Entropy $(S)$ is a state function, meaning its change depends only on the initial and final states, not the path taken. Therefore, $\Delta S_{X \rightarrow Z} = \Delta S_{X \rightarrow Y} + \Delta S_{Y \rightarrow Z}$ is correct.
Work $(w)$ is a path function, meaning its value depends on the path taken. The total work done for a multi-step process is the sum of the work done in each individual step. For the path $X \rightarrow Y \rightarrow Z$, the total work is $w_{X \rightarrow Y \rightarrow Z} = w_{X \rightarrow Y} + w_{Y \rightarrow Z}$.
Comparing the options:
(A) $\Delta S_{X \rightarrow Z} = \Delta S_{X \rightarrow Y} + \Delta S_{Y \rightarrow Z}$ is correct because entropy is a state function.
(B) $w_{X \rightarrow Z} = w_{X \rightarrow Y} + w_{Y \rightarrow Z}$ is incorrect because $w$ is a path function and the work for the direct path $X \rightarrow Z$ is different from the sum of work for the path $X \rightarrow Y \rightarrow Z$.
(C) $w_{X \rightarrow Y \rightarrow Z} = w_{X \rightarrow Y} + w_{Y \rightarrow Z}$ is correct by the definition of path work.
(D) $\Delta S_{X \rightarrow Y \rightarrow Z} = \Delta S_{X \rightarrow Y}$ is incorrect because it ignores the $\Delta S_{Y \rightarrow Z}$ contribution.
Thus, the correct choices are (A) and (C).

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