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 ΔS as change in entropy and was work done]

Engineering

Chemistry

First Law and Various Process

Question

$W_{x \to z} = W_{x \to y} + W_{y \to z}$

${ΔS}_{x \to y \to z} = {ΔS}_{x \to y}$

${ΔS}_{x \to z} = {ΔS}_{x \to y} + {ΔS}_{y \to z}$

$W_{x \to y \to z} = W_{x \to y}$

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For an ideal gas, work (W) is path-dependent, while entropy (ΔS) is a state function. The figure shows two paths from X to Z: one direct and one via Y.

Work is not additive: $W_{x \to z} \neq W_{x \to y} + W_{y \to z}$ because the area under the P-V curve differs for each path.

Entropy is a state function, so the change is path-independent: ${ΔS}_{x \to z} = {ΔS}_{x \to y \to z}$ . However, it is not simply additive for sub-paths unless the process is reversible.

The correct choice is: ${ΔS}_{x \to y \to z} = {ΔS}_{x \to y}$ is incorrect, and $W_{x \to y \to z} = W_{x \to y}$ is also incorrect as work is done on both legs. None of the provided options are correct for an ideal gas.