Understanding the Problem

We are analyzing an ideal gas undergoing a process from state X to state Z via two different paths: one direct path (X→Z) and one indirect path (X→Y→Z). The work done (W) and change in entropy (ΔS) are path-dependent and state functions respectively. We need to determine which of the given options are correct.

Step 1: Analyze Work Done (W)

Work done in a P-V process is given by $W=\int PdV$. Work is a path function, meaning it depends on the path taken.

For path X→Y→Z, the total work done is the sum of work in each step: $W_{X\to Y\to Z}=W_{X\to Y}+W_{Y\to Z}$. However, the direct path X→Z may have a different work value. Since the paths are different, $W_{X\to Z}\ne W_{X\to Y\to Z}$ in general. Looking at the figure, path X→Y is at constant volume (isochoric), so $W_{X\to Y}=0$. Path Y→Z is at constant pressure (isobaric), so $W_{Y\to Z}=P\Delta V$. Therefore, $W_{X\to Y\to Z}=0+P\Delta V=W_{Y\to Z}$. The direct path X→Z is a straight line, and its work is the area under the curve, which is different from the area under the constant pressure path Y→Z. Hence, $W_{X\to Y\to Z}\ne W_{X\to Z}$ and also $W_{X\to Y\to Z}=W_{X\to Y}+W_{Y\to Z}$ is true by definition, but $W_{X\to Z}=W_{X\to Y}+W_{Y\to Z}$ is not necessarily true.

Step 2: Analyze Change in Entropy (ΔS)

Entropy is a state function. The change in entropy depends only on the initial and final states, not on the path taken. Therefore, for any path from X to Z, ${\mathrm{\Delta S}}_{X\to Z}$ is the same.

For the indirect path X→Y→Z, the total entropy change is ${\mathrm{\Delta S}}_{X\to Y\to Z}={\mathrm{\Delta S}}_{X\to Y}+{\mathrm{\Delta S}}_{Y\to Z}$. Since entropy is a state function, this sum must equal the entropy change for the direct path: ${\mathrm{\Delta S}}_{X\to Y\to Z}={\mathrm{\Delta S}}_{X\to Z}$. Therefore, ${\mathrm{\Delta S}}_{X\to Z}={\mathrm{\Delta S}}_{X\to Y}+{\mathrm{\Delta S}}_{Y\to Z}$ is correct.

However, ${\mathrm{\Delta S}}_{X\to Y\to Z}={\mathrm{\Delta S}}_{X\to Y}$ is incorrect because it ignores the entropy change from Y to Z.

Step 3: Evaluate Each Option

Option 1: $W_{X\to Z}=W_{X\to Y}+W_{Y\to Z}$
This is false. Work is path-dependent, and the work for the direct path X→Z is not equal to the sum of works for path X→Y and Y→Z.

Option 2: ${\mathrm{\Delta S}}_{X\to Y\to Z}={\mathrm{\Delta S}}_{X\to Y}$
This is false. The total entropy change for path X→Y→Z includes both steps, not just the first.

Option 3: ${\mathrm{\Delta S}}_{X\to Z}={\mathrm{\Delta S}}_{X\to Y}+{\mathrm{\Delta S}}_{Y\to Z}$
This is true. Entropy is a state function, so the change is additive over paths.

Option 4: $W_{X\to Y\to Z}=W_{X\to Y}$
This is false. The total work for path X→Y→Z is not just the work from X→Y (which is zero) but also includes the work from Y→Z.

Final Answer

Only Option 3 is correct: ${\mathrm{\Delta S}}_{X\to Z}={\mathrm{\Delta S}}_{X\to Y}+{\mathrm{\Delta S}}_{Y\to Z}$

Related Topics and Formulae

State Functions vs. Path Functions

State Functions: Properties that depend only on the initial and final states of the system (e.g., Internal Energy (U), Enthalpy (H), Entropy (S), Gibbs Free Energy (G)).

Path Functions: Properties that depend on the path taken between initial and final states (e.g., Work (W), Heat (Q)).

Work Done in Thermodynamic Processes

For a reversible process, work done by the gas is given by: $W=\int PdV$

• Isochoric process (constant volume): $W=0$
• Isobaric process (constant pressure): $W=P\Delta V$
• Isothermal process (constant temperature, for ideal gas): <math xmlns="http://www.w