An ideal gas goes from state A to state B by three different processes labelled 1, 2 and 3 in the P–V diagram. If Q₁, Q₂, Q₃ denote the heats absorbed and ΔU₁, ΔU₂, ΔU₃ the changes in internal energy along the three paths respectively, which relation is correct?
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Question

An ideal gas goes from state A to state B by three different processes labelled 1, 2 and 3 in the P–V diagram. If Q₁, Q₂, Q₃ denote the heats absorbed and ΔU₁, ΔU₂, ΔU₃ the changes in internal energy along the three paths respectively, which relation is correct?

Options

G1 Physics · Accepted Answer

Step 1: Every path 1, 2 and 3 begins at state A and ends at state B. Step 2: Internal energy is a state function, so its change is independent of the path: ΔU₁ = ΔU₂ = ΔU₃. Step 3: On the P–V diagram, work on each expansion path is area under the respective curve: W = ∫P dV. Step 4: Over the same volume range, curve 1 lies above 2, which lies above 3. Hence W₁ > W₂ > W₃. Step 5: First law Q = ΔU + W, with identical ΔU for all paths, yields Q₁ > Q₂ > Q₃. Final answer: (A).