An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. If Q₁, Q₂, Q₃ indicate the heat absorbed by the gas along the three processes and Δ U₁, Δ U₂, Δ U₃ indicate the change in internal energy along the three processes respectively, then

Question

Accepted Answer

(1) Q₁>Q₂>Q₃ and Δ U₁=Δ U₂=Δ U₃. Idea: Δ U depends only on the end states, while W is the area under the path, so the three heats differ exactly as the three areas do. All three processes run from A to B, so Δ U₁=Δ U₂=Δ U₃. On the diagram path 1 lies above path 2, which lies above path 3, over the whole range of volume. Area under the curve: W₁>W₂>W₃. First law: Q=Δ U+W, with Δ U the same for all three. Q₁>Q₂>Q₃, with Δ U₁=Δ U₂=Δ U₃.

Step-by-step solution

Why the other options are wrong