An ideal gas goes from state A to state B via three different processes as indicated in the P-V diagram. [figure] If Q1, Q2, Q3 indicate the heat absorbed by the gas along the three processes and U1, U2, U3 indicate the change in internal energy along the three processes respectively, then
- Q1 > Q2 > Q3 and U1 = U2 = U3
- Q3 > Q2 > Q1 and U1 = U2 = U3
- Q1 = Q2 = Q3 and U1 > U2 > U3
- Q3 = Q2 = Q1 and U1 > U2 > U3
Answer
(1) Q1 > Q2 > Q3 and U1 = U2 = U3
Solution
Internal energy is a state function and all three processes have the same initial state A and final state B, so U1 = U2 = U3.
Work done by the gas is the area under the P–V curve, and from the diagram the area under 1 > area under 2 > area under 3, so W1 > W2 > W3.
By the first law, Q = U + W with the same U.
Q1 > Q2 > Q3
Answer: (1)