An ideal gas undergoes a four step cycle as shown in the diagram below. During this cycle, heat is absorbed by the gas in
- (A)steps 1 and 2
- (B)steps 1 and 3
- (C)steps 1 and 4✓ Correct
- (D)steps 2 and 4
Answer: (C)
Solution
Process 1 (isobaric expansion): P = constant, V increases → T increases. W = positive, ΔU = positive → ΔQ = positive (heat absorbed).
Process 2 (isochoric): V = constant, P decreases → T decreases. W = 0, ΔU = negative → ΔQ = negative (heat rejected).
Process 3 (isobaric compression): P = constant, V decreases → T decreases. W = negative, ΔU = negative → ΔQ = negative (heat rejected).
Process 4 (isochoric): V = constant, P increases → T increases. W = 0, ΔU = positive → ΔQ = positive (heat absorbed).
Heat is absorbed in steps 1 and 4.
Process 2 (isochoric): V = constant, P decreases → T decreases. W = 0, ΔU = negative → ΔQ = negative (heat rejected).
Process 3 (isobaric compression): P = constant, V decreases → T decreases. W = negative, ΔU = negative → ΔQ = negative (heat rejected).
Process 4 (isochoric): V = constant, P increases → T increases. W = 0, ΔU = positive → ΔQ = positive (heat absorbed).
Heat is absorbed in steps 1 and 4.
JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Single correct option