JEE Advanced 2021 · Paper 1 · Physics Q2 · Dronacharya Exams

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An ideal gas undergoes a four step cycle as shown in the $P-V$ diagram below. During this cycle, heat is absorbed by the gas in

P-V diagram of a rectangular cycle traversed clockwise: step 1 is an isobaric expansion along the top edge (to the right), step 2 an isochoric pressure drop along the right edge (down), step 3 an isobaric compression along the bottom edge (to the left), step 4 an isochoric pressure rise along the left edge (up).

A

steps 1 and 2

B

steps 1 and 3

C

steps 1 and 4

D

steps 2 and 4

Answer and worked solution

Official answer: C

Sign of the heat in each step.

Step 1: isobaric expansion. $T$ rises and the gas does work, so $Q=nC_p\Delta T>0$ : heat is absorbed.
Step 2: isochoric pressure drop. $T$ falls, so $Q<0$ .
Step 3: isobaric compression. $T$ falls, so $Q<0$ .
Step 4: isochoric pressure rise. $T$ rises, so $Q=nC_v\Delta T>0$ : heat is absorbed.

Heat is absorbed in steps 1 and 4. (C)

Topics:Thermodynamic processes First law and internal energy

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