JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Heat & Thermodynamics

Using the given

P - V

diagram, the work done by an ideal gas along the path ABCD is-

Answer: (D)

Solution

Work done by a gas equals the area under the process on a

P

V

diagram, signed according to the direction of traversal.

Reading the diagram, the path runs through

A (2 V_{0}, P_{0}) \to B (3 V_{0}, P_{0}) \to C (3 V_{0}, 2 P_{0}) \to D (V_{0}, 2 P_{0})

A \to B

(isobaric expansion at

P_{0}

W_{A B} = P_{0} (3 V_{0} - 2 V_{0}) = P_{0} V_{0}

B \to C

(isochoric, no volume change):

W_{B C} = 0

C \to D

(isobaric compression at

2 P_{0}

W_{C D} = 2 P_{0} (V_{0} - 3 V_{0}) = - 4 P_{0} V_{0}

Total work along ABCD:

W = P_{0} V_{0} + 0 - 4 P_{0} V_{0} = - 3 P_{0} V_{0}

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option