JEE Main 2025 Jan 23 Shift 2, Physics Q18: Thermodynamic Processes

JEE Main2025Jan 23, Shift 2Physics

Using the given P-V diagram, the work done by an ideal gas along the path ABCD is :

Solution

Work done by the gas equals the area under the curve (with sign given by direction of volume change).

A $\to$ B (isobaric at $P_{0}$ , volume increases by $V_{0}$ ):

$W_{A B} = P_{0} (V_{B} - V_{A}) = P_{0} V_{0} .$

B $\to$ C (isochoric, $V$ constant): $W_{B C} = 0$ .

C $\to$ D (isobaric at $2 P_{0}$ , volume decreases by $2 V_{0}$ ):

$W_{C D} = 2 P_{0} \cdot (- 2 V_{0}) = - 4 P_{0} V_{0} .$

Total: $W_{A B C D} = P_{0} V_{0} + 0 - 4 P_{0} V_{0} = - 3 P_{0} V_{0} .$

Concept behind this question

Concept-wise practice with instant solutions on Fundamenthol.