An ideal gas exists in a state with pressure $\mathrm{P}_{0}$, volume $\mathrm{V}_{0}$. It is isothermally expanded to 4 times of its initial volume ($\mathrm{V}_{0}$), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :

(A) $\mathrm{P}_{0} \mathrm{~V}_{0}(2 \ln 2-0.75)$

(B) $\mathrm{P}_{0} \mathrm{~V}_{0}(\ln 2-0.75)$

(C) $\mathrm{P}_{0} \mathrm{~V}_{0}(\ln 2-0.25)$

(D) $\mathrm{P}_{0} \mathrm{~V}_{0}(2 \ln 2-0.25)$

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Heat & Thermodynamics

An ideal gas exists in a state with pressure

P_{0}

, volume

V_{0}

. It is isothermally expanded to 4 times of its initial volume (

V_{0}

), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is :

(A) $P_{0} V_{0} (2 ln 2 - 0.75)$ ✓ Correct
(B) $P_{0} V_{0} (ln 2 - 0.75)$
(C) $P_{0} V_{0} (ln 2 - 0.25)$
(D) $P_{0} V_{0} (2 ln 2 - 0.25)$

Answer: (A)

Solution

Over one complete cycle the internal energy returns to its starting value, so the net heat exchanged equals the net work done. The isothermal expansion from

V_{0}

4 V_{0}

gives

W_{1} = P_{0} V_{0} ln 4 = 2 P_{0} V_{0} ln 2

. At the end of that step the pressure has fallen to

P_{0} /4

, and the isobaric compression back to

V_{0}

gives

W_{2} = \frac{P _{0}}{4} (V_{0} - 4 V_{0}) = - \frac{3}{4} P_{0} V_{0}

. The isochoric step that restores the pressure does no work. Adding these, the heat exchanged over the cycle is

P_{0} V_{0} (2 ln 2 - 0.75)

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Heat & Thermodynamics

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