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MCQMediumJEE 2025Isothermal & Adiabatic Processes

JEE Physics 2025 Question with Solution

An ideal gas exists in a state with pressure P0P_0, volume V0V_0. It is isothermally expanded to 44 times of its initial volume (V0)\left(V_0\right), 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

    P0V0(2ln⁡2−0.75)P_0 V_0 \left(2 \ln 2 - 0.75\right)

  • B

    P0V0(ln⁡2−0.75)P_0 V_0 \left(\ln 2 - 0.75\right)

  • C

    P0V0(ln⁡2−0.25)P_0 V_0 \left(\ln 2 - 0.25\right)

  • D

    P0V0(2ln⁡2−0.25)P_0 V_0 \left(2 \ln 2 - 0.25\right)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The gas starts at (P0,V0)\left(P_0, V_0\right), expands isothermally to (P1,4V0)\left(P_1, 4V_0\right), is compressed isobarically to (P04,V0)\left(\frac{P_0}{4}, V_0\right), and is then heated isochorically back to the initial state.

Find: The total heat exchanged in the complete cycle.

For a cyclic process, the net change in internal energy is zero:

ΔUcyclic=0\Delta U_{\text{cyclic}} = 0

Therefore, the total heat exchanged equals the total work done by the system:

QT=ω1+ω2+ω3Q_T = \omega_1 + \omega_2 + \omega_3

Step 1: Isothermal expansion from (P0,V0)\left(P_0, V_0\right) to (P1,4V0)\left(P_1, 4V_0\right).

For an isothermal process,

PV=constantPV = \text{constant}

So,

P0V0=P1(4V0)P_0 V_0 = P_1 \left(4V_0\right)

Hence,

P1=P04P_1 = \frac{P_0}{4}

Work done is

ω1=∫V04V0P dV=∫V04V0P0V0V dV=P0V0[ln⁡V]V04V0\omega_1 = \int_{V_0}^{4V_0} P \, dV = \int_{V_0}^{4V_0} \frac{P_0 V_0}{V} \, dV = P_0 V_0 [\ln V]_{V_0}^{4V_0} ω1=P0V0ln⁡4V0V0=P0V0ln⁡4=P0V0(2ln⁡2)\omega_1 = P_0 V_0 \ln \frac{4V_0}{V_0} = P_0 V_0 \ln 4 = P_0 V_0 \left(2 \ln 2\right)

Step 2: Isobaric compression from (P04,4V0)\left(\frac{P_0}{4}, 4V_0\right) to (P04,V0)\left(\frac{P_0}{4}, V_0\right).

Work done is

ω2=∫4V0V0P dV=P1(V0−4V0)\omega_2 = \int_{4V_0}^{V_0} P \, dV = P_1 \left(V_0 - 4V_0\right)

Substituting P1=P04P_1 = \frac{P_0}{4},

ω2=P04(−3V0)=−34P0V0=−0.75P0V0\omega_2 = \frac{P_0}{4} \left(-3V_0\right) = -\frac{3}{4} P_0 V_0 = -0.75 P_0 V_0

Step 3: Isochoric heating from (P04,V0)\left(\frac{P_0}{4}, V_0\right) to (P0,V0)\left(P_0, V_0\right).

For an isochoric process, volume remains constant, so

dV=0dV = 0

Therefore,

ω3=∫V0V0P dV=0\omega_3 = \int_{V_0}^{V_0} P \, dV = 0

Now the total heat exchanged is

QT=ω1+ω2+ω3=2P0V0ln⁡2−0.75P0V0+0Q_T = \omega_1 + \omega_2 + \omega_3 = 2 P_0 V_0 \ln 2 - 0.75 P_0 V_0 + 0 QT=P0V0(2ln⁡2−0.75)Q_T = P_0 V_0 \left(2 \ln 2 - 0.75\right)

Therefore, the total heat exchanged is P0V0(2ln⁡2−0.75)P_0 V_0 \left(2 \ln 2 - 0.75\right), so the correct option is A.

A discrepancy appears in the secondary approach where the isobaric step is briefly written using P0P_0 instead of P04\frac{P_0}{4}, but the final evaluated result still matches the correct expression above.

Cycle Work Shortcut

Given: The process is a closed thermodynamic cycle.

Find: Net heat exchanged.

Use the shortcut that for any complete cycle,

Qnet=WnetQ_{\text{net}} = W_{\text{net}}

because

ΔU=0\Delta U = 0

So only the work in each leg is needed.

For the isothermal expansion,

W1=nRTln⁡4V0V0=P0V0ln⁡4=2P0V0ln⁡2W_1 = nRT \ln \frac{4V_0}{V_0} = P_0 V_0 \ln 4 = 2P_0 V_0 \ln 2

For the isobaric compression, the pressure after isothermal expansion is

P1=P0V04V0=P04P_1 = \frac{P_0 V_0}{4V_0} = \frac{P_0}{4}

Thus,

W2=P1(V0−4V0)=P04(−3V0)=−0.75P0V0W_2 = P_1 (V_0 - 4V_0) = \frac{P_0}{4}(-3V_0) = -0.75 P_0 V_0

For the isochoric step,

W3=0W_3 = 0

Hence,

Qnet=Wnet=W1+W2+W3=P0V0(2ln⁡2−0.75)Q_{\text{net}} = W_{\text{net}} = W_1 + W_2 + W_3 = P_0 V_0 \left(2 \ln 2 - 0.75\right)

Therefore, the correct option is A.

Common mistakes

  • Using the initial pressure P0P_0 for the isobaric compression is incorrect because the gas first expands isothermally to 4V04V_0, so the pressure drops to P04\frac{P_0}{4}. Always compute the intermediate state before applying W=PΔVW = P\Delta V.

  • Assuming the net heat over the cycle is zero is incorrect. In a cyclic process, it is the net internal energy change that is zero, not necessarily the heat exchanged. Use Qnet=WnetQ_{\text{net}} = W_{\text{net}} when ΔUcycle=0\Delta U_{\text{cycle}} = 0.

  • Taking the work in compression as positive is incorrect for the sign convention used here, because the volume decreases from 4V04V_0 to V0V_0. Since ΔV<0\Delta V < 0, the work done by the gas is negative.

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