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JEE Advanced 2026Paper 1
Exam
JEE Advanced
Year
2026
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2026 · Paper 1 · Physics · Thermodynamics

A quasi-static cycle of a monoatomic ideal gas contains an...

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (ab)(\boldsymbol{ab}), followed by an isochoric process (bc)(\boldsymbol{bc}) and an adiabatic process (ca)(\boldsymbol{ca}) as shown in the figure. The volumes of the gas are V1V_1 and V2V_2 at a\boldsymbol{a} and b\boldsymbol{b}, respectively. If the cycle has heat input QinQ_{\text{in}} and output QoutQ_{\text{out}}, then the efficiency of the cycle is defined as η=Qin−QoutQin\eta = \dfrac{Q_{\text{in}} - Q_{\text{out}}}{Q_{\text{in}}}. The correct statement(s) is/are:
[Given: ln⁡2≈0.7\ln 2 \approx 0.7]
A quasi-static cycle of a monoatomic ideal gas contains an isothermal process ( ab ) , followed by an isochoric process

Options

  1. AIf V2/V1=8V_2/V_1 = 8, the heat released in the process bc\boldsymbol{bc} is smaller than the heat absorbed in the process ab\boldsymbol{ab}.
  2. BFor a given value of V2/V1V_2/V_1, η\eta does not depend on the temperature of the isothermal process.
  3. CIf V2/V1=8V_2/V_1 = 8, then the temperature of the gas at a\boldsymbol{a} is 44 times the temperature of the gas at c\boldsymbol{c}.
  4. DIf V2/V1=8V_2/V_1 = 8, then the pressure of the gas at a\boldsymbol{a} is 44 times the pressure of the gas at b\boldsymbol{b}.

Correct answer

A. If V2/V1=8V_2/V_1 = 8, the heat released in the process bc\boldsymbol{bc} is smaller than the heat absorbed in the process ab\boldsymbol{ab}.

Step-by-step solution

For the monoatomic ideal gas, γ=53\gamma = \dfrac{5}{3}.

Process abab is isothermal, so Ta=TbT_a = T_b and Va=V1V_a = V_1, Vb=V2V_b = V_2.

Process bcbc is isochoric, so Vb=Vc=V2V_b = V_c = V_2.

Process caca is adiabatic, so TcVcγ−1=TaVaγ−1T_c V_c^{\gamma-1} = T_a V_a^{\gamma-1}.

TcV22/3=TaV12/3⇒Tc=Ta(V1V2)2/3T_c V_2^{2/3} = T_a V_1^{2/3} \Rightarrow T_c = T_a \left(\dfrac{V_1}{V_2}\right)^{2/3}.

If V2V1=8\dfrac{V_2}{V_1} = 8, Tc=Ta(18)2/3=Ta4⇒Ta=4TcT_c = T_a \left(\dfrac{1}{8}\right)^{2/3} = \dfrac{T_a}{4} \Rightarrow T_a = 4 T_c. Thus, statement (C) is correct.

Heat absorbed in isothermal process abab is Qab=nRTaln⁡(V2V1)Q_{ab} = nRT_a \ln\left(\dfrac{V_2}{V_1}\right).

For V2V1=8\dfrac{V_2}{V_1} = 8, Qab=nRTaln⁡(8)=3nRTaln⁡2≈3×0.7nRTa=2.1nRTaQ_{ab} = nRT_a \ln(8) = 3nRT_a \ln 2 \approx 3 \times 0.7 nRT_a = 2.1 nRT_a.

Heat released in isochoric process bcbc is Qbc=nCv(Tb−Tc)=n(32R)(Ta−Ta4)=98nRTa=1.125nRTaQ_{bc} = nC_v(T_b - T_c) = n\left(\dfrac{3}{2}R\right)\left(T_a - \dfrac{T_a}{4}\right) = \dfrac{9}{8}nRT_a = 1.125 nRT_a.

Since 1.125nRTa<2.1nRTa1.125 nRT_a < 2.1 nRT_a, the heat released in bcbc is smaller than the heat absorbed in abab. Thus, statement (A) is correct.

The efficiency of the cycle is η=1−QoutQin=1−QbcQab\eta = 1 - \dfrac{Q_{\text{out}}}{Q_{\text{in}}} = 1 - \dfrac{Q_{bc}}{Q_{ab}}.

η=1−n(32R)Ta(1−(V1V2)2/3)nRTaln⁡(V2V1)=1−3(1−(V1V2)2/3)2ln⁡(V2V1)\eta = 1 - \dfrac{n \left(\dfrac{3}{2} R\right) T_a \left(1 - \left(\dfrac{V_1}{V_2}\right)^{2/3}\right)}{nRT_a \ln\left(\dfrac{V_2}{V_1}\right)} = 1 - \dfrac{3 \left(1 - \left(\dfrac{V_1}{V_2}\right)^{2/3}\right)}{2 \ln\left(\dfrac{V_2}{V_1}\right)}.

This expression depends only on the volume ratio V2V1\dfrac{V_2}{V_1} and is independent of the temperature TaT_a. Thus, statement (B) is correct.

For the isothermal process abab, PaVa=PbVb⇒PaV1=PbV2⇒Pa=8PbP_a V_a = P_b V_b \Rightarrow P_a V_1 = P_b V_2 \Rightarrow P_a = 8 P_b. Thus, statement (D) is incorrect.

Answer: If V2/V1=8V_2/V_1 = 8, the heat released in the process bc\boldsymbol{bc} is smaller than the heat absorbed in the process ab\boldsymbol{ab}.; For a given value of V2/V1V_2/V_1, η\eta does not depend on the temperature of the isothermal process.; If V2/V1=8V_2/V_1 = 8, then the temperature of the gas at a\boldsymbol{a} is 44 times the temperature of the gas at c\boldsymbol{c}.
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