JEE Challenger
More from Thermodynamics
PhyHardAdv 2026 • Paper 1

Thermodynamic Cycle Efficiency of Monoatomic Gas

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (abab), followed by an isochoric process (bcbc) and an adiabatic process (caca) as shown in the figure. The volumes of the gas are V1V_1 and V2V_2 at aa and bb, 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 = \frac{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]

Question Diagram 1

Options

A

If V2/V1=8V_2/V_1 = 8, the heat released in the process bcbc is smaller than the heat absorbed in the process abab.

Correct
B

For a given value of V2/V1V_2/V_1, η\eta does not depend on the temperature of the isothermal process.

Correct
C

If V2/V1=8V_2/V_1 = 8, then the temperature of the gas at aa is 4 times the temperature of the gas at cc.

Correct
D

If V2/V1=8V_2/V_1 = 8, then the pressure of the gas at aa is 4 times the pressure of the gas at bb.

Topics & Concepts

ThermodynamicsFirst Law of Thermodynamics

Step-by-Step Solution

To analyze the given thermodynamic cycle for a monatomic ideal gas (γ=5/3\gamma = 5/3, Cv=32RC_v = \frac{3}{2}R), let us examine each process step-by-step:

1. Analysis of Processes:

  • Process abab (Isothermal Expansion): The gas expands isothermally at temperature Ta=TbT_a = T_b. PaV1=PbV2  ⟹  Pa=Pb(V2V1)P_a V_1 = P_b V_2 \implies P_a = P_b \left(\frac{V_2}{V_1}\right) The heat absorbed during this process is: Qin=Qab=nRTaln⁡(V2V1)Q_{\text{in}} = Q_{ab} = n R T_a \ln\left(\frac{V_2}{V_1}\right)

  • Process bcbc (Isochoric Cooling): The gas is cooled at constant volume V2V_2 from temperature Tb=TaT_b = T_a to TcT_c. The heat released during this process is: Qout=Qbc=nCv(Tb−Tc)=32nR(Ta−Tc)Q_{\text{out}} = Q_{bc} = n C_v (T_b - T_c) = \frac{3}{2} n R (T_a - T_c)

  • Process caca (Adiabatic Compression): The gas undergoes adiabatic compression from (Pc,V2,Tc)(P_c, V_2, T_c) back to (Pa,V1,Ta)(P_a, V_1, T_a). Using the relation for an adiabatic process, TVγ−1=constantT V^{\gamma-1} = \text{constant}: TcV2γ−1=TaV1γ−1T_c V_2^{\gamma-1} = T_a V_1^{\gamma-1} Since the gas is monatomic, γ=5/3  ⟹  γ−1=2/3\gamma = 5/3 \implies \gamma - 1 = 2/3: Tc=Ta(V1V2)2/3T_c = T_a \left(\frac{V_1}{V_2}\right)^{2/3}

2. Verification of Options:

  • Option C: Given V2V1=8\frac{V_2}{V_1} = 8: Tc=Ta(18)2/3=Ta(123)2/3=Ta4  ⟹  Ta=4TcT_c = T_a \left(\frac{1}{8}\right)^{2/3} = T_a \left(\frac{1}{2^3}\right)^{2/3} = \frac{T_a}{4} \implies T_a = 4 T_c Thus, the temperature at aa is 4 times the temperature at cc. Statement (C) is correct.

  • Option D: Given V2V1=8\frac{V_2}{V_1} = 8: Pa=Pb(V2V1)=8PbP_a = P_b \left(\frac{V_2}{V_1}\right) = 8 P_b Thus, the pressure at aa is 8 times the pressure at bb, not 4 times. Statement (D) is incorrect.

  • Option A: For V2V1=8\frac{V_2}{V_1} = 8: Qab=nRTaln⁡(8)=3nRTaln⁡2≈3×0.7 nRTa=2.1 nRTaQ_{ab} = n R T_a \ln(8) = 3 n R T_a \ln 2 \approx 3 \times 0.7 \, n R T_a = 2.1 \, n R T_a Qbc=32nR(Ta−Tc)=32nR(Ta−Ta4)=32×34nRTa=98nRTa=1.125 nRTaQ_{bc} = \frac{3}{2} n R \left(T_a - T_c\right) = \frac{3}{2} n R \left(T_a - \frac{T_a}{4}\right) = \frac{3}{2} \times \frac{3}{4} n R T_a = \frac{9}{8} n R T_a = 1.125 \, n R T_a Comparing the two heat values: Qbc=1.125 nRTa<2.1 nRTa=QabQ_{bc} = 1.125 \, n R T_a < 2.1 \, n R T_a = Q_{ab} Therefore, the heat released in bcbc is smaller than the heat absorbed in abab. Statement (A) is correct.

  • Option B: The efficiency η\eta of the cycle is given by: η=1−QoutQin=1−32nR(Ta−Tc)nRTaln⁡(V2V1)\eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} = 1 - \frac{\frac{3}{2} n R (T_a - T_c)}{n R T_a \ln\left(\frac{V_2}{V_1}\right)} Substituting Tc=Ta(V1V2)2/3T_c = T_a \left(\frac{V_1}{V_2}\right)^{2/3}: η=1−32nRTa[1−(V1V2)2/3]nRTaln⁡(V2V1)=1−321−(V1V2)2/3ln⁡(V2V1)\eta = 1 - \frac{\frac{3}{2} n R T_a \left[ 1 - \left(\frac{V_1}{V_2}\right)^{2/3} \right]}{n R T_a \ln\left(\frac{V_2}{V_1}\right)} = 1 - \frac{3}{2} \frac{1 - \left(\frac{V_1}{V_2}\right)^{2/3}}{\ln\left(\frac{V_2}{V_1}\right)} Since TaT_a cancels out completely, η\eta depends only on the volume ratio V2V1\frac{V_2}{V_1} and is independent of the isothermal temperature. Statement (B) is correct.

Conclusion:

The correct statements are A, B, and C.

Thermodynamic Cycle Efficiency of Monoatomic Gas | Physics PYQ Solution - JEE Challenger