A poly-atomic molecule (CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A(PA = 10⁵ Pa, VA = 4 × 10⁻⁶ m³) to point B(PB = 5 × 10⁴ Pa, VB = 6 × 10⁻⁶ m³) to point C (PC = 10⁴ Pa, VC = 8 × 10⁻⁶ m³). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is: (a) 500R(ln 3 + ln 4) (b) 450R(ln 4 - ln 3) (c) 500Rln 2 (d) 400Rln 4

Answer: (b). A to B is adiabatic: QAB = 0. B to C is isothermal at T = 450 K (from the figure): QBC = RTln (VC)/(VB) per mole. Q = 450Rln 8/6 = 450Rln 4/3 = 450R(ln 4 - ln 3) So the answer is (b).

A poly-atomic molecule (CV = 3R, CP = 4R, where R is gas constant)… | JEE Main 2025 (29 Jan, Shift 2)

Q25JEE Main 2025 · 29 JAN · SHIFT 2Difficult

P-V Diagrams & Cyclic Processes

A poly-atomic molecule $(C_V = 3R, C_P = 4R$ , where $R$ is gas constant) goes from phase space point $A\left(P_A = 10^{5}\text{ Pa},\ V_A = 4 \times 10^{-6}\text{ m}^3\right)$ to point $B\left(P_B = 5 \times 10^{4}\text{ Pa},\ V_B = 6 \times 10^{-6}\text{ m}^3\right)$ to point $C\ (P_C = 10^{4}\text{ Pa}, V_C = 8 \times 10^{-6}\text{ m}^3$ ). A to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path. The net heat absorbed per unit mole by the system is:

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Answer: (b)

A to B is adiabatic: $Q_{AB} = 0$ .
B to C is isothermal at $T = 450$ K (from the figure): $Q_{BC} = RT\ln\frac{V_C}{V_B}$ per mole.
$Q = 450R\ln\frac{8}{6} = 450R\ln\frac{4}{3} = 450R(\ln 4 - \ln 3)$
So the answer is (b).

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