A poly-atomic molecule ( C v =3 R , C p =4 R . , where R is gas constant) goes from phase space point A ( P A =10 5 Pa , V A =4 × 10 -6 m 3 ) to point B ( P B =5 . .× 10 4 Pa , V B =6 × 10 -6 m 3 ) to point C ( P C =10 4 Pa . , V c =8 × 10 -6 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 :

Physics · Thermodynamics

JEE Main 2025 — 29 January, Evening Shift — Question 22

A poly-atomic molecule $\left(\mathrm{C}_{\mathrm{v}}=3 \mathrm{R}, \mathrm{C}_{\mathrm{p}}=4 \mathrm{R}\right.$ , where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^{5} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^{3}\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5\right.$

$\left.\times 10^{4} \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^{3}\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^{4} \mathrm{~Pa}\right.$ , $\mathrm{V}_{\mathrm{c}}=8 \times 10^{-6} \mathrm{~m}^{3}$ ).

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 :

Option A:
$500 \mathrm{R}(\ln 3+\ln 4)$
Option B:
$450 \mathrm{R}(\ln 4-\ln 3)$
Correct
Option C:
$500 \mathrm{R} \ln 2$
Option D:
$400 \mathrm{R} \ln 4$

Answer: B

Step-by-step solution

$\Delta Q_{A B}=0$ adiabatic $\Delta \mathrm{Q}_{\mathrm{BC}}=\Delta \mathrm{W}_{\mathrm{BC}}$

$=n R T \ell\left(\frac{\mathrm{~V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}\right)=450 \mathrm{R} \ell \mathrm{n}\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right)$

$=450 \mathrm{R} \ln \left(\frac{4}{3}\right)=450 \mathrm{R}(\ln 4-\ln 3)$

$\therefore \Delta \mathrm{Q}=\Delta \mathrm{Q}_{\mathrm{AB}}+\Delta \mathrm{Q}_{\mathrm{BC}}$

$\Delta \mathrm{Q}=450 \mathrm{R}(\ln 4-\ln 3)$

Note : Solution is based on direct data. $B$ and $C$ are not satisfying the condition of isothermal process.

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Different Thermodynamic Processes

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Answer: B

Step-by-step solution

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