JEE Advanced 2025 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

A container holds a mixture of 1 mol of a monatomic ideal gas and 2 mol of a diatomic ideal gas (rigid molecules, no vibration), in thermal…

Single correct option

A container holds a mixture of

1

mol of a monatomic ideal gas and

2

mol of a diatomic ideal gas (rigid molecules, no vibration), in thermal equilibrium. The mixture is slowly taken through a process in which its pressure is directly proportional to its volume, and its temperature rises by

50\ \mathrm{K}

. Take

R=8.31\ \mathrm{J\,mol^{-1}K^{-1}}

. The heat absorbed by the mixture is closest to

A. $2.70\ \mathrm{kJ}$
B. $3.95\ \mathrm{kJ}$
C. $1.87\ \mathrm{kJ}$
D. $3.32\ \mathrm{kJ}$

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Answer: (D) $3.32\ \mathrm{kJ}$

For the mixture of

n=3

mol the total internal energy change is

\Delta U=(n_1C_{v1}+n_2C_{v2})\Delta T=\left(1\cdot\tfrac32R+2\cdot\tfrac52R\right)\Delta T=6.5R\,\Delta T

. In the process

P=\alpha V

we have

T=PV/nR=\alpha V^2/nR

, so

dV/dT=nR/(2\alpha V)

and the work done is

W=\int P\,dV=\tfrac12 nR\,\Delta T=1.5R\,\Delta T

. Therefore

Q=\Delta U+W=8R\,\Delta T=8(8.31)(50)=3324\ \mathrm{J}\approx3.32\ \mathrm{kJ}

. The isochoric value

6.5R\Delta T=2.70\ \mathrm{kJ}

ignores the work, the isobaric value

9.5R\Delta T=3.95\ \mathrm{kJ}

uses

W=nR\Delta T

, and

1.87\ \mathrm{kJ}

comes from using work alone with a wrong factor.

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