JEE Advanced 2023 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

A mixture of 1 mol of a monatomic ideal gas and 1 mol of a diatomic ideal gas (rigid, no vibration) is kept at 300 K. It is compressed adiabatically…

Numerical answer

A mixture of

1

mol of a monatomic ideal gas and

1

mol of a diatomic ideal gas (rigid, no vibration) is kept at

300\ \mathrm{K}

. It is compressed adiabatically and reversibly to one quarter of its initial volume. Take

R=\dfrac{25}{3}\ \mathrm{J\,mol^{-1}K^{-1}}

. The work done on the gas, in kJ, is

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Answer: 10

For a mixture of ideal gases the heat capacities add:

C_V=\tfrac32R+\tfrac52R=4R

and

C_P=\tfrac52R+\tfrac72R=6R

for the two moles together. Hence the effective index is

\gamma=C_P/C_V=3/2

. In a reversible adiabatic process

TV^{\gamma-1}=

constant, so

T_2=300\times4^{1/2}=600\ \mathrm{K}

. The work done on the gas equals the increase in internal energy,

W=C_V\Delta T=4R\times300=4\times\dfrac{25}{3}\times300=10000\ \mathrm{J}=10\ \mathrm{kJ}

. A check using

W=\dfrac{nR(T_2-T_1)}{\gamma-1}

with

n=2

and the effective

\gamma

gives

2\times\tfrac{25}{3}\times300/0.5=10000\ \mathrm{J}

, the same.

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