JEE Advanced 2019 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

A closed insulated cylinder with a frictionless piston contains a mixture of 1 mol of a monatomic ideal gas and 1 mol of a rigid diatomic ideal gas…

Single correct option

A closed insulated cylinder with a frictionless piston contains a mixture of

1

mol of a monatomic ideal gas and

1

mol of a rigid diatomic ideal gas at

300\ \mathrm{K}

. The mixture is compressed quasi-statically and adiabatically to one quarter of its initial volume. Take

R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}

. The work done on the gas is

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Answer: (B) $9.96\ \mathrm{kJ}$

For the mixture, total

C_V=\tfrac32R+\tfrac52R=4R

for the two moles, i.e.

\bar C_V=2R

per mole of mixture, and

\bar C_P=3R

, so

\gamma=\bar C_P/\bar C_V=3/2

. In a reversible adiabatic process

TV^{\gamma-1}

is constant, hence

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

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

W=n_{tot}\bar C_V\Delta T=2\times2R\times300=1200R=9.96\times10^3\ \mathrm{J}

. Treating the gas as purely monatomic (

\gamma=5/3

) gives

11.35\ \mathrm{kJ}

, using

C_P

in place of

C_V

gives

14.94\ \mathrm{kJ}

, and halving the moles gives

4.98\ \mathrm{kJ}

. The correct answer is therefore

9.96\ \mathrm{kJ}

, option (B).

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake