JEE Advanced 2019 · Paper 2 · Physics Q5

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PhysicsMultiple correct+4 / -1Question 5 of 54

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure $P_0$ , volume $V_0$ , and temperature $T_0$ . If the gas mixture is adiabatically compressed to a volume $V_0/4$ , then the correct statement(s) is/are,

(Given $2^{1.2} = 2.3$ ; $2^{3.2} = 9.2$ ; $R$ is gas constant)

The work

|W|

done during the process is

13RT_0

The average kinetic energy of the gas mixture after compression is in between

18RT_0

and

19RT_0

The final pressure of the gas mixture after compression is in between

9P_0

and

10P_0

Adiabatic constant of the gas mixture is 1.6

Answer and worked solution

Official answer: A, C, D

(D) γ of the mixture.
$C_V=5\cdot\tfrac32R+1\cdot\tfrac52R=10R,\qquad C_P=5\cdot\tfrac52R+1\cdot\tfrac72R=16R,\qquad\gamma=\frac{16}{10}=1.6.$
True.

Final state. Compressing to $V_0/4$ :
$T=T_0\,4^{0.6}=2^{1.2}T_0=2.3T_0,\qquad P=P_0\,4^{1.6}=2^{3.2}P_0=9.2P_0.$
So (C) is true.

(A) $|W|=\Delta U=10R(1.3T_0)=13RT_0$ . True.

(B) The total kinetic energy is $U=10R\times2.3T_0=23RT_0$ , which is not between $18RT_0$ and $19RT_0$ . False.

(A), (C), (D)

Topics:Specific heats of gases Thermodynamic processes Work done by a gas

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JEE Advanced 2019 · Paper 2, Physics question 5