A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure $\mathrm{P}_{0}$, volume $\mathrm{V}_{0}$ and temperature $\mathrm{T}_{0}$. If the gas mixture is adiabatically compressed to a volume $\mathrm{V}_{0} / 4$, then the correct statement(s) is/are, (Give $2^{1.2}=2.3 ; 2^{3.2}=9.2 ; \mathrm{R}$ is gas constant)

(A) The final pressure of the gas mixture after compression is in between $9 \mathrm{P}_{0}$ and $10 \mathrm{P}_{0}$; (C) The work |W| done during the process is $13 \mathrm{RT}_{0}$; (D) Adiabatic constant of the gas mixture is 1.6

(B) The average kinetic energy of the gas mixture after compression is in between $18 \mathrm{RT}_{0}$ and $19 \mathrm{RT}_{0}$

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Heat & Thermodynamics

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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,
(Give

2^{1.2} = 2.3; 2^{3.2} = 9.2; R

is gas constant)

(A)The final pressure of the gas mixture after compression is in between $9 P_{0}$ and $10 P_{0}$ ✓ Correct
(B)The average kinetic energy of the gas mixture after compression is in between $18 RT_{0}$ and $19 RT_{0}$
(C)The work |W| done during the process is $13 RT_{0}$ ✓ Correct
(D)Adiabatic constant of the gas mixture is 1.6✓ Correct

Answer: (A), (C), (D)

Solution

First, calculate the effective

C_{v}

for the mixture:

(C_{v})_{m} = \frac{5 \times \frac{3 R}{2} + 1 \times \frac{5 R}{2}}{6} = \frac{5 R}{3}

Then

γ_{m} = \frac{C _{p}}{C _{v}} = \frac{8 R /3}{5 R /3} = \frac{8}{5} = 1.6

, so option (D) is correct.

For adiabatic process:

P_{0} V_{0}^{γ} = P (V_{0} /4)^{γ}

, giving

P = P_{0} (4)^{8/5} = 9.2 P_{0}

, which is between

9 P_{0}

and

10 P_{0}

, so option (A) is correct.

Using ideal gas law to find final temperature:

T = \frac{9.2}{4} T_{0} = 2.3 T_{0}

.

Average KE =

5 \times \frac{3}{2} R T + 1 \times \frac{5}{2} R T = 10 R T = 10 R (2.3 T_{0}) = 23 R T_{0}

, so option (B) is incorrect.

Work done:

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{P _{0} V _{0} - 9.2 P _{0} ( V _{0} /4 )}{3/5} = - 13 R T_{0}

, so

∣ W ∣ = 13 R T_{0}

, option (C) is correct.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Heat & Thermodynamics

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