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Physics · Thermodynamics

JEE Advanced 2019 — Paper 2 — Question 8

A mixture of ideal gas contains 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure

P0P_{0}, volume V0V_{0}, and temperature T0T_{0}. If the gas mixture is adiabatically compressed to a volume V04\frac{V_{0}}{4}, then the

correct statement(s) is /are, (Given 21.2=2.3;23.2=9.2;R2^{1.2}=2.3 ; 2^{3.2}=9.2 ; \mathrm{R} is gas constant)

  1. Option A:

    Adiabatic constant of the gas mixture is 1.6

    Correct
  2. Option B:

    The final pressure of the gas mixture after compression is in between 9P09 P_{0} and 10P010 \mathrm{P}_{0}

    Correct
  3. Option C:

    The work ∣W∣|\mathrm{W}| done during the process is 13RT013 \mathrm{RT}_{0}

  4. Option D:

    The average kinetic energy of the gas mixture after compression is in between 18RT018 \mathrm{RT}_{0} and 19RT019 \mathrm{RT}_{0}

Answer: A, B

Step-by-step solution

5+1γm−1=n1γ1−1+n2γ2−1\frac{5+1}{\gamma_{m}-1}=\frac{n_{1}}{\gamma_{1}-1}+\frac{n_{2}}{\gamma_{2}-1}

⇒6γm−1=553−1+175−1=10\Rightarrow \quad \frac{6}{\gamma_{m}-1}=\frac{5}{\frac{5}{3}-1}+\frac{1}{\frac{7}{5}-1}=10 γm=1.6\gamma_{\mathrm{m}}=1.6

P0V0γm=P=V04Oγm⇒P=P0×23.2=9.2P0P_{0} V_{0}^{\gamma_{m}}=P=\frac{V_{0}}{4} O^{\gamma_{m}} \Rightarrow P=P_{0} \times 2^{3.2}=9.2 P_{0}

W=P0 V0−(9.2P0)×V040.6=−1.3P0 V00.6=1366RT0=13RT0\mathrm{W}=\frac{\mathrm{P}_{0} \mathrm{~V}_{0}-\left(9.2 \mathrm{P}_{0}\right) \times \frac{\mathrm{V}_{0}}{4}}{0.6}=\frac{-1.3 \mathrm{P}_{0} \mathrm{~V}_{0}}{0.6}=\frac{13}{6} 6 \mathrm{RT}_{0}=13 \mathrm{RT}_{0}

T0 V0Vm−1=TVVm−1⇒ T=T0□ V0 V Vm−1l0.=T0(4)0.6=2.3 T0\mathrm{T}_{0} \mathrm{~V}_{0}^{\mathrm{V}_{\mathrm{m}}-1}=\mathrm{TV}^{\mathrm{V}_{\mathrm{m}}-1} \Rightarrow \mathrm{~T}=\mathrm{T}_{0} \frac{\square \mathrm{~V}_{0}}{\frac{\mathrm{~V}}{\mathrm{~V}_{\mathrm{m}}-1}} \mathrm{l}^{0 .}=\mathrm{T}_{0}(4)^{0.6}=2.3 \mathrm{~T}_{0}

Average kinetic energy of the gas mixture

=5×32RT+1×52RT=10RT=23RT0=5 \times \frac{3}{2} \mathrm{RT}+1 \times \frac{5}{2} \mathrm{RT}=10 \mathrm{RT}=23 \mathrm{RT}_{0}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes