A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure P₀ volume V₀ and temperature T₀.If the gasmixture is adiabatically compressed to a volume V₀/4, then the correct statement(s) is/are,(Given 2¹.2}=2.3;2³.2}9.2;R is gas constant) [JEE Adv 2019 P2]

The work $\left|W\right|$ done during the process is $13RT_0$
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
The average kinetic energy of the gas mixture after compression is in between $18RT_{0}$ and $19RT_{0}$

Solution:

P_{1}=\; \;\frac{5RT_{0}}{V_{0}}

P_{2}=\; \;\frac{RT_{0}}{V_{0}}

P_{1}+P_{2}=P_{0}=\; \;\frac{6RT_{0}}{V_{0}}

\gamma_{\; \text{mixture} \;}=\; \;\frac{5C_{P}^{1}+(1)C_{P}^{11}}{5C_{V}^{1}+(1)C_{V}^{11}}

=\; \;\frac{5\times\; \;\frac{5R}{2}+(1)\left(\; \;\frac{7R}{2}\right)}{5\times\; \;\frac{3R}{2}+(1)\left(\; \;\frac{5R}{2}\right)}

=\; \;\frac{32}{20}=1.6

PV^{\gamma}=\text{consant}

P_{0}V_{0}^{1.6}=P\left[\frac{V_{0}}{4}\right]^{1.6}

P=P_{0}(4)^{1.6}=P_{0}(2^{3.2})

P=9.2P_{0}

W=\; \;\frac{P_{2}V_{2}-P_{1}V_{1}}{1-\gamma}=\; \;\frac{(9.2P_{0})\; \;\frac{V_{0}}{4}-P_{0}V_{0}}{1-1.6}

=\; \;\frac{(2.3-1)P_{0}V_{0}}{0.6}=\; \;\frac{1.3\times 6RT_{o}}{0.6}

W=13RT_{0}

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