Question

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 gas mixture is adiabatically compressed to a volume V₀/4, 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 9P₀ and 10P₀
• B. The average kinetic energy of the gas mixture after compression is in between 18RT₀ and 19RT₀
• C. The work |W| done during the process is 13RT₀
• D. Adiabatic constant of the gas mixture is 1.6

Answer 1

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

$n_{1} = 5$ moles $C_{\vee_1} = \frac{3R}{2}\,\,P_{0} \vee_{0} T_{0}$

$n_{2} = 1$ mole $C_{\vee_2} = \frac{5R}{2}$

$\left(C_{\vee}\right)_{m} = \frac{n_{1}C_{\vee_1}+n_{2}C_{\vee_2}}{n_{1}+n_{2}} = \frac{5\times\frac{3R}{2}+1\times\frac{5R}{2}}{6} = \frac{5R}{3}$

$\gamma_{n} = \frac{\left(C_{p}\right)_{m}}{\left(C_{\vee}\right)_{m}} = \frac{8}{5}$

$\therefore $ Option 4 is correct

$\left(C_{p}\right)_{m} = \frac{5R}{3}+R = \frac{8R}{3}$

$\left(1\right) P_{0}\vee_{0}^{\gamma} = P\left(\frac{\vee_{0}}{4}\right)^{\gamma}\,\Rightarrow \,P = P_{o}\left(4\right)^{8/5} = 9.2 P_{0}$ which is between $9P_{0}$ and $10P_{0}$

$\left(2\right)$ Average K.E. $= 5 × \frac{3}{2}RT + 1\times \frac{5RT}{2}$

$= 10RT$

To calculate T

$\frac{P_{0}\vee_{0}}{T_{0}} = 9.2P_{0}\times\frac{\vee_{0}}{4\times T}$

so $\quad\quad T = \frac{9.2}{4}T_{0}$

Now average $KE = 10 R × 9.2 \frac{T_{0}}{4} = 23RT_{0}$

$\left(3\right) W = \frac{P_{1}\vee_{1}-P_{2}\vee_{2}}{\gamma-1}$

$= \frac{P_{0}\vee_{0}-9.2P_{0}\times\frac{\vee_{0}}{4}}{3/5} = -13RT_{0}$