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

${\left(C_p\right)}_m=\frac{5R}{3}+R=\frac{8R}{3}$

$\left(1\right)P_0{\vee }_0^{\gamma }=P{\left(\frac{{\vee }_0}{4}\right)}^{\gamma }\Rightarrow P=P_o{\left(4\right)}^{8/5}=9.2P_0$ which is between $9P_0$ and $10P_0$

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

$=10RT$

To calculate T

$\frac{P_0{\vee }_0}{T_0}=9.2P_0\times \frac{{\vee }_0}{4\times T}$

so $T=\frac{9.2}{4}{T}_0$

Now average $KE=10R\times 9.2\frac{T_0}{4}=23R{T}_0$

$\left(3\right)W=\frac{P_1{\vee }_1-P_2{\vee }_2}{\gamma -1}$

$=\frac{P_0{\vee }_0-9.2P_0\times \frac{{\vee }_0}{4}}{3/5}=-13R{T}_0$