Question

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,
(Given $2^{1.2}=2.3;2^{3.2}=9.2;R$ is gas constant)

• A. The final pressure of the gas mixture after compression is in between $9P_0$ and $10P_0$
• B. The average kinetic energy of the gas mixture after compression is in between $18R{T}_0$ and $19R{T}_0$
• C. Adiabatic constant of the gas mixture is $1.6$
• D. The work |W| done during the process is $13R{T}_0$

Answer 1

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

The work |W| done during the process is $13R{T}_0$

$n_1=5$ moles $C_{V_1}=\frac{3R}{2}$

$n_2=1$ mole $C_{V_2}=\frac{5R}{2}$

${\left(C_V\right)}_m=\frac{n_1{C}_{V_1}+n_2{C}_{V_2}}{n_1+n_2}=\frac{5\times \frac{3R}{2}+1\times \frac{5R}{2}}{6}=\frac{5R}{3}$

${\gamma }_n=\frac{{\left(C_p\right)}_m}{{\left(C_v\right)}_m}=\frac{8}{5}$
So, option (3) is correct.

$P_0{V}_0^{8/5}=P_2{\left(\frac{V_0}{4}\right)}^{8/5}$

$P_2=9.2P_0$
Option 'A' is also correct.

(2) $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$T_2=T_1(2{)}^{6/5}=2.3T_0$

Average kinetic energy of gas mixture = $n{C}_{V_{mix}}{T}_2$ $=23R{T}_0$

(4) $W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$

$W=\frac{P_0{V}_0-9.2P_0\frac{V_0}{4}}{3/5}=-13R{T}_0$

$\therefore |W|=13R{T}_0$