Question

A gaseous mixture enclosed in a vessel of volume $2m^3$ consists of $2$ moles of a gas A with ${\gamma }_A=\frac{C_P}{C_V}=\frac{5}{3}$, another gas B with ${\gamma }_B=\frac{7}{5}$ at a temperature ${43}^{\circ }C$. The gram molecular weights of A & B are $32\text{gm}$ and $4\text{gm}$ respectively. Assume gases to be non - reacting. The gaseous mixture follows $P{V}^{20/14}=\text{constant}$ in adiabatic process. Find the number of moles of the gas B in the gaseous mixture.

• A. $n_B=4\text{moles}$
• B. $n_B=2\text{moles}$
• C. $n_B=7\text{moles}$
• D. $n_B=10\text{moles}$

Answer 1

The correct option is D $n_B=10\text{moles}$

Given
Volume of vessel $=2m^3$
moles of $A=2$
moles of $B=n_B$ (say)
${\gamma }_A=\frac{5}{3}\left\{\text{monoatomic}\right\}$
${\gamma }_B=\frac{7}{5}\left\{\text{diatomic}\right\}$
Molar mass of$A,M_A=32\text{gm}$
Molar mass of $B,M_B=4\text{gm}$
Also, $P{V}^{20/14}=\text{constant}$
$\Rightarrow {\gamma }_{mixture}=\frac{20}{14}$
We know that
${\gamma }_{\text{mixture}}=\frac{C_P\text{mixture}}{C_V\text{mixture}}$
$\Rightarrow \frac{20}{14}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}$
$=\frac{2\times \frac{5R}{2}+n_B\times \frac{7R}{2}}{2\times \frac{3R}{2}+n_B\times \frac{5R}{2}}\Rightarrow \frac{20}{14}=\frac{10+7n_B}{6+5n_B}$
$\Rightarrow 120+100n_B=140+98n_B$
$\Rightarrow 2n_B=20$
$\Rightarrow n_B=10\text{moles}$