Question

A gaseous mixture enclosed in a vessel consists of $1g$ mole of gas $A$ with $({\gamma }_1=\frac{5}{3})$ and another gas $B$ with $({\gamma }_2=\frac{7}{5})$ at a temperature $T$. The gases $A$ and $B$ do not react with each other and assume to be ideal. Find the number of gram moles of the gas $B$, if $\gamma$ for the gaseous mixture is $\left(\frac{19}{13}\right)$.

• A. ${\mu }_2=2gmol$
• B. ${\mu }_2=2.2gmol$
• C. ${\mu }_2=2.1gmol$
• D. ${\mu }_2=2.6gmol$

Answer 1

The correct option is A ${\mu }_2=2gmol$

As for ideal gas $C_p-C_v=R$ and $\gamma =\left(\frac{C_p}{C_v}\right)$,
$\gamma -1=\frac{R}{C_v}$ or $C_v=\frac{R}{(\gamma -1)}$
${\left(C_v\right)}_1=\frac{R}{\left(\frac{5}{3}\right)-1}=\frac{3}{2}R$
${\left(C_v\right)}_2=\frac{R}{\left(\frac{7}{5}\right)-1}=\frac{5R}{2}$
${\left(C_v\right)}_{mix}=\frac{R}{\left(\frac{19}{13}\right)-1}=\frac{13}{6}R$
Now from the conservation of energy, i.e., $\Delta U=\Delta U_1+\Delta U_2$, we get
$({\mu }_1+{\mu }_2){\left(C_v\right)}_{mix}\Delta T=[{\mu }_1{\left(C_v\right)}_1+{\mu }_2{\left(C_v\right)}_2]\Delta T$
${\left(C_v\right)}_{mix}=\frac{{\mu }_1{\left(C_v\right)}_1+{\mu }_2{\left(C_v\right)}_2}{{\mu }_1+{\mu }_2}$
$\frac{13}{6}R=\frac{1\times \frac{3}{2}R+{\mu }_2\times \frac{5}{2}R}{1+{\mu }_2}=\frac{(3+5{\mu }_2)R}{2(1+{\mu }_2)}$
Or $13+13{\mu }_2=9+15{\mu }_2$, i.e., ${\mu }_2=2gmol$