Question

A gaseous mixture enclosed in a vessel consists of $1\text{gm mole}$ of a gas $\text{A}$ with $\gamma =5/3$ and another gas $\text{B}$ with $\gamma =7/5$ at a temperature $T$. The gases $\text{A}$ and $\text{B}$ do not react with each other and are assumed to be ideal. Find the number of $\text{gm moles}$ of the gas $\text{B}$ if $\gamma$ for the gaseous mixture is $\frac{19}{13}$.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

For an ideal gas,

$C_P-C_V=R....\left(1\right)$

and $\gamma =\frac{C_P}{C_V}....\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we have,

$\gamma -1=\frac{R}{C_v}$

$\Rightarrow C_V=\frac{R}{(\gamma -1)}$
Substituting the value of $\gamma$ in the above equation,

${\left(C_v\right)}_A=\frac{R}{\left(5/3\right)-1}=\frac{3}{2}R$

${\left(C_V\right)}_B=\frac{R}{\left(7/5\right)-1}=\frac{5}{2}R$

and ${\left(C_v\right)}_{mix}=\frac{R}{\left(19/13\right)-1}=\frac{13}{6}R$

Now from conservation of energy,

$\Delta U=\Delta U_A+\Delta U_B,$

$\Rightarrow ({\mu }_A+{\mu }_B){\left(C_v\right)}_{mix}\Delta T=[{\mu }_A{\left(C_v\right)}_A+{\mu }_B{\left(C_v\right)}_B]\Delta T$

$\Rightarrow {\left(C_v\right)}_{mix}=\frac{{\mu }_A{\left(C_v\right)}_A+{\mu }_B{\left(C_v\right)}_B}{{\mu }_A+{\mu }_B}$

Substituting the values in the above equation,

$\frac{13}{6}R=\frac{1\times \frac{3}{2}R+n\times \frac{5}{2}R}{1+n}$

$\Rightarrow \frac{13}{6}R=\frac{(3+5n)R}{2(1+n)}$

$\Rightarrow 13+13n=9+15n$

$\therefore n=2\text{gm moles}$

Option (b) is the correct answer.