Question

$2$ moles of monoatomic gas is mixed with $1$ mole of a diatomic gas. Then $\gamma$ for the mixture is

• A. 1.4
• B. 1.55
• C. 1.62
• D. 1.67

Answer 1

Answer: (B)

For monoatomic gas
$C_{V}=\frac{3}{2} R,$
From Mayer's formula
$C_{p}-C_{V} =R$
$\Rightarrow C_{p} =R+\frac{3}{2} R=\frac{5}{2} R$
For diatomic gas
$C_{V}=\frac{5}{2} R $,
and $C_{p}=R+C_{V}=R+\frac{5}{2} R=\frac{7}{2} R$
For the mixture
$Q =n C_{V} \Delta t,$ we have
$3 C_{V} \Delta t =2 \times \frac{3}{2} R \times \Delta t+1 \times \frac{5}{2} R \times \Delta t $
$\Rightarrow C_{V} =\frac{11}{6} R$
The value of $C_{p}$ for mixture is
$2 \times \frac{5}{2} R \times \Delta t+1 \times \frac{7}{2} R \times \Delta t=3 C_{p} \Delta t $
$\Rightarrow C_{p}=\frac{17 R}{6} $
$\therefore \gamma=\frac{C_{p}}{C_{V}}=\frac{17 R}{6} \times \frac{6}{11 R}=1.55$