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
$3C_V\Delta t=2\times \frac{3}{2}R\times \Delta t+1\times \frac{5}{2}R\times \Delta t$
$\Rightarrow C_V=\frac{11}{6}R$
The value of $C_p$ for mixture is
$2\times \frac{5}{2}R\times \Delta t+1\times \frac{7}{2}R\times \Delta t=3C_p\Delta t$
$\Rightarrow C_p=\frac{17R}{6}$
$\therefore \gamma =\frac{C_p}{C_V}=\frac{17R}{6}\times \frac{6}{11R}=1.55$