If one mole of an ideal gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas…
If one mole of an ideal gas $\left(\gamma=\frac{5}{3}\right)$ is mixed with one mole of a diatomic gas $\left(\gamma=\frac{7}{5}\right)$. The value of $\gamma$ for the mixture is
$1.50$
$1.53$
$3.07$
$1.40$
Solution
As we know, $C_v=\frac{3}{5} R T ; C_p=\frac{5 R T}{2}$ for monoatomic gas;
$C_v=\frac{5}{2} R T ; C_p=\frac{7}{2} R T$ for diatomic gas
Thus, for mixture of 1 mole each,
$C_v=\frac{\frac{3}{2} R T+\frac{5}{2} R T}{2}$ and $C_p=\frac{\frac{5}{2} R T+\frac{7}{2} R T}{2}=3 R T$
Therefore, $\frac{C_v}{C_p}=\frac{3 R T}{2 R T}=1.5$