A gaseous mixture consists of $4 \mathrm{~g}$ oxygen and $4 \mathrm{~g}$ of helium. The ratio…
A gaseous mixture consists of $4 \mathrm{~g}$ oxygen and $4 \mathrm{~g}$ of helium. The ratio $\frac{C_p}{C_V}$ of mixture is $\left(C_p\right.$ and $C_V$ are molar specific heats of the mixture at constant pressure and at constant volume respectively).
$\frac{29}{13}$
$\frac{47}{18}$
$\frac{47}{29}$
$\frac{18}{13}$
Solution
For a gas mixture,
$r_{\text {mixture }}=\frac{C_{p_{(\mathrm{mix})}}}{C_{V(\text { mix })}}$
$=\frac{\left(\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}\right)}{\left(\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2}\right)}$
$\Rightarrow \quad r_{\text {mix }}=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1 C_{V_1}+n_2 C_{V_2}}$ ...(i)
Now, in given mixture we have $4 \mathrm{~g}$ of oxygen $\left(\mathrm{O}_2\right)$ and $4 \mathrm{~g}$ of helium (He).
Now, number of moles of oxygen, $n_1=\frac{m}{M}$
$=\frac{\text { Sample mass }}{\text { Molar mass }}$
$n_1=\frac{4}{32}=\frac{1}{8}$
Similarly number of moles of helium, $n_2=\frac{4}{4}=1$
Also oxygen is a diatomic gas so,
$C_{p_1}=\frac{7}{2} R$ and $C_{V_1}=\frac{5}{2} R$
Helium is monoatomic, hence
$C_{p_2}=\frac{5}{2} R, C_{V_2}=\frac{3}{2} R$
Substituting these values in eq. (i) we get,
$r_{\text {mix }}=\frac{\frac{1}{8} \times \frac{7}{2} R+1 \times \frac{5}{2} R}{\frac{1}{8} \times \frac{5}{2} R+1 \times \frac{3}{2} R}$
$=\frac{\frac{7}{16}+\frac{5}{2}}{\frac{5}{16}+\frac{3}{2}}$
or $\quad r_{\text {mix }}=\frac{7+40}{5+24}=\frac{47}{29}$