Given,
Moles of helium is, $n_{1} = \frac{m}{M} = \frac{16}{4} = 4$
Moles of oxygen, $n_{2} = \frac{m}{M} = \frac{16}{32} = \frac{1}{2}$

Specific heat at constant volume for mixture is given by,
$C_{v} = \frac{n_{1} C_{v_{1}} + n_{2} C_{v_{2}}}{n_{1} + n_{2}} . . . (1)$

For helium

$n_{1} = \frac{16}{4} = 4$ and $γ_{1} = \frac{5}{3}$

$C_{v_{1}} = \frac{R}{γ_{1} - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{3}{2} R$

For oxygen,
$n_{2} = \frac{16}{32} = \frac{1}{2}$ and $γ_{2} = \frac{7}{5}$
$C_{v_{2}} = \frac{R}{γ_{2} - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{5}{2} R$
Using expression $(1)$ , we have
$\Rightarrow C_{v} = \frac{4 \times \frac{3}{2} R + \frac{1}{2} \times \frac{5}{2} R}{4 + \frac{1}{2}} = \frac{6 R + \frac{5}{4} R}{\frac{9}{2}}$
$\Rightarrow C_{v} = \frac{29 R \times 2}{9 \times 4} = \frac{29 R}{18}$ Now $C_{v} = \frac{R}{γ - 1}$
$\Rightarrow γ - 1 = \frac{R}{\frac{29}{18} R}$
$\Rightarrow \frac{C_{p}}{C_{v}} = \frac{18}{29} + 1$
$\Rightarrow \frac{C_{p}}{C_{v}} = \frac{18 + 29}{29} = 1.62$

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