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Considering the gases to be ideal, the value of $\gamma = \frac{{{C_p}}}{{{C_v}}}$ for a gaseous mixture consisting of $3\ moles$ of carbon dioxide and $2\ moles$ of oxygen will be $({\gamma _{{O_2}}} = 1.4\,,{\gamma _{C{O_2}}} = 1.3)$
$1.37$
$1.33$
$1.55$
$1.63$
Solution
$\begin{array}{ll}{\nu_{1}=2} & {O_{2}\left(\gamma_{1}=1.4\right)} \\ {\nu_{2}=3} & {C O_{2}\left(\gamma_{2}=1.3\right)}\end{array}$
Let us assume adiabatic exponent of $O_{2}$ and $C O_{2}$ are $\gamma_{1}$ and $\gamma_{2}$
Internal energy of system$:$
$U=\nu_{1} \frac{R}{\eta-1} T+\nu_{2} \frac{R}{\gamma_{2}-1} T=\left(\nu_{1}+\nu_{2}\right) \frac{R T}{r-1}$
or, $\frac{\nu_{1}}{\gamma_{1}-1}+\frac{\nu_{2}}{\gamma_{3}-1}=\frac{\mu_{1}+\nu_{2}}{r-1}$
or, $\frac{\nu_{1}\left(\gamma_{2}-1\right)+\nu_{2}\left(\gamma_{1}-1\right)}{\left(\gamma_{1}-1\right)\left(\gamma_{2}-1\right)}=\frac{\nu_{1}+\nu_{2}}{(r-1)}$
or, $r=\frac{\left(\gamma_{1}-1\right)\left(\gamma_{2}-1\right)\left(\nu_{1}+\nu_{2}\right)}{\nu_{1}\left(\gamma_{2}-1\right)+\nu_{2}\left(\gamma_{1}-1\right)}+1$
or, $r=\frac{\nu_{1} \gamma_{1}\left(\gamma_{2}-1\right)+\nu_{2} \gamma_{2}\left(\gamma_{1}-1\right)}{\nu_{1}\left(\gamma_{2}-1\right)+\nu_{2}\left(\gamma_{1}-1\right)}=1.33=\gamma$
