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A gas mixture contains $n_1$ moles of a monoatomic gas and $n_2$ moles of gas of rigid diatomic molecules. Each molecule in monoatomic and diatomic gas has 3 and 5 degrees of freedom respectively. If the adiabatic exponent $\left(\frac{C_p}{C_V}\right)$ for this gas mixture is 1.5 , then the ratio $\frac{n_1}{n_2}$ will be
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1
For a gas mixture,
$$
C_V=\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \text { and } C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}
$$
For monoatomic gas,
$$
n=n_1, C_{V_1}=\frac{3}{2} R, \quad C_{p_1}=\frac{5}{2} R
$$
For diatomic gas,
$$
n=n_2, C_{V_2}=\frac{5}{2} R \text { and } C_{p_2}=\frac{7}{2} R
$$
Now given for gas mixture,
$$
\begin{gathered}
\quad \frac{C_p}{C_V}=1.5 \\
\text { So, } 1.5=\frac{\frac{n_1}{n_2}\left(\frac{5}{2}\right)+\frac{7}{2}}{\frac{n_1}{n_2}\left(\frac{3}{2}\right)+\frac{5}{2}} \quad \frac{9}{2}\left(\frac{n_1}{n_2}\right)+\frac{15}{2}=\left(\frac{n_1}{n_2}\right) 5+7 \\
\Rightarrow \frac{n_1}{n_2}=1
\end{gathered}
$$
$$
C_V=\frac{n_1 C_{V_1}+n_2 C_{V_2}}{n_1+n_2} \text { and } C_p=\frac{n_1 C_{p_1}+n_2 C_{p_2}}{n_1+n_2}
$$
For monoatomic gas,
$$
n=n_1, C_{V_1}=\frac{3}{2} R, \quad C_{p_1}=\frac{5}{2} R
$$
For diatomic gas,
$$
n=n_2, C_{V_2}=\frac{5}{2} R \text { and } C_{p_2}=\frac{7}{2} R
$$
Now given for gas mixture,
$$
\begin{gathered}
\quad \frac{C_p}{C_V}=1.5 \\
\text { So, } 1.5=\frac{\frac{n_1}{n_2}\left(\frac{5}{2}\right)+\frac{7}{2}}{\frac{n_1}{n_2}\left(\frac{3}{2}\right)+\frac{5}{2}} \quad \frac{9}{2}\left(\frac{n_1}{n_2}\right)+\frac{15}{2}=\left(\frac{n_1}{n_2}\right) 5+7 \\
\Rightarrow \frac{n_1}{n_2}=1
\end{gathered}
$$
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