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(D) For a monatomic gas,the degrees of freedom $f_1 = 3$. For a diatomic gas,the degrees of freedom $f_2 = 5$.The adiabatic constant $\gamma$ for a mi

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$3$

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(D) For a monatomic gas,the degrees of freedom $f_1 = 3$. For a diatomic gas,the degrees of freedom $f_2 = 5$.The adiabatic constant $\gamma$ for a mixture is given by $\gamma_{mix} = \frac{C_{p,mix}}{C_{v,mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}}$.Given $n_1 = 1$ (monatomic) and $n_2 = n$ (diatomic).$C_{v1} = \frac{3}{2}R$,$C_{p1} = \frac{5}{2}R$.$C_{v2} = \frac{5}{2}R$,$C_{p2} = \frac{7}{2}R$.Substituting these into the formula:$\gamma_{mix} = \frac{1(\frac{5}{2}R) + n(\frac{7}{2}R)}{1(\frac{3}{2}R) + n(\frac{5}{2}R)} = \frac{5 + 7n}{3 + 5n}$.Given $\gamma_{mix} = \frac{13}{9}$,we have $\frac{5 + 7n}{3 + 5n} = \frac{13}{9}$.Cross-multiplying: $9(5 + 7n) = 13(3 + 5n)$.$45 + 63n = 39 + 65n$.$45 - 39 = 65n - 63n$.$6 = 2n$,which gives $n = 3$.

$A$ mixture of two non-reactive ideal gases is enclosed in a vessel consisting of one mole of a monatomic gas '$A$' and 'n' moles of a diatomic gas '$B$' at a temperature '$T$'. If the adiabatic constant of the gaseous mixture is $\frac{13}{9}$,then the value of 'n' is:

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