Equal volumes of monoatomic and diatomic gase | vedclass

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(D) For a monoatomic gas,${C_v} = \frac{3}{2}R$ and ${C_p} = \frac{5}{2}R$. For a diatomic gas,${C_v} = \frac{5}{2}R$ and ${C_p} = \frac{7}{2}R$. Sinc

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(D) For a monoatomic gas,${C_v} = \frac{3}{2}R$ and ${C_p} = \frac{5}{2}R$. For a diatomic gas,${C_v} = \frac{5}{2}R$ and ${C_p} = \frac{7}{2}R$. Since the gases are at the same temperature and pressure,equal volumes contain an equal number of moles $(n_1 = n_2 = 1)$. The molar heat capacity at constant volume for the mixture is ${C_{v,mix}} = \frac{n_1{C_{v,1}} + n_2{C_{v,2}}}{n_1 + n_2} = \frac{\frac{3}{2}R + \frac{5}{2}R}{2} = 2R$. The molar heat capacity at constant pressure for the mixture is ${C_{p,mix}} = \frac{n_1{C_{p,1}} + n_2{C_{p,2}}}{n_1 + n_2} = \frac{\frac{5}{2}R + \frac{7}{2}R}{2} = 3R$. Therefore,the ratio of specific heats is $\gamma = \frac{{C_{p,mix}}}{{C_{v,mix}}} = \frac{3R}{2R} = 1.5$.

Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed. The ratio of specific heats of the mixture $({C_p}/{C_v})$ will be

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