For a gas if gamma = 1.4,then atomicity,{C_p} | vedclass

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(C) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.Given $\gamma = 1.4$,we have $1.4 = 1 + \frac{2}{f}$

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Diatomic,$\frac{7}{2}R, \frac{5}{2}R$

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(C) The adiabatic index is given by $\gamma = 1 + \frac{2}{f}$,where $f$ is the degree of freedom.Given $\gamma = 1.4$,we have $1.4 = 1 + \frac{2}{f}$.This implies $\frac{2}{f} = 0.4$,so $f = \frac{2}{0.4} = 5$.$A$ gas with $f = 5$ degrees of freedom is diatomic.The molar specific heat at constant volume is ${C_v} = \frac{f}{2}R = \frac{5}{2}R$.The molar specific heat at constant pressure is ${C_p} = {C_v} + R = \frac{5}{2}R + R = \frac{7}{2}R$.Therefore,the gas is diatomic,${C_p} = \frac{7}{2}R$,and ${C_v} = \frac{5}{2}R$.

List-$I$	List-$II$
$A$. Triatomic rigid gas	$I$. $\frac{C_P}{C_V} = \frac{5}{3}$
$B$. Diatomic non-rigid gas	$II$. $\frac{C_P}{C_V} = \frac{7}{5}$
$C$. Monoatomic gas	$III$. $\frac{C_P}{C_V} = \frac{4}{3}$
$D$. Diatomic rigid gas	$IV$. $\frac{C_P}{C_V} = \frac{9}{7}$

For a gas if $\gamma = 1.4$,then atomicity,${C_p}$ and ${C_v}$ of the gas are respectively

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