The specific heats,C_P and C_V of a gas of di | vedclass

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(D) For gas $A$: The molar heat capacity at constant volume is $C_V = 22 \, J\, mol^{-1}\, K^{-1}$.Using the relation $C_V = \frac{f}{2}R$,where $R \a

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$A$ has a vibrational mode but $B$ has none

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(D) For gas $A$: The molar heat capacity at constant volume is $C_V = 22 \, J\, mol^{-1}\, K^{-1}$.Using the relation $C_V = \frac{f}{2}R$,where $R \approx 8.314 \, J\, mol^{-1}\, K^{-1}$,we get $f = \frac{2 C_V}{R} = \frac{2 	imes 22}{8.314} \approx 5.29$.Since $f$ is slightly greater than $5$ (the degrees of freedom for a rigid diatomic molecule),it indicates the presence of $1$ vibrational mode.For gas $B$: The molar heat capacity at constant volume is $C_V = 21 \, J\, mol^{-1}\, K^{-1}$.Using $f = \frac{2 C_V}{R} = \frac{2 	imes 21}{8.314} \approx 5.05$.This value is approximately $5$,which corresponds to a rigid diatomic molecule with no vibrational modes.Thus,$A$ has a vibrational mode but $B$ has none.

The specific heats,$C_P$ and $C_V$ of a gas of diatomic molecules,$A$,are given (in units of $J\, mol^{-1}\, K^{-1}$) by $29$ and $22$,respectively. Another gas of diatomic molecules,$B$,has the corresponding values $30$ and $21$. If they are treated as ideal gases,then:

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$Assertion :$ For a gas atom,the number of degrees of freedom is $3$.
$Reason :$ $\frac{C_P}{C_V} = \gamma $

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