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(D) Given that the molar specific heat of a gas is $\frac{5}{2} R$.We know that for a gas with $f$ degrees of freedom:$C_V = \frac{fR}{2}$ and $C_P =

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can be monoatomic or rigid diatomic

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(D) Given that the molar specific heat of a gas is $\frac{5}{2} R$.We know that for a gas with $f$ degrees of freedom:$C_V = \frac{fR}{2}$ and $C_P = \left(1 + \frac{f}{2}\right) R$.Case $1$: If the given specific heat is $C_V$,then $\frac{fR}{2} = \frac{5}{2} R$,which implies $f = 5$. $A$ gas with $f = 5$ is a rigid diatomic gas.Case $2$: If the given specific heat is $C_P$,then $\left(1 + \frac{f}{2}\right) R = \frac{5}{2} R$. This simplifies to $1 + \frac{f}{2} = 2.5$,so $\frac{f}{2} = 1.5$,which implies $f = 3$. $A$ gas with $f = 3$ is a monoatomic gas.Since the problem does not specify whether the value is $C_P$ or $C_V$,the gas can be either monoatomic or rigid diatomic.

The molar specific heat of a gas as given from the kinetic theory is $\frac{5}{2} R$. If it is not specified whether it is $C_P$ or $C_V$,one could conclude that the molecules of the gas

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