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(C) The total kinetic energy of a gas in a moving container with respect to the ground is the sum of the kinetic energy of the center of mass and the

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$\frac{1}{2} M v^2 + \frac{5}{2} n R T$

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(C) The total kinetic energy of a gas in a moving container with respect to the ground is the sum of the kinetic energy of the center of mass and the internal kinetic energy of the gas molecules.$1$. The kinetic energy of the center of mass of the gas with respect to the ground is given by $K_{cm} = \frac{1}{2} M v^2$,where $M$ is the total mass of the gas and $v$ is the velocity of the container.$2$. The internal kinetic energy of a diatomic gas (which has $f = 5$ degrees of freedom) with respect to the center of mass is given by $U = n \frac{f}{2} R T$.$3$. Substituting $f = 5$,we get $U = \frac{5}{2} n R T$.$4$. Therefore,the total kinetic energy with respect to the ground is $K_{total} = K_{cm} + U = \frac{1}{2} M v^2 + \frac{5}{2} n R T$.

$A$ light container having a diatomic gas enclosed within is moving with velocity $v$. The mass of the gas is $M$ and the number of moles is $n$. The kinetic energy of the gas with respect to the ground is:

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