We know that, for one molecule of ideal gas,

$U = \text{Kinetic energy} = \frac{f}{2}RT$

where $f$ is the number of degrees of freedom, $R$ is the gas constant and $T$ is the absolute temperature.

So we get that $$\mathrm{KE}\propto \frac{f}{2}T$$

But, according to the Kinetic interpretation of temperature, All the gases with the same temperature have the same average kinetic energy per molecule irrespective of atomicity, but degrees of freedom depend on atomicity, so how is this statement valid? Is it that they are considering only translational KE? If so then why does rotational Kinetic energy not impact the temperature of the gas?


1 Answer 1


If so then why does rotational Kinetic energy not impact the temperature of the gas?

It doesn't impact the kinetic temperature of an ideal gas (which depends only on translational kinetic energy), but rotational kinetic energy does impact the internal kinetic energy of an ideal gas because of the dependency of the specific heat on degrees of freedom.

For any ideal gas, any process, the change in internal energy (which is considered totally kinetic) is given by

$$\Delta U=C_{v}\Delta T$$

Where $\Delta T$ is the change in kinetic temperature. But $C_v$ depends on the gas molecule. For a monatomic gas


Reflecting that there are only three (translational) degrees of freedom associated with a monatomic gas (considered as point masses).

For a diatomic gas


Reflecting two additional degrees of freedom associated with rotation. And for a polyatomic gas


Reflecting an additional rotational degree of freedom.

Hope this helps.


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