Before moving on to the question, let me clarify what I mean by the terms "total translational kinetic energy" and "average kinetic energy of a molecule". The total translational kinetic energy $K$ of all the molecules of the gas is

$$K=\sum\frac 1 2 mv^2=\frac 1 2 mN\frac{\sum v^2}{N}=\frac 1 2 Mv^2_\mathrm{rms}$$

where $m$ is the mass of a single molecule, $v$ is its velocity, $N$ is the total number of molecules in the container, $M$ is the total mass of the gas sample, and $v_\mathrm{rms}$ is the root-mean-square (RMS) speed.

The average kinetic energy of a molecule is

$$K/N=\frac 1 2 \frac M N v^2_\mathrm{rms}=\frac 1 2 mv^2_\mathrm{rms}$$

where the symbols have the same meaning as of the previous case.

It is said that, at a particular temperature, different gases have the same average kinetic energy.

$$\frac 1 2 m_1v_1^2=\frac 1 2 m_2v_2^2$$

where $m_1$, $m_2$ are the masses and $v_1$, $v_2$ are the rms speeds of the two gases respectively.

As we discussed at the beginning of the question the average kinetic energy of a single molecule is much different from the total translational kinetic energy of all the molecules of the gas. I don't understand why the average kinetic energy of a single molecule must be the same for all gases at the same temperature instead of the total kinetic energy of all the molecules.

I understood that the absolute temperature $T$ of a given gas is proportional to the square of the RMS speed $v_\mathrm{rms}$ of its molecules as per the following equation:

$$T=\left(\frac{273.16~\mathrm K}{v^2_\mathrm{tr}}\right)v^2_\mathrm{rms}$$

where $v^2_\mathrm{tr}$ is the RMS speed of the molecules at $273.16~\mathrm K$ (triple point of water). Further, both the total translational kinetic energy and the average kinetic energy of a molecule are proportional to the square of the RMS speed. Due to this similarity, I'm unable to see which factor is responsible for the same average kinetic energy instead of the same total kinetic energy at the same temperature for two different gases.

So, why do different gases have the same average kinetic energy at the same temperature instead of the same total kinetic energy?

While searching this site for this doubt, I came across these questions - Temperature and kinetic energy of molecules, Does the same temperature imply the same translational kinetic energy?, If two gasses are in thermal equilbrium, do their molecules have same amount of Kinetic energy?, and Intuitive explanation why rate of energy transfer depends on difference in energy between two materials?. But I didn't find any distinction between the two types of kinetic energy discussed in the question and hence the reason for the choice of average kinetic energy over the total translational kinetic energy.

Initial part of the question is based on the chapter "Kinetic Theory of Gases" from the book "Concepts of Physics" by Dr. H.C.Verma.

  • 1
    $\begingroup$ Please not that they do have to have from thermodynamics the same average kinetic energy in an ideal gas, but their average velocity will be different, for different mass molecules. $\endgroup$
    – anna v
    Commented Apr 1, 2020 at 12:30
  • $\begingroup$ Did you ever find a satisfactory answer to this? $\endgroup$ Commented Oct 6, 2020 at 9:39
  • $\begingroup$ @Buraian: Unfortunately, no. That was why I haven't accepted any of the answers. The already existing answers might solve the problem as they have positive scores. If that's the case, I just need to read them again carefully and accept the best one. $\endgroup$
    – Vishnu
    Commented Oct 6, 2020 at 10:00
  • $\begingroup$ I thought of your question a bit, and I concluded that even if average if s equal then it need not be that the total is equal. The average of $2$ and $ 8$ is 5 but so is the average of $0$ and $ 10$. The relation only holds for molecular level $\endgroup$ Commented Oct 6, 2020 at 10:08
  • $\begingroup$ But, why it holds at a molecular level, I don't know.. I think you should edit your question to explicitly state that you are confused about the molecular picture. I think what I said now resolves the confusion between why molecular is taken over bulk picture $\endgroup$ Commented Oct 6, 2020 at 10:09

5 Answers 5


The thermoynamic equilibrium in a gas (here a monoatomic perfect gas as you consider only translation and ignore interactions) , as long as the temperature is fixed, is characterized by the Boltzmann distribution of velocities, stating that the density of probability in the space of velocities is proportional to $\exp(-E/k_BT)$ (where $k_B$ is the Boltzmann constant), and here $E$ is simply the translational kinetic energy $E=\frac12 m \vec{v}^2$. An important consequence is the so called "equipartition of energy" :

$$ \langle E\rangle \equiv \frac12 m \langle \vec{v}^2\rangle \equiv \frac12 m v_{rms}^2 = \frac32 k_B T.$$

where the 3 comes from the 3 directions of the space.

This holds independently of the mass $m$ of the gas constituants, and is still true for a diluted diatomic gas like in air. Hence, one has actually:

$$\frac12 m_1 v_{rms,1}^2 = \frac12 m_2 v_{rms,2}^2=\frac32 k_B T,$$

and the relation for the temperature that you give is the "thermodynamic definition of temperature". It is also the fundamental principle of the "gas themometer" used by metrologist for low temperature measurements.

Furthermore, for a perfect gas, and for a given pression and temperature, the number $N$ of molecules depends only on the volume $V$ and not on the molecular weight, and the total translational kinetic energy for this volume is the same for any gas.

Nevertheless, this does not imply that the total kinetic energy is the same, as for poly atomic gas, you have to consider not only the motion of the barycenter (as we have done above) but also the kinetic energy associated to the relative motion of the atoms in the molecules, making the laws more complex.


The molecules in a gas are in constant collision. Conservation of energy means that this will randomise the distribution about the mean energy level, which must therefore be the same for both gases. If this is not sufficient argument, the zeroth law of thermodynamics will ensure that the gases are in thermal equilibrium.


It would not make sense to say that the total energy has to be the same for different gases. If that were true, I could say that the total energy of 1 kg of air should be the same as the total energy of, say, 5000 kg of water vapour, which is of course absurd. Total energy is an extensive quantity: it depends on the system size. Average energy, on the other hand, is an intensive quantity, so it makes sense to compare it between systems of different size.


From kinetic theory of gases, we know that,

Pressure $(P)=\frac{1}{3}\frac{nM}{V}v^2$ , $ \space\space\space\space\space -(1)$


$n=$ number of moles, $M=$ molar mass, $V=$ volume, $v=$ root-mean-squared velocity.

And, Kinetic Energy $(K) = \frac{1}{2}mv^2 = \frac{1}{2}(nM)v^2 =\frac{3}{2}PV $, (from $(1)$).$\space\space\space\space\space -(2)$

But, we define absolute temperature scale as,

$\frac{P}{P_{tp}}=\frac{T}{273.16 K}$, where $T=$ temperature.

From here, we get, $P=P_{tp} \space (\frac{T}{273.16 \space K})$

Using $(2)$, we get,

$K=\frac{3}{2}[P_{tp} \space (\frac{T}{273.16 \space K})]\space V$

Now, $P_{tp}=$ constant, $\Rightarrow$ if $V=$ constant, $K\propto T$.

Hence, we arrive at the required expression.

You may think of it in the way that increasing temperature implies that you are supplying heat energy to the substance. As a result, for a fixed a volume, $T \equiv K_{avg}$. Hope this clears your doubt!


What you describe is the difference between temperature and heat. If you declare that two gases have the same total kinetic energy, this implies that they have the same heat. The mass of the molecule is related to the heat capacity (for ideal gases).

Edit/answer to comment: this is mainly a mistranslation/everyday use of the word heat in my mothertongue, which probably doesn't work in english. More precisely, the total kinetic energy equals the total heat the gas could theoretically transfer to a 0K object with infinit heat capacity, assuming no phase transitions or volume change. I think this is helpfull nevertheless.

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    $\begingroup$ It sounds like you are using an unconventional definition of heat. What is your definition? $\endgroup$
    – garyp
    Commented Apr 1, 2020 at 11:39
  • 1
    $\begingroup$ In this context heat is a non scientific term as it could be translated in various thermodynamics potentials, including intenal energy I or enthalpy H, or their free variants (when temperature is constant) F an G. In any case the heat is not really defined by the state, but only for a given transformation. $\endgroup$
    – Jhor
    Commented Apr 1, 2020 at 13:28

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