The heat capacity ratio is defined as the ratio of the heat capacities at constant pressure and volume respectively:

$$\gamma = \frac {C_p}{C_V}\tag{1}$$

Then obviously

$$\gamma = \frac{n\,C_p\,\Delta T}{n\,C_V\,\Delta T} = \frac{\Delta Q}{\Delta U}\tag{2}$$

This means that $1/\gamma$ is the fraction of the heat $(\Delta Q)$ given or taken from the gas, which is spent in increasing or decreasing the internal energy $(\Delta U)$ of the gas, respectively.

I saw online that the value of $\gamma$ was the same for every monoatomic gas $(5/3).$ The same goes for diatomic $(7/5)$ and triatomic $(4/3).$ Why is it that this ratio is the same for these gases?

I think it has something to do with how only a certain fraction of the heat given or taken is contributed to by the change in internal energy, but I can't exactly find the right reason.

Another way this question can be asked is: why is the fraction of the heat given or taken that goes into the change in internal energy same for molecules with the same atomicity?

  • 2
    $\begingroup$ See also Heat capacity of gases $\endgroup$
    – Poutnik
    Commented May 21, 2022 at 5:17
  • $\begingroup$ Note that those values of $\gamma$ are only approximations, theoretical predictions from statistic mechanics. Two gases with the same number of atoms can have significantly different heat capacities in reality. $\endgroup$ Commented May 21, 2022 at 6:31
  • $\begingroup$ So, for two real gases with the same atomicity, since there is a presence of potential energy, which may differ with change in number of electrons or protons, the value of internal energy may not be the same and thus the heat capacity ratio. Did I understand correctly, @electronpusher? $\endgroup$ Commented May 21, 2022 at 13:24
  • 1
    $\begingroup$ It is rather due different quantization steps of vibration energies due different atomic masses. They lead to different temperature ranges of progressive involving of vibration contribution to heat capacity. The same for rotation contribution for very low temperatures ( see also the page bottom on the referred hyperphysics link). $\endgroup$
    – Poutnik
    Commented May 21, 2022 at 14:52

1 Answer 1


If we deal with an ideal gas then $C_p=C_V+R $ then $\gamma=1+R/C_V$ which then begs the question if $C_V$ is different for different molecules why is $\gamma$ constant and the same for molecules with the same number of atoms.

The answer is that for this to be true the temperature has to be sufficiently high that the slope of the internal energy with temperature, which is $C_V$ has become constant, i.e. $C_V=\left(\frac{dU}{dT}\right)_V$. As energy levels are discrete for $C_V$ to be constant with temperature energy levels have to be populated at a constant rate, i.e. the slope of $U$ vs. $T$ is constant. Only then will $\gamma$ be the same for molecules with equal number of atoms. (In practice it may not always be possible to reach this temperature as the molecule may decompose first.)

At v. low temperatures as the average energy can be less than a quantum then $C_V\to 0$ but as $T$ increases, by the Boltzmann distribution the average internal energy $U$ increases as more rotational and vibrational levels are populated and so $C_V$ increases. Eventually $C_V$ becomes constant but for many molecule this may be far above room temperature. So this means that experimentally $\gamma$ is usually greater than predicted from classical thermodynamics, but instead statistical mechanics has to be used. For Nitrogen, which has widely spaced quanta, at room temperature $\gamma \approx 1.4$ but classical prediction is 1.29, but for iodine with far more closely spaced levels the classical value is predicted.

In the classical limit (i.e. many levels populated) the heat capacity has a value $R/2$ for each degree of freedom, so is $C_V=3R/2$ for (x,y,z) translation and so $\gamma=1+2/3=1.667$ for atoms, and for a diatomic $C_V(trans) = 3R/2$, $C_V(vib) = RT$, $ C_V(rotation)=RT$ making $\gamma =1.29$. You can extend this for $n$ vibrations where $C_V=(3+n)RT$. (Many Phys. Chem. textbook will give a fuller explanation and detailed calculations.)


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