Let me show you my thought process and let's simplify it to a single atom instead of 1 mole. $k_B$ is double the amount of energy per one degree of freedom of a particular gas per atom (or $\tfrac23$ of heat capacity of a single atom) required to heat it up by 1 K. And let's call heat capacity of a single atom for isochoric process $\gamma_V$ and for isobaric process $\gamma_p$. Then \begin{equation} \gamma_V = \frac{\nu}{2}k_B \end{equation} where $\nu$ is the number of degrees of freedom. If we want to find out the heat capacity of a single atom for isobaric process, it's \begin{equation} \gamma_p = \gamma_V + \frac{W}{N} \end{equation} where $\frac{W}{N}$ is the amount of work done per atom due to isobaric expansion. Work done by ideal gas is done by transforming heat into kinetic energy via elastic collision and can be done only by the velocity component that is perpendicular to the surface it hits, which means only one degree of freedom is going to "transfer" the heat into work by collision. And if the definition of $k_B$ I provided is true, that would mean \begin{equation} \frac{W}{N} = \tfrac{1}{2}k_B \end{equation} or, for one mole \begin{equation} \frac{W}{n} = \frac{1}{2}R \end{equation} Substituting it into the equation for heat capacity of isobaric process (the second one) yields \begin{equation} \gamma_p = \gamma_V + \frac12 k_B \end{equation} or, for one mole \begin{equation} c_p = c_V + \frac12 R \end{equation}

I am aware of how Mayer's relation is derived, but I still don't know what's wrong with my thought process here. It's probably that $\frac{W}{n} = R$, but why?

  • 1
    $\begingroup$ The argument is a mess of mismatched units. (The gas constant is an energy per mole? Work is added to a specific heat capacity?) Please consider editing to address this problems, as the question currently can't be understood if the premises are wrong. $\endgroup$ Mar 15, 2023 at 20:22
  • $\begingroup$ @Chemomechanics Sorry for the inconvenience. Hopefully it makes sense now, except for the part where my incorrect reasoning or premise leads to the incorrect factor of $\frac12$, which is what I am trying to uncover. $\endgroup$
    – Henry05
    Mar 16, 2023 at 10:03
  • $\begingroup$ Many of the equations are still dimensionally inconsistent. $\endgroup$ Mar 16, 2023 at 14:55
  • $\begingroup$ I've found this question tantalising. I've completely rewritten my answer. $\endgroup$ Mar 18, 2023 at 19:55
  • $\begingroup$ ... And have now added another paragraph. $\endgroup$ Mar 20, 2023 at 6:50

1 Answer 1


Your $\gamma_v$ takes account of the way the molecules store random energy: translation, rotation (vibration). On the other hand $\gamma_p-\gamma_v$ tells you the extra heat you have to put in, per molecule, per kelvin rise at constant pressure. That extra heat doesn't increase the molecules' random energy, but enables the gas to do work. $$\text{Work increment}=p\Delta V= p\left(\frac{Nk_B\Delta T}p\right)=Nk_B\Delta T$$ So extra heat input, per molecule, per kelvin rise = $k_B$.

This follows inescapably from the ideal gas equation, $pV=Nk_B T$, irrespective of any consideration of degrees of freedom.

However, we can bring in the kinetic theory pressure formula, $pV=\frac 13 Nm\overline{c^2}$ to check that $\frac 12 m\overline{c^2}=\frac 32 k_BT$. So we have...

$\gamma_v$ (your notation)$= \tfrac 12 k_B$ per degree of freedom, per kelvin rise.

You are concerned that $\gamma_P-\gamma_V$ is $k_B$ rather than $k_B/2$, because you consider gas pressure (on which the work calculation relies) to be due only to molecular motion in a direction normal to a wall. But this 'one-dimensionality' is taken account of by the factor of $\frac 13$ in the formula $pV=\frac 13 Nm\overline{c^2}$. And this equation, as we've seen is consistent with a work increment of $2k_B$ per molecule per kelvin and an internal energy increment of $k_B$ per kelvin per degree of freedom.

I suspect that your uneasiness over the $2k_B$ is because you have not applied the 'degrees of freedom' criterion rigorously enough. The work done per kelvin may indeed have a one-dimensionality about it, but that doesn't put it on a level with internal energy due to molecular motion in one direction. It isn't a degree of freedom.

Maybe it will help to point out that the gas pressure (and hence the work done by the expanding gas) is due to the change in momentum of molecules hitting the moving wall, and the total change in momentum per unit time of the stream of molecules is twice the momentum per unit time of the stream approaching the wall.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.