The Maxwell distribution can be written in terms of translational energies $\epsilon_{tr}=\frac{1}{2}mv^2$, solving for v, $v=\left(\frac{2\epsilon_{tr}} {m}\right)^{1/2}$, differentiating: $$dv=\left(\frac{2}{m}\right)^{1/2}\frac{1}{2 }\epsilon_{tr}^{-1/2}d\epsilon_{tr}=\left(\frac{1}{2m}\right)^{1/2}\frac{d\epsilon_{tr}} {\epsilon_{tr}^{1/2}}$$ Substituting vy dv into $\frac{dN_v}{N}=\left(\frac{m}{2\pi mkT}\right)^ {3/2}e{\frac{-mv^2}{2kT}}4\pi v^2dv$ $$\frac{dN_{\epsilon_{tr}}}{N}=\left( \frac{m}{2\pi mkT}\right)^{3/2}e^{\frac{-m(2\epsilon_{tr}/m)}{2kT}}4\pi \left(\ frac{2\epsilon_{tr}}{m}\right)\left(\frac{1}{2m}\right)^{1/2}\frac{d\epsilon_{tr}}{\epsilon_{tr }^{1/2}}$$ Simplifying: $$\frac{dN_{\epsilon_{tr}}}{N}=2\pi \left(\frac{1}{\pi kT}\right)^{3/2}e^{\frac{-\epsilon_{tr}}{kT}\epsilon_{tr}^{1/2}}d\epsilon_{tr}$$ $dN_{\epsilon_{tr}}$ represents the number of molecules whose translational energy is comp rendered between $\epsilon_{tr}$ and $\epsilon_{tr}+d\epsilon_{tr}$

The mean translational energy is the same for all gases at the same temperature.