# Why is the temperature of a gas proportional to the average kinetic energy of its particles?

I'm studying the kinetic theory of gases I managed to derive that pressure is inversely proportional to volume and directly proportional to the average kinetic energy of the gas particles. Similarly, volume is inversely proportional to pressure and also directly proportional to the average kinetic energy. But why does the temperature increase with the average kinetic energy? This is assumed as a postulate but I cannot understand why.

• You are confusing effect with cause. It's in reverse,- due to increased average kinetic energy of molecules,- gas temperature increases. I.E. temperature is sort of measure of how fast molecules are moving around. Commented Sep 8, 2020 at 18:15
• But why does it increase? Commented Sep 8, 2020 at 18:20
• Reasons are very different and depends on exact situation. For example temperature can increase by transferring heat to a gas,- in the form of hotter other gas convection or by means of passing radiation energy or simply compressing gas in smaller volume, so that mean free path of molecules decreases and due to that average speed increases. Commented Sep 8, 2020 at 18:48

It depends on what temperature is. Two systems are in thermal equilibrium when the fractional change of their multiplicities $$\Omega$$ with energy $$E$$, $$\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E}$$, are equal to each other. Let us call this quantity $$\beta$$.

For a classical ideal gas of $$N$$ independent particles the number of accessible states $$\Omega$$ is proportional to the surface of a hypersphere in a phase space with $$3N$$ dimensions. The radius of that sphere is proportional to the square root of the kinetic energy $$\sqrt{E}$$, so that $$\Omega(E) \propto E^\frac{3N-1}{2}.$$

This is enough to see that for the ideal classical gas $$\beta=\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E} = \frac{3N-1}{2} E^{-1}$$ which is equal to $$\frac{3N}{2} E^{-1}$$ because $$N$$ is on the order of Avogadro's number.

From kinetic theory, the product $$pV= \frac{2}{3}E$$.

Combining these two expressions we find the equation of state of the ideal gas $$\beta pV = N.$$

Comparing this with the empirical ideal gas law we see that $$\beta = \frac{1}{k_B T}.$$

• I don't understand many of the quantities you mentioned. Is there a simpler way? Commented Sep 8, 2020 at 23:37
• @TomAvery This was very condensed, but it is the simplest way. The more familiar way involves entropy $S = k_B \ln \Omega$ and its derivative with respect to energy, but it comes down to the same argument. Or there are derivations using the Zustandssumme which are more abstract.
– user137289
Commented Sep 9, 2020 at 7:58
• @TomAvery This answer also uses not more of what is in the Sackur-Tetrode equation than what is necessary to derive the general gas equation.
– user137289
Commented Sep 9, 2020 at 9:33
• Could you explain it with Newtonian mechanics? Commented Sep 9, 2020 at 9:45
• @TomAvery This is almost completely classical. The statistics with the concept of a countable multiplicity $\Omega$ is more intuitive in quantum mechanics, but it can also be phrased as the volume of phase space, similar to the derivation of the Maxwell-Boltzmann distribution of velocities (just in more dimensions).
– user137289
Commented Sep 9, 2020 at 10:09

To increase the temperture of a specific amount you need energy, this energie shows itself as kinetic energy , or how would you increase themperature