# Relationship between energy and temperature in ideal gas

I have encountered this old question and since I was bored I figured I use statisical mechanics to derive the relationship $E=\frac{3}{2}kT$ for a single particle in an ideal gas without using outside help. As a disclaimer, it has been a while since I had statistical mechanics courses, so I am a bit rusty. I have encountered some strange results I cannot explain and am curious about.

The problem: Using microcanonical ensemble for a single particle gives $E=(1/2)kT$. Using canonical ensemble for a single particle gives $E=(3/2)kT$.

Using canonical enseble for $N$ particles, the energy per particle $E/N$ is $E/N=(3/2)kT$. Using microcanonical ensemble for $N$ particles, the energy per particles is $E/N=(3/2-1/N)kT$. Clearly the latter goes to $(3/2)kT$ in the $N\rightarrow\infty$ limit.

Questions:

• An ideal gas is noninteracting, so there should be no difference between using a single-particle ensemble and a multiparticle ensemble, right? Why is it that the canonical ensemble gives the same results in both cases, but the microcanonical is different?

• Why do the microcanonical and canonical ensembles give different results, aside from the $N\rightarrow\infty$ limit? In particular, the microcanonical ensemble, while probably the furtherst removed from experiment, is the "purest" ensemble from a theoretical standpoint, the one that comes directly from the postulates of statistical mechanics. Is the formula $E=(3/2)kT$ valid only when $T$ is fixed? Or what?

The calculations I have used are presented below:

Ideal gas using microcanonical ensemble:

Assume we have an ideal gas in a box of volume $V$, at energy $E$ (at an uncertainty $(E,E+\delta E)$), and since the particles of an ideal gas are noninteracting anyways, we consider $N=1$.

The number of microstates is taken to be the phase volume between $E$ and $E+\delta E$, where $\delta E$ is a first-order "infinitesimal".

This can be calculated by calculating $\Omega(E'<E)$ the number of microstates whose energy is less than $E$. This is the following phase volume: $$\Omega(E'<E)=\int_Vd^3x\int_Bd^3p,$$ where $B$ is the ball in momentum space given by $B=\{(p_x,p_y,p_z):\ p^2<2mE\}$, so $$\Omega(E'<E)=Vg(2mE)^{3/2},$$ where $g$ is an irrelevant constant. The number of microstates (of energy $E$) is then $$\Omega(E,E+\delta E)=\frac{\partial}{\partial E}\Omega(E'<E)\delta E=\frac{3}{2}gV(2mE)^{1/2}2m\delta E=3gVm(2mE)^{1/2}\delta E.$$

The entropy is $S=k\ln\Omega$ so $$S=k\left[\ln(3gVm\delta E)+\frac{1}{2}\ln(2m)+\frac{1}{2}\ln E\right],$$ the reciprocal of temperature is then $$\frac{1}{T}=\frac{\partial S}{\partial E}=\frac{k}{2}\frac{1}{E},$$ from which we get $$E=\frac{1}{2}kT.$$

This is missing a factor of 3. Using a canonical ensemble on the other hand gives the proper answer:

Ideal gas using canonical ensemble:

We now consider the ideal gas fixed at temperature $T$, at volume $V$ and at particle number $N=1$.

The canonical partition function is $$Z=V\left(\frac{2m\pi}{\beta}\right)^{3/2},$$ the expectation value of energy is $$E=-\frac{\partial}{\partial\beta}\ln Z=\frac{\partial}{\partial\beta}\frac{3}{2}\ln\beta=\frac{3}{2}\frac{1}{\beta}=\frac{3}{2}kT,$$ which is the result I expect.

Now, I have no idea why the microcanonical ensemble gives only the third of the expected result but I considered using an $N$-particle gas instead of a single particle gas. The canonical ensemble gave $E=(3N/2)kT$, so the energy of a single particle is $E/N=(3/2)kT$, the same as before, so I will not do this calculation here. The microcanonical ensemble was different however.

$N$-particle ideal gas using microcanonical ensemble:

The number of microstates whose energy is less than $E$ is now given as $$\Omega(E'<E)=V^Ng(2mE)^{3N/2},$$ where $g$ is a different constant, but still irrelevant. The number of microstates (of energy $E$) is then $$\Omega(E,E+\delta E)=\frac{\partial\Omega(E'<E)}{\partial E}\delta E=3NmgV^N(2mE)^{(3N/2)-1}\delta E,$$ the entropy is $$S=k\ln\Omega=k\left[\left(\frac{3N}{2}-1\right)\ln(2mE)+c\right],$$ the reciprocal temperature is $$\frac{1}{T}=\frac{\partial S}{\partial E}=k\left(\frac{3N}{2}-1\right)\frac{1}{E},$$ $$E=\left(\frac{3N}{2}-1\right)kT.$$ The energy for a single particle is then $$\frac{E}{N}=\left(\frac{3}{2}-\frac{1}{N}\right)kT.$$ In the limit $N\rightarrow\infty$, this reduces to $E=(3/2)kT$.

• The $3/2$ and $1/2$ derive from the degrees of freedom of the system, do they not? It is perfectly okay to have $E = 1/2 kT$ for a 1D ideal gas. It's equivalent to saying that each DoF has $1/2 kT$. – honeste_vivere Jan 9 '17 at 20:34
• @honeste_vivere That's true but the specification of the system was the same for both ensembles, so they should yield the same answer. More precisely, I used three dimensions for both calculations. – Bence Racskó Jan 9 '17 at 20:40