Ultra-relativistic gas
Nice question. Sometimes we get used to a certain fact, such as equipartition with $(1/2)kT$ per degree of freedom, that we forget that it's not always true, or what assumptions are required in order to make it true. I had to refresh my memory on how equipartition works.
Basically the $(1/2)kT$ form of the equipartition theorem is a special case that only works if the energy consists of terms that are proportional to the squares of the coordinates and momenta. The 1/2 comes from the exponent in these squares.
The WP article on equipartition has a discussion of this. There is a general equipartition theorem that says that
$$\langle x \frac{\partial E}{\partial x} \rangle = kT,$$
where $x$ could be either a coordinate or a conjugate momentum. If $E$ has a term proportional to $x^m$, the partial derivative has a factor of $m$ in it. In the ultrarelativistic case, where $E\propto\sqrt{p_x^2+p_y^2+p_z^2}$, you don't actually have a dependence on the momenta (momentum components) that breaks down into terms proportional to a power of each momentum. However, I think it's pretty easy to see why we end up with the result we do, because in one dimension, we have $|\textbf{p}|=|p_x|$, which does have the right form, with an exponent of 1.
First note that the different energy relationship doesn't change the equation of state for the gas. For a classical gas with no interactions, the Hamiltonian doesn't depend on the position, so we can immediately see that the partition function $Z\sim V^N$ and therefore $$p = \frac{\partial}{\partial V}(kT\log{Z})=\frac{NkT}{V}$$ So an ultra-relativistic gas behaves just like an ideal gas for many purposes. The difference in energy can be interpreted in terms of degrees of freedom. The $N$ atoms are in both cases free (no interactions) and carry no rotational or vibrational modes. Equipartition of energy then tells us that for the classical ideal gas each mode carries an energy $\frac{1}{2}kT$, while in the ultra-relativistic case the energy is twice as big, $kT$. This is very well explained on this Wikipedia page. The reason for the difference is essentially that $H\sim p^2$ in the ideal case, while $H\sim p$ in the ultra-relativistic case.
As to the physical consequences, the equation for the energy tells you exactly what you need: at the same temperature, an ultra-relativistic gas will have twice as much energy than an ideal gas. The reason why it's exactly two is explained by the equipartition theorem.