Will a neutrino gas always stay at the same temperature? Does the neutrino gas pervading the universe always have the same temperature? Does it still have the temperature as the neutrino gas that emerged from the big bang?  Neutrinos don't interact, so how could the gas be cooled down (apart from space expansion)?
 A: In thermal equilibrium, if we assume that the entropy conserved by expanding the universe, we can obtain the temperature of the photon gas:
$$T_\gamma = T_\gamma^0 (\frac{4}{11})^{\frac{1}{3}}(1+z)$$
where $T_\gamma^0$ is the initial temperature of photon gas and z is the redshift. Therefore, over time, the gas temperature decreases.

Another hand we can show that the temperature of neutrino gas equal to:
$$T_\nu = (\frac{4}{11})^{\frac{1}{3}}T_\gamma$$
A: The neutrino background cools as the universe expands. The reason for this is the de Broglie relation which relates the de Broglie wavelength to the momentum $p = h/\lambda$. As the universe expands, so does the de Broglie wavelength and thus the momentum decreases as the inverse of the scale factor of the universe.
As the universe expands, the neutrinos have a Fermi-Dirac distribution that looks like
$$F(p,T) = \frac{1}{\exp(pc/kT) + 1}\ ,$$
and this relates temperature to momentum.
As the momentum decreases as $a^{-1}$, then this evolving distribution can be characterised as having a temperature that also scales as $a^{-1}$, or $(1+z)^{-1}$, where $z$ is redshift. This is exactly the same scaling as the temperature of the CMB and so one can calculate that the temperature of the CMB and the temperature of the C$\nu$B ought to have a fixed ratio of $T_\gamma/T_\nu \simeq 1.41$ and thus $T_\nu \simeq 1.95$ K at the present epoch.
However, in order to calculate other properties of the C$\nu$B such as the rms speed or average energy, then this "temperature" $T_\nu$ can only be safely used while the neutrinos are ultrarelativistic with $k_B T_{\nu}\gg m_\nu c^2$. Once cooling gets to $k_B T_\nu \sim m_\nu c^2$, which for a neutrino rest-mass energy of say 0.1 eV, means $T_\nu \sim 1000$ K, then the neutrinos become non-relativistic. What this means is that although the neutrino momentum distribution is essentially "frozen" in its relativistic form and scales with the universal expansion as explained above, the same is not true for the energy or speed distributions of the neutrinos.
i.e. Although you can assign a "temperature" to the C$\nu$B of $\sim 1.95$ K today, the speed and kinetic energy distributions are not characterised by distributions at those temperatures. Indeed, once the neutrinos become non-relativistic then their kinetic energies scale as $p^2/2m_\nu$ and thus as $a^{-2}$ or $(1+z)^{-2}$.
Thus it is important to be careful about what you mean by the "temperature" of the C$\nu$B. For example, you cannot use $T_\nu \sim 1.95$ K to calculate an rms speed for present-day neutrinos in the C$\nu$B as $\sqrt{3k_BT/m_\nu}$. e.g. Using $\sqrt{3k_BT_\nu/m_\nu}$ would give you an rms speed of 21000 km/s for $m_\nu c^2 = 0.1$ eV, whereas a correct calculation would give more like 1600 km/s - a significant difference!
For this reason, I think most cosmologists prefer to work with the energy density of the C$\nu$B, which although one still has to deal with the transition between relativistic and non-relativistic neutrino species, at least has a well-defined meaning.
