Mixing 2 non-reacting gas of different temperature

Question

Can we theoritically calculate the final temperature of 2 non-chemically reacting gases(say A and B) by modelling their collision as elastic collision and finding the net root mean square velocity. I think the equations will be lengthy but what I wish to know is, " Do all the gas molecules randomly collide with each other and attain a net resultant velocity which we measure as changed temperature? If possible, please suggest just how one can do such calculations."

Answer 1

I'll assume that the gases are ideal and monatomic. By conservation of energy before and after mixing, \begin{equation} K_A + K_B = K_f, \end{equation} where \begin{equation} K_A = \frac{1}{2}m_A \sum_{i=1}^{N_A} v_{A,i}^2 \end{equation} is the kinetic energy of gas $A$ before mixing ($N_A$ is the number of particles of gas $A$), and similarly for gas $B$; $K_f$ is the final, total kinetic energy of the mixture.

Initially, the gases will have velocities given by the Maxwell-Boltzmann distribution, so that \begin{equation} K_A = \frac{1}{2}m_A \sum_{i=1}^{N_A} v_{A,i}^2 = \frac{1}{2}m_A N_A \left<v_A^2\right>, \end{equation} where the angle brackets denote the average according to the Maxwell-Boltzmann distribution at $T_A$, the initial temperature of gas $A$. The (non-normalized) Maxwell-Boltzmann (MB) weight is given by $\exp\left(-\frac{1}{2} \frac{m_A v_A^2}{kT}\right)$, such that when the average is taken (in 3D) with the proper normalization, \begin{equation} K_A = \frac{3}{2} N_A k T_A. \end{equation} All of this applies to $B$ as well.

Considering the elastic collision of a particle of gas $A$ colliding with one of gas $B$ (now in 1D to simplify), the change in velocity of a particle of $B$ is given by \begin{equation} \Delta v_B = \frac{2(v_{Ao}-v_{Bo})}{1+\frac{m_B}{m_A}}, \end{equation}
where the subscript $o$ indicates initial velocity before collision. From this, we can try to find some kind of average velocity exchange given that the initial velocities obeyed the MB distribution. The average of $\Delta v_{Bo}$ vanishes since the integrand is odd. So, we can take $\left<\left(\Delta v_{Bo}\right)^2\right>$, viewing this as a measure of kinetic energy exchange (upon multiplying by $m_B$). Doing this, we find that \begin{equation} \left<\left(\Delta v_{Bo}\right)^2\right> \propto \frac{T_A}{m_A}+\frac{T_B}{m_B}. \end{equation} The trouble is that this expression is strictly positive, and its overall sign is not determined by Newtonian mechanics. It's just as likely for the kinetic energy to be exchanged in one direction as it is the other---from hot to cold, or from cold to hot. To determine the evolution of a particular system with Newton's laws, we need to specify a set of initial conditions, which then fully determines the future system. This will provide only one possible region of phase space, whereas the equilibrium distribution will involve many more possible configurations.

If we assume, however, that the microscopic collisions will lead to a new MB distribution where both gases are at the same temperature, then the energy conservation condition as described previously allows determination of the final temperature through the average values of kinetic energy. This could be shown faster using the equipartition theorem. To my knowledge, this is as close to a microscopic description as can be hoped for. Without entropic considerations, equations of motion are not enough to predict thermal equilibrium.