# Rigorous reason behind internal energy change being zero while mixing

A container is divided in two parts: one part contains oxygen gas $$(n_1$$ moles, at temperature $$T_1)$$ and the other part contains helium gas $$(n_2$$ moles, at temperature $$T_2).$$ The partition separating them is removed. Find the final temperature of the mixture $$(T).$$

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question. I realized the question has explicitly mentioned the gases, i.e. it gave us information about their degrees of freedom (since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $$C_V.$$ At the time, I hypothesized $$\Delta{U}_\mathrm{sys} = 0.$$ Thus:

$$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2) = 0$$ $$T = \frac{5n_1T_1 + 3n_2T_2}{5n_1 + 3n_2}$$

Which was indeed one of the options, and in fact, the correct one.

However, is there any rigorous method/reasoning that tells us why exactly is it that $$\Delta{U}_\mathrm{sys} = 0?$$ I can't think of a satisfactory explanation. I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

• Could it be similar to free expansion of gases? – Safdar Faisal Jul 6 '20 at 8:08

The process in question is presumably adiabatic, since we are told that the temperature changes and are not given any information about the temperature of the surroundings (adiabaticity can also sometimes be assumed if gases are mixed rapidly). Secondly, the final volume of the system (the entire container) is the same as the initial volume (and in addition we can assume that each gas expands without performing any work, i.e. the expansions can be assumed to be free). That means that neither heat nor work exchanged with the system during the process, which means that in accordance with the first law $$\Delta U = 0$$.
• It depends on how you define the system. Here the system is the entire container, so we are interested in the change of volume of the entire container. The container is assumed rigid so $\Delta V=0$, therefore no total work. – Buck Thorn Jul 6 '20 at 9:42
• I didn't mean that $\Delta U$ is not zero. It is zero,for both a real gas and for an ideal gas. What I meant was that, if the mixture of gases is not ideal, then the internal energy of the mixture is a function not only of the temperature, but also the pressure and the mole fractions of the components. – Chet Miller Jul 6 '20 at 19:10