Animated Solution for Physics - Thermodynamics: Three perfect gases at absolute temperatures $T_1,T_2$ and $T_3$ are mixed. The masses of molecules are $m_1,m_2$ and $m_3$ and the number of molecules are $n_1,n_2$ and $n_3$ respectively. Assuming no loss of energy, the final temperature of the mixture is

Mixing Perfect Gases

Imagine you are in a laboratory with three separate, perfectly insulated containers. Each container holds a different "perfect gas" at its own unique absolute temperature ($T_1,T_2,T_3$) and contains a specific number of molecules ($n_1,n_2,n_3$). The problem also hands you the masses of the individual molecules ($m_1,m_2,m_3$).

Now, you open the valves and let all three gases mix together into one large, insulated chamber. The question asks for the final equilibrium temperature of this mixture.

Why Mass Doesn't Matter

Before we dive into the math, let's address the elephant in the room: the masses $m_1,m_2$, and $m_3$. Why did the examiner give them to us?

The internal energy of an ideal (or perfect) gas is purely a function of its temperature and the number of particles it contains. It is a measure of the average kinetic energy per molecule. Whether a molecule is light like hydrogen or heavy like xenon, at the same temperature, they possess the exact same average translational kinetic energy. Therefore, the mass of the individual molecules is completely irrelevant to finding the final temperature. We can safely ignore $m_1,m_2$, and $m_3$.

Conservation of Energy

The phrase "assuming no loss of energy" is our golden ticket. It tells us that the system is thermally isolated. No heat escapes to the surroundings, and no work is done. According to the First Law of Thermodynamics, the total internal energy of the system must remain constant.

Mathematically, the sum of the initial internal energies of the three separate gases must equal the total internal energy of the final mixture:

Let's recall the formula for the internal energy of a gas. For a gas with $n$ molecules at an absolute temperature $T$, the internal energy $U$ is given by:

Here, $f$ represents the degrees of freedom of the gas molecules, and $k$ is the Boltzmann constant. Since the problem refers to them simply as "perfect gases" without specifying their atomicity, we assume they are of the same type and thus share the same degree of freedom, $f$.

Let's substitute this formula into our conservation of energy equation:

The total number of molecules in the mixture is simply the sum of the molecules from each gas: $n_{\text{total}}=n_1+n_2+n_3$. Substituting this in, we get:

A Weighted Average

Notice the beauty of this equation. The term $\frac{f}{2}k$ appears in every single term on both the left and right sides. We can elegantly divide the entire equation by $\frac{f}{2}k$, completely eliminating the degrees of freedom and the Boltzmann constant from our calculation:

Now, all that is left is to isolate our target variable, the final temperature $T$:

This result is incredibly intuitive. The final temperature is simply a weighted average of the initial temperatures, where the "weight" is the number of molecules of each gas. A gas with a massive number of molecules will have a much stronger influence on the final temperature than a gas with only a few molecules.

Different Atomicities

What if the gases were not identical? Suppose Gas 1 was monatomic ($f_1=3$) and Gas 2 was diatomic ($f_2=5$). In that scenario, we could no longer cancel out the $f$ terms. The conservation of energy equation would look like this:

Solving for the final temperature in such cases requires calculating the effective degree of freedom for the mixture, which adds a wonderful layer of complexity to the problem. Always ensure you check whether the gases share the same atomicity before blindly canceling terms!