Two ideal polyatomic gases at temperatures T1 and T2 are mixed so that there is no loss of energy. I

Q: Why do we use $k_B$ instead of $R$ in the internal energy formula?

Because the problem specifies $n_1$ and $n_2$ as the 'number of molecules', not the number of moles. The Boltzmann constant $k_B$ is used for individual molecules, while the universal gas constant $R$ is used for moles. However, since the constant cancels out entirely, using $R$ would yield the exact same final expression.

Q: Does the mass of the gas molecules matter for the final temperature?

No, the mass of the individual molecules does not appear in the final equation. The internal energy of an ideal gas depends only on its degrees of freedom and its absolute temperature, not on the mass of its constituent particles.

Q: What if the gases react chemically when mixed?

This problem strictly assumes that the gases are ideal and non-reacting. If a chemical reaction occurred, chemical potential energy would be converted into thermal energy (or vice versa), and the simple $\Delta U = 0$ equation for thermal energy would no longer hold.

Q: Why is the final temperature not just the simple average of $T_1$ and $T_2$?

Because different gases have different heat capacities. A gas with more molecules or more degrees of freedom requires more energy to change its temperature. Therefore, it acts as a 'thermal anchor' and pulls the final equilibrium temperature closer to its own initial temperature.

Q: What does 'no loss of energy' imply physically?

It implies that the container holding the gases is perfectly insulated from the outside environment. No heat escapes the container, and no work is done on or by the surroundings. Therefore, the total internal energy of the gas mixture remains perfectly constant.

Question

Two ideal polyatomic gases at temperatures $T_1$ and $T_2$ are mixed so that there is no loss of energy. If $f_1$ and $f_2$, $m_1$ and $m_2$, $n_1$ and $n_2$ be the degrees of freedom, masses, number of molecules of the first and second gas respectively, the temperature of mixture of these two gases is

Options:

(A) $\frac{n_1T_1 + n_2T_2}{n_1 + n_2}$

(B) $\frac{n_1f_1T_1 + n_2f_2T_2}{n_1f_1 + n_2f_2}$

(C) $\frac{n_1f_1T_1 + n_2f_2T_2}{f_1 + f_2}$

(D) $\frac{n_1f_1T_1 + n_2f_2T_2}{n_1 + n_2}$

Sigma Percentile · Accepted Answer

Have you ever wondered what happens when two completely different gases, minding their own business at different temperatures, are suddenly forced to share the same space? It’s a chaotic microscopic dance! Molecules of all shapes and sizes colliding, transferring energy, and eventually settling down into a harmonious equilibrium. This problem is a classic exploration of that exact scenario. It’s not just about plugging numbers into a formula; it’s about understanding the fundamental laws that govern the universe—specifically, the conservation of energy.

### The Art of Mixing Gases
Imagine you are standing in a lab with two sealed containers. In the first container, you have a gas with $n_1$ molecules. These aren't just any molecules; they have a specific complexity, denoted by their degrees of freedom, $f_1$. They are buzzing around at a temperature $T_1$. In the second container, you have a different gas with $n_2$ molecules, $f_2$ degrees of freedom, and a temperature $T_2$.

When we connect these two containers and let the gases mix, a thermal symphony begins. The hotter gas will transfer kinetic energy to the colder gas through billions of microscopic collisions. But here is the beautiful part: the problem states that there is **no loss of energy**. The containers are perfectly insulated. The universe outside these containers doesn't get a single joule of this energy.

### The Principle of Energy Conservation
Because the system is perfectly isolated, the First Law of Thermodynamics tells us that the total internal energy must remain constant. The energy before the mixing must exactly equal the energy after the mixing.

Mathematically, we write this as:
$$U_{	ext{initial}} = U_{	ext{final}}$$

This simple equation is the anchor for our entire solution. No matter how complex the gases are, they must obey this rule.

### Decoding Internal Energy
Before we can use our conservation equation, we need to know how to calculate the internal energy of a gas. The internal energy $U$ of an ideal gas is directly tied to how many ways its molecules can move—its degrees of freedom $f$—and its absolute temperature $T$.

For a gas with $n$ molecules, the internal energy is given by the Equipartition Theorem:
$$U = \frac{f}{2} n k_B T$$
where $k_B$ is the Boltzmann constant.

Notice that we are using the number of molecules $n$ and the Boltzmann constant $k_B$. If the problem had given us the number of moles, we would have used the universal gas constant $R$. Both approaches are perfectly valid and lead to the exact same result, but it's crucial to match your constants to your variables!

### Setting Up the Master Equation
Let's calculate the total energy before we open the valve. It's simply the sum of the energies of the two isolated gases:
$$U_{	ext{initial}} = \frac{f_1}{2} n_1 k_B T_1 + \frac{f_2}{2} n_2 k_B T_2$$

Now, what happens after they mix? They eventually reach a thermal equilibrium—a single, common temperature $T$. The total energy of this mixture is the sum of the energies of the two gases at this new temperature:
$$U_{	ext{final}} = \frac{f_1}{2} n_1 k_B T + \frac{f_2}{2} n_2 k_B T$$

### The Elegance of Algebra
Now, we bring it all together by equating the initial and final states:
$$\frac{f_1}{2} n_1 k_B T_1 + \frac{f_2}{2} n_2 k_B T_2 = \frac{f_1}{2} n_1 k_B T + \frac{f_2}{2} n_2 k_B T$$

I know this equation looks a bit intimidating, but let's take a breath and look closer. Do you see the $\frac{k_B}{2}$ term? It's everywhere! It's in every single term on both the left and the right side of the equation. This means we can divide the entire equation by $\frac{k_B}{2}$ and watch it beautifully vanish:
$$f_1 n_1 T_1 + f_2 n_2 T_2 = f_1 n_1 T + f_2 n_2 T$$

Suddenly, the physics problem has transformed into a simple algebra problem. Our goal is to find the final temperature $T$. Let's factor $T$ out of the right side:
$$f_1 n_1 T_1 + f_2 n_2 T_2 = (f_1 n_1 + f_2 n_2) T$$

### Why Degrees of Freedom Matter
Let's pause for a moment and think about why the degrees of freedom $f$ appear in our final answer. Why isn't the final temperature just a simple average based on the number of molecules?

Imagine mixing a monatomic gas like Helium ($f=3$) with a complex polyatomic gas ($f=6$). The monatomic gas can only store energy by moving in straight lines (translation). The polyatomic gas, however, can translate, rotate, and vibrate. It has many more "pockets" to store energy.

When you add heat to the polyatomic gas, a lot of that energy goes into making the molecules spin and vibrate, rather than just making them fly faster. Since temperature is only a measure of the *translational* kinetic energy, a polyatomic gas requires much more total energy to raise its temperature by one degree compared to a monatomic gas. It has a higher heat capacity.

Therefore, when mixing these two gases, the polyatomic gas acts like a massive thermal anchor. It will stubbornly resist changing its temperature, pulling the final equilibrium temperature closer to its own initial temperature. This physical reality is beautifully captured in our final mathematical expression, where the temperature $T$ is weighted by the product $n \cdot f$. The math isn't just symbols; it's a perfect reflection of the physical world!

### The Final Takeaway
To isolate $T$, we simply divide both sides by the bracketed term $(f_1 n_1 + f_2 n_2)$:
$$T = \frac{n_1 f_1 T_1 + n_2 f_2 T_2}{n_1 f_1 + n_2 f_2}$$

And there we have it! The final equilibrium temperature is a weighted average of the initial temperatures. A gas with more molecules or more ways to store energy (higher $f$) will have a stronger influence on the final temperature. This perfectly matches option (b).

Select Answer:

Visualized Solution

The Mixing Setup

Conservation of Energy

Internal Energy Formula

Initial Internal Energy

Final Internal Energy

Equating Energies

Simplifying the Equation

Factoring the Final Temperature

Final Answer

The Sigma Insight: Kinetic Theory of Gases

The Art of Mixing Gases Imagine you are standing in a lab with two sealed containers

The Principle of Energy Conservation Because the system is perfectly isolated, the First Law of Thermodynamics tells us that the total internal energy must remain constant

Decoding Internal Energy Before we can use our conservation equation, we need to know how to calculate the internal energy of a gas

Setting Up the Master Equation Let's calculate the total energy before we open the valve

The Elegance of Algebra

Why Degrees of Freedom Matter Let's pause for a moment and think about why the degrees of freedom $f$ appear in our final answer

The Final Takeaway

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