Three closed vessels A, B and C at the same temperature T and contain gases which obey the Maxwellia

Q: Doesn't the presence of $N_2$ molecules slow down the $O_2$ molecules due to collisions?

In an ideal gas, collisions are perfectly elastic. While individual molecules change speeds during collisions, the overall statistical distribution of speeds for each type of gas remains constant and depends only on temperature and its own molar mass.

Q: Does the total mass of the mixture affect the average speed?

No, the average speed of a specific gas in a mixture depends only on the molar mass of *that specific gas*, not the average molar mass of the mixture.

Q: What if the question asked for the average speed of the *mixture* in Vessel C?

That would be a different calculation! You would need to find the average molar mass of the mixture and use that in the formula, or calculate the weighted average of the individual speeds.

Q: Why is the formula $\sqrt{8RT/\pi M}$ and not $\sqrt{3RT/M}$?

$\sqrt{8RT/\pi M}$ is the formula for the *average* speed, while $\sqrt{3RT/M}$ is the formula for the *root mean square* (RMS) speed. The question specifically asks for the average speed.

Q: Would the answer change if the quantities of $O_2$ and $N_2$ were not equal?

Not at all! The velocity distribution of $O_2$ is completely independent of the partial pressure or number of moles of $N_2$ present, as long as the temperature is constant.

Question

Three closed vessels $A$, $B$ and $C$ at the same temperature $T$ and contain gases which obey the Maxwellian distribution of velocities. Vessel $A$ contains only $	ext{O}_2$, $B$ only $	ext{N}_2$ and $C$ a mixture of equal quantities of $	ext{O}_2$ and $	ext{N}_2$. If the average speed of the $	ext{O}_2$ molecules in vessel $A$ is $v_1$, that of the $	ext{N}_2$ molecules in vessel $B$ is $v_2$, the average speed of the $	ext{O}_2$ molecules in vessel $C$ is
(where, $M$ is the mass of an oxygen molecule)

Options:

(A) $\frac{v_1 + v_2}{2}$

(B) $v_1$

(C) $(v_1 v_2)^{1/2}$

(D) $\sqrt{3kT/M}$

Sigma Percentile · Accepted Answer

The behavior of gases in a mixture is one of the most elegant concepts in kinetic theory. Imagine a crowded room where people are walking around. Some are moving fast, some are moving slow. If you introduce a new group of people into the room, does the original group suddenly change their natural walking pace? In the world of ideal gases, the answer is a resounding no.

Let's dive into the physics behind this beautiful independence.

### The Maxwellian Distribution

The problem states that the gases obey the **Maxwellian distribution of velocities**. According to this statistical model, the speeds of molecules in an ideal gas are not uniform; they follow a specific probability distribution. However, we can define an **average speed** for the molecules, which is given by the formula:

$$v_{	ext{avg}} = \sqrt{\frac{8kT}{\pi M}}$$

Here, $k$ is the Boltzmann constant, $T$ is the absolute temperature, and $M$ is the mass of a single molecule of the gas.

Take a close look at this equation. What does it tell us? It reveals that the average speed of a gas molecule depends on exactly two physical quantities:
1. The temperature of the gas ($T$).
2. The mass of the specific molecule ($M$).

It does **not** depend on the pressure, the volume, or the presence of any other types of molecules in the container.

### Analyzing the Vessels

Let's apply this logic to our specific setup.

In **Vessel A**, we have pure oxygen ($	ext{O}_2$) at a temperature $T$. The average speed of these oxygen molecules is given as $v_1$. Mathematically, this is:

$$v_1 = \sqrt{\frac{8kT}{\pi M_{	ext{O}_2}}}$$

Now, let's look at **Vessel C**. This vessel contains a mixture of oxygen ($	ext{O}_2$) and nitrogen ($	ext{N}_2$). Crucially, Vessel C is maintained at the exact same temperature $T$ as Vessel A.

The question asks for the average speed of the **oxygen molecules** in Vessel C. Let's call this $v_C$. Using our master equation for the oxygen molecules in this mixture:

$$v_C = \sqrt{\frac{8kT}{\pi M_{	ext{O}_2}}}$$

### The Grand Conclusion

Comparing the two expressions, it is mathematically obvious that:

$$v_C = v_1$$

The presence of nitrogen molecules in Vessel C is completely irrelevant to the speed distribution of the oxygen molecules. In an ideal gas mixture, the molecules are assumed to be point masses that do not exert intermolecular forces on each other (other than perfectly elastic collisions). Therefore, each gas component behaves as if it were alone in the container, maintaining its own independent Maxwellian velocity distribution.

This principle is deeply connected to Dalton's Law of Partial Pressures, which states that each gas in a mixture exerts a pressure independent of the others. Similarly, each gas maintains a kinetic energy and speed distribution independent of the others.

So, the average speed of the $	ext{O}_2$ molecules in Vessel C remains exactly $v_1$.

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Visualized Solution

$Visualizing the Setup$

$Average Speed Formula$

$Analyzing Vessel A$

$Analyzing Vessel C$

$Comparing Speeds$

$Conclusion$

$The Way Forward$

The Sigma Insight: Kinetic Theory of Gases

The Maxwellian Distribution

Analyzing the Vessels

The Grand Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Average Speed Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Vessel A

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing Vessel C

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Speeds

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward