An ideal gas in a closed container is slowly heated. As its temperature increases, which of the foll

Q: Why doesn't the mean free path depend on temperature? I thought heating makes molecules expand.

Heating a gas makes it *want* to expand, which increases the pressure if the volume is fixed. However, the mean free path $\lambda = \frac{V}{\sqrt{2}\pi d^2 N}$ only depends on the physical volume $V$ and the number of molecules $N$. Since the container is rigid and closed, neither $V$ nor $N$ can change, so the average distance between molecules remains exactly the same.

Q: Doesn't the formula $\lambda = \frac{kT}{\sqrt{2}\pi d^2 p}$ show that mean free path depends on temperature $T$?

It looks like it does, but you have to be careful! In a closed container, as temperature $T$ increases, the pressure $p$ also increases proportionally (since $p = \frac{nRT}{V}$). The ratio $\frac{T}{p}$ remains perfectly constant. Therefore, $\lambda$ remains constant.

Q: What is the difference between mean free path and mean collision time?

Mean free path is a **distance**—it's the average length a molecule travels before hitting another. Mean collision time is a **duration**—it's the average time it takes to travel that distance. If you drive the same distance but at a higher speed, it takes less time. Similarly, heating the gas increases the speed, reducing the collision time.

Q: Why do we use average velocity instead of rms velocity for calculating collision time?

Strictly speaking, collision frequency calculations in kinetic theory are derived using the Maxwell-Boltzmann distribution of speeds, which naturally brings out the mean (average) relative speed between molecules, not the rms speed. However, since both $v_{\text{avg}}$ and $v_{\text{rms}}$ are proportional to $\sqrt{T}$, the proportionality $\tau \propto \frac{1}{\sqrt{T}}$ holds true regardless of which speed you use for a rough estimate.

Q: What would happen if the container was open instead of closed?

If the container was open to the atmosphere, the pressure $p$ would remain constant. As you heat the gas, it would expand, and molecules would leave the container (so $N$ decreases while $V$ is constant, or $V$ increases for a fixed $N$). In that case, using $\lambda \propto \frac{T}{p}$, since $p$ is constant, the mean free path $\lambda$ would actually increase!

Question

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true?
A. The mean free path of the molecules decreases.
B. The mean collision time between the molecules decreases.
C. The mean free path remains unchanged.
D. The mean collision time remains unchanged.

Options:

(A) B and C

(B) A and B

(C) C and D

(D) A and D

Sigma Percentile · Accepted Answer

### The Dance of Gas Molecules: Heating a Closed Container

Imagine you are observing a microscopic world inside a rigid, closed container filled with an ideal gas. The molecules are in a constant state of chaotic motion, darting around and colliding with each other and the walls of the container. Now, imagine we start slowly heating this container. What happens to the microscopic parameters of these molecules? Let's dive into the physics of the mean free path and the mean collision time to find out.

### Analyzing the Mean Free Path

The **mean free path** ($\lambda$) is the average distance a gas molecule travels before it collides with another molecule. It is a measure of how 'crowded' the gas is. The formula for the mean free path is given by:

$\lambda = \frac{V}{\sqrt{2}\pi d^2 N}$

Let's break down these variables. $V$ is the volume of the container, $N$ is the total number of gas molecules, and $d$ is the diameter of a single molecule.

Here is the crucial catch: our container is **closed and rigid**. Because it is rigid, the volume $V$ cannot change. Because it is closed, no gas can escape or enter, meaning the total number of molecules $N$ is strictly constant. The diameter $d$ of the molecules is an intrinsic property and obviously doesn't change with temperature.

Since every single term on the right side of the equation is a constant, the mean free path $\lambda$ must also be a constant. It does not depend on the temperature $T$ in a closed system. Therefore, as we heat the gas, the mean free path remains completely unchanged. This makes Statement C correct.

### The Impact on Mean Collision Time

Now, let's shift our focus to the **mean collision time** ($	au$). This is the average time that elapses between two successive collisions for a molecule. Using basic kinematics (time = distance / speed), we can write:

$	au = \frac{\lambda}{v_{	ext{avg}}}$

We already established that the distance $\lambda$ is constant. But what about the average speed $v_{	ext{avg}}$? According to the kinetic theory of gases, the average speed of gas molecules is directly tied to the absolute temperature $T$:

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

This tells us that $v_{	ext{avg}} \propto \sqrt{T}$. As we heat the container and the temperature rises, the molecules gain kinetic energy and start zipping around much faster.

If we substitute this proportionality back into our collision time equation, we get:

$	au \propto \frac{1}{\sqrt{T}}$

Because the molecules are moving faster but have to cover the exact same average distance ($\lambda$) between collisions, they will cover that distance in less time. Consequently, the mean collision time $	au$ decreases as the temperature increases. This makes Statement B correct.

### Final Conclusion

By carefully analyzing the constraints of a closed container, we deduced that the mean free path remains constant while the mean collision time decreases due to the increased thermal velocity of the molecules. Both statements B and C are true, making option (a) the correct choice. This problem beautifully illustrates how macroscopic constraints (like a closed volume) dictate microscopic behaviors!

Select Answer:

Visualized Solution

Visualizing the System

Formula for Mean Free Path

Analyzing the Variables

Conclusion for Mean Free Path

Formula for Mean Collision Time

Temperature Dependence of Speed

Conclusion for Mean Collision Time

Final Answer

The Sigma Insight: Kinetic Theory of Gases

The Dance of Gas Molecules

Analyzing the Mean Free Path

The Impact on Mean Collision Time

Final Conclusion

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