Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic we

Q: Why do we use the radius instead of the diameter in the proportionality?

The formula for the mean free path uses the collision cross-section, which depends on the square of the diameter ($d^2$). Since the diameter is simply twice the radius ($d = 2r$), $d^2 = 4r^2$. When we take the ratio of two mean free paths, the constant factor of 4 cancels out completely, leaving us with a direct inverse proportionality to $r^2$.

Q: Why did we use the mean speed instead of the root-mean-square (rms) speed?

For derivations involving collision frequency and mean free path, the mean speed $\langle v \rangle = \sqrt{\frac{8kT}{\pi m}}$ is technically the correct statistical average to use because we are interested in the average relative velocity between colliding particles. However, since both $\langle v \rangle$ and $v_{rms}$ are proportional to $\sqrt{\frac{T}{m}}$, using either would yield the exact same final ratio!

Q: What if the temperature was different for both gases?

If the temperatures were different, $T$ would no longer be a constant that cancels out. You would have to keep it in the proportionality: $\tau \propto \frac{\sqrt{m}}{r^2 \sqrt{T}}$. You would then plug in the respective temperatures into your final ratio calculation.

Q: Why is the calculated answer not in the given options?

Sometimes, official competitive exams like JEE contain errors in the question statement, the provided data, or the options. In this specific case, the mathematically rigorous answer is approximately 1.09, which doesn't match any option. Such questions are usually declared as 'bonus' and marks are awarded to all students.

Q: Does the number density $n$ depend on the type of gas?

According to Avogadro's law, equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules. Therefore, the number density $n = \frac{N}{V} = \frac{P}{kT}$ depends only on pressure and temperature, not on the identity of the gas. Furthermore, the problem explicitly states that both gases have the same number density.

Question

Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to

Options:

(A) 4.67

(B) 2.3

(C) 3.67

(D) 1.83

Sigma Percentile · Accepted Answer

### The Microscopic Chaos

Imagine you are standing in a massive, bustling room filled with people. If you try to run in a straight line, you won't get very far before bumping into someone. This is exactly what life is like for a gas molecule! The distance you manage to travel between two consecutive bumps is called the **mean free path**, and the time it takes you to travel that distance is the **mean free time** or **relaxation time**.

In this fascinating problem, we are comparing the chaotic lives of two different noble gases: Argon and Xenon. Both are kept in identical rooms (same temperature and same number density), but they have very different physical builds. Argon is the smaller, lighter runner, while Xenon is the bulkier, heavier one. Our mission is to find the ratio of their mean free times.

### Unpacking the Mean Free Path

Let's first look at the distance between collisions. The mean free path $\lambda$ is given by the beautiful formula:

$$ \lambda = \frac{1}{\sqrt{2}\pi n d^2} $$

Here, $n$ is the number density (how crowded the room is) and $d$ is the diameter of the molecule (how wide the runner is). Notice that the heavier Xenon atom, with its larger radius, presents a bigger target. It will inevitably bump into things more often, meaning its mean free path will be shorter.

### The Need for Speed

But distance is only half the story. To find the time between collisions, we also need to know how fast these atoms are zipping around. The average thermal speed $\langle v angle$ of a gas molecule is dictated by the temperature and its mass:

$$ \langle v angle = \sqrt{\frac{8kT}{\pi m}} $$

Because both gases are at the same temperature $T$, their kinetic energies are comparable. However, the lighter Argon atoms will be darting around much faster than the sluggish Xenon atoms.

### The Master Equation

Now, we bridge the gap. Time is simply distance divided by speed. Therefore, the mean free time $	au$ is:

$$ 	au = \frac{\lambda}{\langle v angle} = \frac{\frac{1}{\sqrt{2}\pi n d^2}}{\sqrt{\frac{8kT}{\pi m}}} $$

This equation looks terrifying, but let's take a breath. We are only interested in the *ratio* between Argon and Xenon. Since both gases share the same temperature $T$ and number density $n$, all those messy constants ($\pi$, $k$, $\sqrt{2}$, $8$) will completely vanish when we divide them!

We are left with a remarkably elegant proportionality:

$$ 	au \propto \frac{\sqrt{m}}{d^2} $$

And since the diameter $d$ is just twice the radius $r$, the factor of $4$ also cancels out, leaving us with:

$$ 	au \propto \frac{\sqrt{m}}{r^2} $$

### Final Calculation

Let's set up our final ratio for Argon over Xenon:

$$ \frac{	au_{	ext{Ar}}}{	au_{	ext{Xe}}} = \frac{\sqrt{m_{	ext{Ar}}}}{\sqrt{m_{	ext{Xe}}}} 	imes \left(\frac{r_{	ext{Xe}}}{r_{	ext{Ar}}}ight)^2 $$

Notice how the radius ratio is flipped because it's in the denominator! Now, we carefully substitute our given values:

$$ \frac{	au_{	ext{Ar}}}{	au_{	ext{Xe}}} = \sqrt{\frac{40}{140}} 	imes \left(\frac{0.1}{0.07}ight)^2 $$

Simplifying the fractions, we get:

$$ \frac{	au_{	ext{Ar}}}{	au_{	ext{Xe}}} = \sqrt{\frac{2}{7}} 	imes \left(\frac{10}{7}ight)^2 $$

Evaluating these numbers:

$$ \frac{	au_{	ext{Ar}}}{	au_{	ext{Xe}}} \approx 0.5345 	imes 2.0408 \approx 1.09 $$

**The final ratio is approximately 1.09.**

Interestingly, if you look at the options provided in the original JEE exam, none of them match this mathematically rigorous result! This happens occasionally in high-stakes exams, and such questions are typically awarded as bonus marks to all students. Regardless of the options, the physics we explored today remains flawlessly elegant.

Animated Solution for Physics - Thermodynamics: Two gases-argon (atomic radius 0.07 nm, atomic weight 40) and xenon (atomic radius 0.1 nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to

Select Answer:

Visualized Solution

Visualizing the Gases

Mean Free Path

Mean Thermal Speed

Mean Free Time

Proportionality Relation

Setting Up the Ratio

Substituting Values

Calculation

Final Answer

The Sigma Insight: Kinetic Theory of Gases

The Microscopic Chaos

Unpacking the Mean Free Path

The Need for Speed

The Master Equation

Final Calculation

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