To calculate the ratio of the mean free path to the diameter of the argon atom, we'll use the following dataTo calculate the ratio of the mean free path to the diameter of the argon atom, we'll use the following

• Mean free path, $\lambda = 5 \times 10^{-7}$ m
• Diameter of an argon atom, $d = 0.34$ nm

First, convert the diameter from nanometers to meters:

$d = 0.34 \times 10^{-9} \, \text{m}$

Now, calculate the ratio $\frac{\lambda}{d}$:

$\frac{\lambda}{d} = \frac{5 \times 10^{-7} \, \text{m}}{0.34 \times 10^{-9} \, \text{m}} = \frac{5 \times 10^{-7}}{0.34 \times 10^{-9}}$

$\frac{\lambda}{d} \approx 1.47 \times 10^3$

Rounding to the nearest integer:

$\frac{\lambda}{d} \approx 1471$

Interpretation

This result indicates that the mean free path of an argon atom is much larger than the diameter of the atom itself. The ratio of approximately 1471 suggests that the intermolecular distance is indeed much larger compared to the size of the atom at ambient conditions. This supports the idea that gas molecules are far apart from each other relative to their size, which is consistent with the properties of gases under standard conditions.

---

Would you like more details or have any other questions? Here are 5 related questions you might find interesting:

1. How would the mean free path change if the pressure is doubled?
2. What is the relationship between the mean free path and temperature?
3. How does the diameter of a molecule affect the mean free path?
4. Can you explain the concept of mean free path in the context of the kinetic theory of gases?
5. How does the mean free path relate to the viscosity of a gas?

Tip: The mean free path is inversely proportional to the number density of particles, which means it decreases as the pressure increases or the temperature decreases.