Calculating Mean Free Path & Avg. Separation of Oxygen Molecules

[kidsmoker]

Homework Statement

(a)What is the mean free path for oxygen molecules at 300K and atmospheric
pressure (105 Pa) and the average frequency of collisions for a particular molecule?
(The diameter of an oxygen molecule is 0.29 nm).

(b)What is the mean free path of oxygen molecules at an altitude of 300 km, where
the pressure is only 10-11 atmospheres? (Assume the temperature is 300 K). Comment
on the implications of your result for simulating conditions at these altitudes in the
laboratory.

(c) Compare the ratio of λ to the average molecular separation for oxygen at sea level
and at an altitude of 300 km.

Homework Equations

$$\lambda = \frac{kT}{4*\pi*\sqrt{2}*r^{2}p}$$

The Attempt at a Solution

I've done parts (b) and (c) I think, and got answers of 1.11*10^-9m and 1.11*10^9m for the mean free paths. I have no idea how to do part (c) though. Just thinking about it, you would expect the ratio to stay roughly 1, since the average molecular separation and mean free path should increase alongside one another. How do you get an estimate of the average separation though?!

Thanks.

 

[Redbelly98]

Can you get the number of molecules per m³, given temperature and pressure? That's helpful in determining the average molecular separation.

 

[nlightner]

I would approach this problem by first understanding the concept of mean free path. Mean free path is the average distance a molecule travels between collisions with other molecules. It is dependent on the temperature, pressure, and size of the molecule. In this case, we are dealing with oxygen molecules at 300K and atmospheric pressure.

To calculate the mean free path and average separation for oxygen molecules, we can use the equation \lambda = \frac{kT}{4*\pi*\sqrt{2}*r^{2}p}, where \lambda is the mean free path, k is Boltzmann's constant, T is the temperature, r is the radius of the molecule, and p is the pressure.

(a) For oxygen molecules at 300K and atmospheric pressure (105 Pa), we can plug in the values and solve for \lambda: \lambda = \frac{(1.38*10^{-23} J/K)*(300 K)}{4*\pi*\sqrt{2}*(0.29*10^{-9} m)^{2}*(105 Pa)} = 7.11*10^{-8} m or 71.1 nm. This means that on average, an oxygen molecule will travel 71.1 nm between collisions with other molecules.

To calculate the average frequency of collisions, we can use the equation \nu = \frac{1}{\lambda}, where \nu is the frequency. Plugging in the value for \lambda, we get \nu = \frac{1}{7.11*10^{-8} m} = 1.41*10^{7} collisions per second. This means that a particular oxygen molecule will experience 1.41*10^{7} collisions per second.

(b) For oxygen molecules at an altitude of 300 km, where the pressure is only 10-11 atmospheres, we can use the same equation to calculate the mean free path: \lambda = \frac{(1.38*10^{-23} J/K)*(300 K)}{4*\pi*\sqrt{2}*(0.29*10^{-9} m)^{2}*(10^{-11} atm)} = 1.11*10^{-9} m or 1.11 nm. This means that at this altitude, an oxygen molecule will travel only 1.11 nm between collisions with other molecules.

(c) To compare the ratio of \lambda to the average molecular separation