# Mean free path and speed as a function of pressure and temp

• jhartc90
In summary, at an altitude of 18,000 ft. above mean sea level, the atmospheric pressure is approximately half of what it is at sea level (760 Torr) and the temperature decreases by 70C. This would result in a change in the speed distribution of air molecules, given by the formula $P(v)=4\pi [\frac{m}{2 \pi kT}]^{3/2} (v^2) exp(\frac{-mv^2}{2kt})$. For the mean free path of air molecules, the formula $\lambda_{mfp}=\frac{kt}{\pi d^2 P (2)^{1/2}}$ can be used, taking into account the
jhartc90

## Homework Statement

I know that at 18,000 ft. above mean sea level, the atm. pressure ~half of what it is at seal level (760 Torr). The temperature also decreases by 70C.

a. How would this change the speed distribution of the air molecules quanitatively?
b.How would it change the mean free path of the air molecules quantitatively?

## Homework Equations

For a) all I can seem to find is this:
$$c=\int_0^\infty vP(v)dv=(8kT/\pi m)^{1/2}$$

But since I am interested in the speed distribution, should I use:

$$P(v)=4\pi [\frac{m}{2 \pi kT}]^{3/2} (v^2) exp(\frac{-mv^2}{2kt})$$

For b) all I can seem to find is this:

$$\lambda_{mfp}=\frac{kt}{\pi d^2 P (2)^{1/2}}$$

Since the pressure is half, that would seem to double the mfp correct? and then i need to take into account the change in temp?

Thus the change would be

$$\frac{T1}{T2} \frac{P2}{P1}$$
is this correct?

## The Attempt at a Solution

For b) If I plug the numbers in I get like 0.66 which to me, might be correct. This is the ratio of mfp_1 to mfp_2 (lower pressure and temperature). Thus 0.66 * mfp_1 / mfp_2 = 0.65 so mfp_2 is larger (which I would expect)

I'm not sure if my approach to b) is correct, and for a), I am not very sure how to solve it. Do I just plug the temperatures into the equation and see the difference, noting that v also depends on temperature?

b: correct.

a: I'm not sure what else you can do quantitatively apart from writing down the formula. The distribution changes as given by this formula.

## What is the mean free path?

The mean free path is the average distance a molecule travels between collisions in a gas. It is a measure of how far a molecule can travel without encountering another molecule.

## How does pressure affect the mean free path?

As pressure increases, the number of molecules in a given space also increases. This leads to a decrease in the mean free path, as there is a higher likelihood of collisions between molecules.

## How does temperature affect the mean free path?

As temperature increases, the molecules in a gas move faster, resulting in a larger mean free path. This is because the higher speeds of the molecules make them less likely to collide with each other.

## What is the relationship between mean free path and speed?

The mean free path and speed are inversely proportional. As the mean free path increases, the speed decreases and vice versa. This is because higher speeds lead to fewer collisions and a longer distance traveled between collisions.

## Can the mean free path be calculated?

Yes, the mean free path can be calculated using the following equation: mean free path = (0.0269 * V) / (n * d^2), where V is the molar volume of the gas, n is the number of molecules per unit volume, and d is the diameter of the molecules.

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