# Graham's Law of Effusion, Gases Diffusion and Mean Free Path Average

## Graham's Law of Effusion

**The rate of effusion can be quantified.**

- Consider two gases with molar masses
**M**_{1} and **M**_{2},
- since the rate of effusion (
*r*) is directly proportional to (*u*),
- where
*u*) is given by the equation: *u* =
√ 3RT/M
- The relative rate of effusion (
*r*_{1}/*r*_{2}) will be equal to
√ M2/M1

### Exercises on Graham's Law of Effusion

*Exercise on Calculating the Ratio of the Effusion Rates*

Check your answers here:
*Solution to the Exercise on Calculating the Ratio of the Effusion Rates*

## Gases Diffusion and Mean Free Path Average

**Definition**:

"Diffusion of a gas is the spread of the gas through space".

**Diffusion** is thus __faster__ for __light__ gas molecules.

**Diffusion** is slowed by gas molecules colliding with eachother. Because of these collisions of the gas molecules, their direction of motion is always changing.

Average distance traveled by a gas molecule between collisions is called **"mean free path"**.

### Example

The **mean free path** of gas molecules at about 100 km altitude is about 10^{6} times that at earth's surface.

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Date of last modification: Summer , 2019