Molecular Effusion and Diffusion

Learning Objectives

  1. Explore the nature of gas movement: molecular effusion and diffusion.
  2. Examine and apply Graham’s law of effusion.

Effusion

The movement of gas molecules can be divided into a few different types. Effusion is the movement of gas molecules from one container to another via a tiny hole. Typically the container to which the gas is moving is kept under lower pressure.

Figure 6.## Molecular Effusion

Figure 6.10 Molecular Effusion

In 1846, the Scottish chemist Thomas Graham found that the rate of effusion of a gas (the amount of gas transferred between containers in a certain amount of time) is inversely proportional to the square root of its molar mass. This means that gases with a lighter molecular weight have higher effusion rates.

Figure 6.## Thomas Graham who proposed Graham's law of effusion in 1846.

Figure 6.11 Thomas Graham who proposed his law of effusion in 1846.[1]

This finding is summarized in Graham’s law of effusion:

$$\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}}$$ =$$\sqrt{\frac{M_2}{M_1}}\$$

This finding can be rationalized by thinking through the process of effusion on the molecular level. For a gas molecule to successfully move from one container to another, it must hit and pass through the tiny hole present in the container. Gases with higher rms speed are more likely to hit and pass through the hole so effusion is dependent on rms speed:

$$\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}}$$ =$$\frac{\sqrt{\frac{3RT}{M_1}}}{\sqrt{\frac{3RT}{M_2}}}$$ =$$\sqrt{\frac{M_2}{M_1}}\$$

 

Example 20

An unknown halogen (diatomic) gas effuses at a rate that is approximately 1.89 times the rate of I2 gas at the same temperature. Determine the molar mass and identity of this unknown gas.

Solution

$$\frac{{\rm Rate\ of\ Effusion\ (Gas\ 1)}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}}$$=$$\sqrt{\frac{M_2}{M_1}}\$$

$$\frac{{\rm Rate\ of\ Effusion\ (Gas\ }{{\rm I}}_2{\rm )}}{{\rm Rate\ of\ Effusion\ (Gas\ 2)}}$$ = $$\sqrt{\frac{M_2}{M{{\rm I}}_2}}\$$
$$\frac{{\rm 1}}{1.89}$$ = $$\sqrt{\frac{M_2}{253.80\ \text{g/mol}}}\$$

0.279 =$$\frac{M_2}{253.80\ \text{g/mol}}\$$

M2 = 71.1 g/mol. Therefore the unknown gas is Cl2

Diffusion

Another type of gas movement is called diffusion; it is the movement of gas molecules through one or more additional types of gas via random molecular motion. Similar to effusion, gases with lower molecular weights (which have a higher rms speed) diffuse faster than gases with higher molecular weights. However, in diffusion, movement is much more complicated as collisions occur between molecules that change the direction and speed of the molecules. As a result of these collisions, the path a molecule travels in diffusion is made up of numerous straight, short segments. The term mean free path is used to describe the average distance travelled by a molecule between collisions.

Figure 6.## Molecular Diffusion

Figure 6.12 Molecular Diffusion

Figure 6.## Stylized depiction of the path travelled by a gas particle during diffusion. Other particles omitted for clarity.

Figure 6.13 Stylized depiction of the path travelled by a gas particle during diffusion. Other particles have been omitted for clarity.

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Video Source: Diffusion by keyj (https://viuvideos.viu.ca/media/Diffusion+/0_zel419p5)
[2]

Key Takeaways

  • Effusion is the movement of gas molecules from one container to another through a tiny hole.
  • Rates of effusion can be compared at the same temperature using Graham’s law.
  • Diffusion  is the movement of gas molecules through one or more other types of gas via random molecular motion.
  • Both the rates of effusion and diffusion are influenced by the molecular weight of the gas particle.
  1. Thomas Graham Lithogram Public Domain ↵
  2. Image: ParticleMeanFreePath.png by DaisyDaisy/Public Domain ↵