Content code
c1037
Slug (identifier)
gas-diffusion-and-effusion-graham-s-law
Parent content
Grades
Secondary V
Topic
Chemistry
Tags
gas diffusion and effusion
effusion
Graham's Law
Content
Contenu
Content
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Diffusion is the process by which a substance mixes with one or more other substances through the movement of its constituent particles.

Effusion is the process by which a gas passes through a small hole in a wall.

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According to the kinetic theory of gases, gas particles displace in a straight line and frequently collide. Moreover, they also collide with the walls of the container in which they find themselves. This displacement explains two types of phenomenon: diffusion and effusion. Moreover, the rate at which these phenomena occur can be quantified using Graham's Law.

Links
Title (level 2)
Diffusion and Effusion
Title slug (identifier)
diffusion-and-effusion
Contenu
Corps

After a while, the random distribution of the particles results in a uniform distribution throughout the container in which they are contained. This phenomenon is called diffusion. In other words, it's a gradual process whereby a gas spreads throughout the space available to it. A gas can also mix with other gases through the motion of its particles. Diffusion explains why an atmospheric pollutant can spread over large areas, or why certain fragrances invade a part and then disappear.

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Representation of the diffusion phenomenon: a gas concentrated in one place (left) is distributed uniformly throughout the available space (right) by diffusion.

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Some containers, rather than being hermetically sealed, have porous walls, i.e. walls with tiny holes. Gas particles can then, by chance of collision, pass through these small openings and escape into the environment. This is known as effusion. A slowly deflating helium balloon is an example of effusion. Oxygen and carbon dioxide passing through the walls of blood capillaries is another.

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Effusion, the passage of gas particles through a membrane, occurs more rapidly with a large opening (right) than with a small one (left).

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The velocity of particle effusion or diffusion is not the same for all gases. This rate depends mainly on the mass of the particles. Indeed, at a given temperature, all particles have the same amount of kinetic energy. This energy depends on the mass and velocity of the particles. As different gases are made up of particles of different masses, their average velocity will be different. Thus, particles with higher masses diffuse more slowly than light particles, because they travelled more slowly.

Content
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Velocity distribution curve as a function of mass: a heavier gas like dinitrogen travels more slowly than a lighter gas like hydrogen.

Title (level 2)
Graham's Law
Title slug (identifier)
graham-s-law
Contenu
Content
Corps

Graham's Law states that the diffusion or effusion velocities of two gases at the same temperature and pressure conditions are inversely proportional to the square root of their molar masses.

Corps

In 1840, the Scottish chemist Thomas Graham posited that, for equivalent conditions of temperature and pressure, the kinetic energies of two gases were equal. Since kinetic energy is a function of particle mass and velocity, he was able to derive the following law:

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|\displaystyle \frac{Velocity_{1}}{Velocity_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}|
where
|Vitesse_{1}| = gas diffusion or effusion velocity 1 (in m/s or mL/s)
|Vitesse_{2}| = gas diffusion or effusion velocity 2 (in m/s or mL/s)
|M_{1}| = molar mass of gas 1 (in g/mol)
|M_{2}| = molar mass of gas 2 (in g/mol)


Note: In Graham's Law, the molar masses of gases 1 and 2 could be replaced by their masses in grams.

Corps

This law states that, under identical temperature and pressure conditions, the diffusion velocities of two gases are inversely proportional to the square roots of their molar masses. This law can also be applied to the velocity of gas effusion. Graham's law is consistent with the kinetic theory of gases, since the lighter the particles in a gas, the faster they travelled. Lighter gas particles diffuse faster into another gas than heavier ones.

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What is the molar mass of an unknown gas Z that diffuses at 700 mL/min in a diffusion device, while methane, |CH_{4}|, diffuses at 350 mL/min in the same device?

  1. Identification of problem data or values
    |Vitesse_{1}| = 700 mL/min for the unknown gas Z
    |Vitesse_{2}| = 350 mL/min for the |CH_{4}|
    |M_{1} = x| for the unknown gas Z
    |M_{2}| = 16 g/mol for the |CH_{4}|
  2. Molar mass calculation
    |\displaystyle \frac{Vitesse_{1}}{Vitesse_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}|
    |\displaystyle \frac{700 mL/min}{350 mL/min}=\sqrt{\frac{16 g/mol}{x}}|
    |x=4 g/mol|
Solution
Corps

Answer: The molar mass of the unknown gas Z is 4 g/mol.

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