Gas Diffusion and Effusion

 

Learning Objective

  • Explain the concepts of diffusion and effusion.

Key Points

    • Gaseous particles are in constant random motion.
    • Gaseous particles tend to undergo diffusion because they have kinetic energy.
    • Diffusion is faster at higher temperatures because the gas molecules have greater kinetic energy.
    • Effusion refers to the movement of gas particles through a small hole.
    • Graham’s Law states that the effusion rate of a gas is inversely proportional to the square root of the mass of its particles.

Terms

  • diffusionmovement of particles from an area of high concentration to one of low concentration
  • mean free paththe average distance traveled by a particle between collisions with other particles
  • Effusionmovement of gas molecules through a tiny hole

Diffusion

According to Kinetic Molecular Theory, gaseous particles are in a constant state of motion, moving at random speeds and in many different directions. Because of their kinetic energy at temperatures above absolute zero, all particles undergo diffusion.

Diffusion refers to the process of particles moving from an area of high concentration to one of low concentration. The rate of this movement is a function of temperature, viscosity of the medium, and the size (mass) of the particles. Diffusion results in the gradual mixing of materials, and eventually, it forms a homogeneous mixture.

DiffusionParticles moving from areas of high concentration to areas of low concentration.

Effusion

Not only do gaseous particles move with high kinetic energy, but their small size enables them to move through small openings as well; this process is known as effusion. For effusion to occur, the hole’s diameter must be smaller than the molecules’ mean free path (the average distance that a gas particle travels between successive collisions with other gas particles). The opening of the hole must be smaller than the mean free path because otherwise, the gas could move back and forth through the hole.

Effusion is explained by the continuous random motion of particles; over time, this random motion guarantees that some particles will eventually pass through the hole.

Interactive: Diffusion & TemperatureExplore the role of temperature on the rate of diffusion. Set the temperature, then remove the barrier, and measure the amount of time it takes the blue molecules to reach the gas sensor. When the gas sensor has detected three blue molecules, it will stop the experiment. Compare the diffusion rates at low, medium and high temperatures. Trace an individual molecule to see the path it takes.

Diffusion and Molecular MassExplore the role of a molecule’s mass with respect to its diffusion rate.

Graham’s Law

Scottish chemist Thomas Graham experimentally determined that the ratio of the rates of effusion for two gases is equal to the square root of the inverse ratio of the gases’ molar masses. This is written as follows:

[latex]\frac{\text{rate of effusion gas 1}}{\text{rate of effusion gas 2}}=\sqrt{\frac{M_2}{M_1}}[/latex]

where M represents the molar mass of the molecules of each of the two gases.

The gases’ effusion rate is directly proportional to the average velocity at which they move; a gas is more likely to pass through an orifice if its particles are moving at faster speeds.

Example

What is the ratio of the rate of effusion of ammonia, NH3, to that of hydrogen chloride, HCl?

[latex]\frac{\text{Rate}_{NH_3}}{\text{Rate}_{HCl}}=\sqrt{\frac{36.46\;g/mol}{17.03\;g/mol}}=1.46[/latex]

The NH3 molecules effuse at a rate 1.46 times faster than HCl molecules.

Derivation of Graham’s Law

Graham’s Law can be understood as a consequence of the average molecular kinetic energy of two different gas molecules (marked 1 and 2) being equal at the same temperature. (Recall that a result of the Kinetic Theory of Gases is that the temperature, in degrees Kelvin, is directly proportional to the average kinetic energy of the molecules.) Therefore, equating the kinetic energy of molecules 1 and 2, we obtain:

[latex]\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2 [/latex]

[latex]m_1v_1^2 = m_2v_2^2[/latex]

[latex] \frac{v_1^2}{v_2^2}=\frac{m_2}{m_1}[/latex]

[latex] \frac{v_1}{v_2}=\frac{\sqrt{m_2}}{\sqrt{m_1}}[/latex]

The rate of effusion is determined by the number of molecules that diffuse through the hole in a unit of time, and therefore by the average molecular velocity of the gas molecules.