DAT General Chemistry: Effusion, Diffusion, and Graham's Law

DAT High-Yield Concept

Kinetic Energy & Temperature: Absolute temperature ( $T$ in Kelvin) directly sets the average molecular kinetic energy ( $KE_{\text{avg}} = \frac{3}{2}RT$ ). All ideal gases at the same temperature have the exact same average kinetic energy, regardless of molar mass.
Molecular Speed: Lighter gas particles move faster on average than heavier gas particles at the same temperature ( $u_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ ).
Effusion vs. Diffusion: Effusion is gas escape through a tiny pinhole into a vacuum; diffusion is the movement/mixing of gas particles through space or another gas.
Graham's Law: Rate of effusion or diffusion is inversely proportional to the square root of molar mass ( $\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}}$ ).

Temperature and Average Molecular Kinetic Energy

Absolute temperature in Kelvin is a direct measure of average translational kinetic energy. According to Kinetic Molecular Theory (covered in depth in its own lesson), the average kinetic energy per mole of any ideal gas depends solely on temperature:

KE_{\text{avg}} = \frac{3}{2} R T = \frac{1}{2} M u_{\text{rms}}^2

Because average kinetic energy is fixed at a given temperature, a sample of helium ( $\ce{He}$ ) and a sample of sulfur hexafluoride ( $\ce{SF6}$ ) at $\pu{298 K}$ possess identical average kinetic energies. However, because individual molar mass ( $M$ ) differs, lighter gas molecules must travel at higher root-mean-square speeds ( $u_{\text{rms}}$ ) to possess that same average kinetic energy:

u_{\text{rms}} = \sqrt{\frac{3 R T}{M}}

Same temperature, same average kinetic energy, different speeds — At a fixed temperature the molecules of a gas are spread over a wide range of speeds rather than sharing one speed, and the average translational kinetic energy is set by the absolute temperature alone. Because that average is the same for all three gases here, the light helium atoms must move much faster than the heavy carbon dioxide molecules, and their distribution is broad and low while the heavy gas gives a narrow, tall one. The same mass dependence is what makes a light gas effuse through a pinhole faster than a heavy one.

Distinguishing Effusion and Diffusion

Diffusion: The process by which gas molecules disperse throughout a space or mix with another gas as a result of random thermal motion and frequent intermolecular collisions.

Effusion: The process by which gas molecules escape from a container to an evacuated chamber through a microscopic hole without undergoing collisions within the aperture.

Both processes depend on the average speed of gas molecules, but their physical environments differ fundamentally.

The image in the left shows effusion, where the image on the right shows diffusion. Effusion occurs through a hole smaller than the mean path of the particles in motion where diffusion occurs through a hole in which multiple particles can flow through simultaneously. (Source: Astrang13 · CC BY-SA 3.0 · Wikimedia Commons)

Schematic drawing of diffusion over a semipermeable membrane . Created by Quasar Jarosz ( talk ) 19:50, 7 August 2009 (UTC) from File:Diffusion.en.jpg (Source: Diffusion.en.jpg : derivative work: Quasar Jarosz ( talk ) · Public domain · Wikimedia Commons)

Property	Diffusion	Effusion
Mechanism	Gradual mixing across a open space or semipermeable membrane	Escape through a pinhole smaller than the mean free path
Collisions	Frequent intermolecular collisions slow net transport	Molecules pass individually without colliding in the hole
Key Driving Factor	Concentration gradient and random thermal motion	Molecular collision rate against container wall pinhole

Graham's Law of Effusion and Diffusion

Because rate of effusion is directly proportional to average molecular speed ( $u_{\text{rms}}$ ), Thomas Graham derived that the rate of effusion of a gas is inversely proportional to the square root of its molar mass ( $M$ ):

\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}}

When comparing the time ( $t$ ) required for equal molar amounts or volumes of two gases to effuse through the same orifice, time is inversely proportional to rate, yielding:

\frac{t_A}{t_B} = \sqrt{\frac{M_A}{M_B}}

DAT Memory Trick: Lighter gas = Faster speed = Shorter effusion time! Always place the molar masses under the square root in inverse order relative to rates (

\text{Rate}_A / \text{Rate}_B = \sqrt{M_B / M_A}

Worked Calculation Example

Problem: An unknown gas effuses through a tiny pinhole at half the rate of methane ( $\ce{CH4}$ , molar mass $\pu{16.04 g/mol}$ ) under identical temperature and pressure conditions. Calculate the molar mass of the unknown gas.

Step-by-Step Solution:

Express the rate relationship: $\frac{\text{Rate}_{\ce{CH4}}}{\text{Rate}_{\text{unknown}}} = 2.00$ .
Apply Graham's Law: $2.00 = \sqrt{\frac{M_{\text{unknown}}}{M_{\ce{CH4}}}} = \sqrt{\frac{M_{\text{unknown}}}{\pu{16.04 g/mol}}}$ .
Square both sides: $(2.00)^2 = 4.00 = \frac{M_{\text{unknown}}}{\pu{16.04 g/mol}}$ .
Solve for molar mass: $M_{\text{unknown}} = 4.00 \times \pu{16.04 g/mol} = \pu{64.16 g/mol}$ (corresponding to sulfur dioxide, $\ce{SO2}$ ).

Common Exam Trap: Don't invert rates and times! The rate ratio is $\sqrt{M_B / M_A}$ , whereas the effusion time ratio is $\sqrt{M_A / M_B}$ . Heavy gases effuse slower and require more time.

Critical Thinking & Application Questions

If absolute temperature quadruples from $\pu{200 K}$ to $\pu{800 K}$ , by what factor does the root-mean-square speed ( $u_{\text{rms}}$ ) of a gas increase?
Why do real gases diffuse through air significantly slower than their calculated root-mean-square speeds (which are several hundred meters per second)?
Gaseous diffusion was used during the Manhattan Project to enrich uranium isotopes using uranium hexafluoride ( $\ce{^{235}UF6}$ vs. $\ce{^{238}UF6}$ ). Which gas effuses faster, and what is the theoretical single-stage rate ratio?

References

Sources returned by Google Search grounding during lesson verification.

9.4 Effusion and Diffusion of Gases - Chemistry 2e - OpenStax — openstax.org
9.5 The Kinetic-Molecular Theory - Chemistry 2e - OpenStax — openstax.org

Effusion, Diffusion, and Graham's Law

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