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Chemistry · Ch 6 — Gaseous State

Graham's Law of Diffusion

6.4.1

Graham's Law of Diffusion

Gas molecules constantly move to occupy all the space available to them. Diffusion is the mixing of one gas's molecules through another gas -- driven by molecules migrating from a region of higher concentration to one of lower concentration until the two gases are uniformly mixed (Figure 6.7, left). Effusion is a related but distinct process: a gas escaping from its container through a very small hole into a vacuum or lower-pressure region (Figure 6.7, right), without necessarily mixing with another gas on the other side.

Thomas Graham found that the rate of either process is inversely proportional to the square root of the gas's molar mass:

rate of diffusion∝1M\text{rate of diffusion}\propto\frac{1}{\sqrt{M}}

This is Graham's law of diffusion/effusion. For two gases A and B diffusing under the same conditions,

rArB=MBMA(6.15)\frac{r_A}{r_B}=\sqrt{\frac{M_B}{M_A}}\qquad(6.15)

and if the two gases happen to be at different pressures PA,PBP_A,P_B,

rArB=PAPBMBMA(6.16)\frac{r_A}{r_B}=\frac{P_A}{P_B}\sqrt{\frac{M_B}{M_A}}\qquad(6.16)

Graham's law is not just a curiosity -- it is the physical basis of one of the most consequential separations in applied chemistry: enriching the fissile isotope U-235 out of a mixture with the far more abundant U-238, by repeated gaseous effusion of uranium hexafluoride. …

Figure 6.7Diffusion and effusion of gases

What this figure shows. Two side-by-side cutaway cylinders. The left one, labelled 'Diffusion', is divided by a partly-open partition into 'Gas 1' (orange dots) and 'Gas 2' (blue dots); a curved arrow through the gap shows the two gases mixing into each other's compartment. The right one, labelled 'Effusion', has a gas-filled compartment (pink dots) connected through a single small hole to an evacuated ('Vacuum') compartment, with a single dot shown passing through the hole -- illustrating effusion as escape of individual molecules through a tiny opening into empty s …

Misc 6.4.1-worked-grahamWorked example: molar mass of an unknown gas from its diffusion rate relative to nitrogen

Worked out. An unknown gas diffuses at 0.5 times the rate of N2_2 (M = 28 g mol−1^{-1}) under the same conditions. rateunknownrateN2=MN2Munknown\dfrac{\text{rate}_{unknown}}{\text{rate}_{N_2}}=\sqrt{\dfrac{M_{N_2}}{M_{unknown}}}, so 0.5=28Munknown0.5=\sqrt{\dfrac{28}{M_{unknown}}}. Squaring: 0.25=28Munknown0.25=\dfrac{28}{M_{unknown}}, giving Munknown=280.25=112 g mol−1M_{unknown}=\dfrac{28}{0.25}=112\ \text{g mol}^{-1}. …

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