A British physicist, James Clerk Maxwell (1831 – 1879) showed the distribution of molecular speeds as shown in the figure below. The speed distribution is dependent on temperature. Most of the gas particles are close to the average speed which corresponds to the maximum of the curve. Note, as the temperature is increased, the curve broadens.

The root-mean-square (rms) speed, μ, is equal to the speed of a molecule having an average molecular kinetic energy. The equation is:

\(\displaystyle μ\;=\;\sqrt{\frac{3RT}{M_m}}\)
 

where R = \( 8.314\frac{kg⋅m^2}{s^2⋅K⋅mol}\), T is in Kelvin, and M_m is in kJ/mol. If there were two gases, the one with the higher molar mass would have the lower rms speed.

The rms speeds can be very high. For example, H₂, at 30°C

\(\displaystyle μ\;=\;\sqrt{\frac{3\times\;8.314\frac{kg⋅m^2}{s^2⋅K⋅mol}\times\;303\;K}{0.0020159\frac{kg}{mol}}}\;=\;1.94\times10^3\frac{m}{s}\)
 
The rms speed is over 4000 mi/hr.

We can define effusion as a process by which a gas escapes through a small hole in its container into an evacuated space. Graham’s Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Rate of Effusion \(\propto\sqrt{\frac{1}{M_m}}\)

A lighter gas moves more quickly and therefore has a higher rate of effusion than a heavier gas at the same temperature. In the figure below, we see that helium, a lighter gas, has a higher rate of effusion than Argon, a heavier gas.

We can use Graham’s law of effusion to compare the effusion rates of two gases:

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;gas\;A}}{\mathrm{rate\;of\;effusion\;of\;gas\;B}}\;=\;\sqrt{\frac{M_mB}{M_mA}}\)

This equation can be used to compare the rate of effusion of helium and argon.

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;of\;helium}}{\mathrm{rate\;of\;effusion\;of\;argon}}\;=\;\sqrt{\frac{39.948\frac{g}{mol}}{4.0026\frac{g}{mol}}}\;=\;3.2\)

For a given amount of time, the helium atoms effuse through the hole 3.2 times faster than argon atoms.

The molar mass of an unknown gas can be determined using Graham’s Law. We can compare the effusion rate of a known gas to that of an unknown gas and find the molar mass of the unknown gas. Recall, a lighter gas will effuse much faster than a heavier gas, but the time, t, it takes for a gas to effuse is inversely proportional to the effusion rate.

\(\displaystyle \frac{\mathrm{rate\;of\;effusion\;gas\;A}}{\mathrm{rate\;of\;effusion\;gas\;B}}\;=\;\frac{t_B}{t_A}\;=\;\sqrt{\frac{M_B}{M_A}}\)

Diffusion is the mixing of gas particles caused by their motion. Due to collisions between gas particles, the rate of diffusion is slower than that of effusion. One example of diffusion is when you are baking a pie in the oven. You might be sitting in another room and you smell the pie. You get up and go into the kitchen, open the oven, and see the pie is starting to burn. It takes a while for gas particles to travel because when the particles collide, the particles do not follow a straight line.

Worksheet: Effusion of Gases

Exercises

Exercise 1. Calculate the rms speed of N₂ at 98.2°C.

Check Solution/Answer to Exercise 1

Exercise 2. Uranium hexafluride, UF₆, sublimes at 57°C under normal atmospheric pressure. What is the rms speed of UF₆ at 57°C? Report your answer in m/s.

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Exercise 3. What is the molar mass of a gas that diffuses 1.98 times faster than Xe gas.

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Exercise 4. What is the molar mass of a gas that diffuses 1.87 times slower than Kr gas?

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Exercise 5. Calculate the ratio of the rates of effusion of N₂ and N₂O under the same conditions.

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Exercise 6. A given volume of N₂ takes 64.6 s to effuse. Another gas, under the same conditions, took 88.2 s to effuse. Calculate the molar mass of this gas.

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