Root-mean-square velocities of gas molecules are sometimes directly useful, but the comparison of velocities explains the results of, and is useful in, studies of effusion of molecules through a small hole in a container or diffusion of molecules through porous barriers.

The comparison between two gases is most conveniently expressed as:

$\large \frac{v_{rms_1}}{v_{rms_2}} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}} $

**Note : **This equation gives the velocity ratio in terms of either the molar mass ratio or the ratio of densities.

The ratio of root-mean-square velocities is also the ratio of the rates of effusion, the process by which gases escape from containers through small holes, and the ratio of the rates of diffusion of gases.

This equation is called Graham’s law of diffusion and effusion because it was observed by Thomas Graham well before the kinetic-molecular theory of gases was developed.

As an empirical law, it simply stated that the rates of diffusion and of effusion of gases varied as the square root of the densities of the gases. Graham’s law is the basis of many separations of gases. The most significant is the separation of the isotopes of uranium as the gases 238UF6 and 235UF6. Fluorine has only one isotope, so the separation on the basis of molar mass is really a separation on the basis of isotopic mass.

Diffusion is the spontaneous intermingling of one substance with another. This occurs with perfumes and aftershaves and also with the fragrance of irate skunks. Diffusion is over when the molecules of the fragrance are evenly spread within a container, be it a room or a gas flask. Another way to look at it is to describe the process using partial pressures. Diffusion is over when the partial pressures of all the gases involved become identical in all parts of the container. When a fragrance is more or less concentrated in a corner of a room, its partial pressure is higher there.

Another necessary term is the gradient. A gradient is used to describe how the concentration or partial pressures change from one place to another. When sugar is added to coffee it sinks to the bottom. If left alone, (and the coffee stayed hot), the sugar would dissolve and gradually spread on its own accord throughout the coffee. (Of course, we stir it because the time involved here is quite long.) At the bottom of the cup we have a lot of sugar. At the top of the cup we have zero dissolved sugar.

A concentration gradient is set up from bottom to top. Sugar will gradually move down this gradient until it is evenly spread out. When a freshly skunked pet comes into the room, it brings with it a high concentration of skunk perfume molecules. Where the pet enters the house there will be a high concentration. Everywhere else in the house there will be a zero concentration of skunk molecules. The point is, one of the great facts about natural processes is that “Nature abhors a vacuum.” Another way to say this is that nature abhors gradients. Given the chance, gradients in nature tend to disappear, some rapidly and some only over eons of time, until there is as much an evenness as possible .

The effusion of a gas is its movement through an extremely tiny opening into a region of lower pressure. The term diffusion really only speaks to the direction of gas movement. Effusion speaks for not only the direction but the rate that a change occurs.

Thomas Graham, studied the rates at which various gases effuse, and he found that the denser the gas is, the slower it effuses. The exact relationship between rate and gas density, d, is called Graham’s Law of Effusion.

$\large r_A \times \sqrt{d_A} = r_B \times \sqrt{d_B}$

Finding the densities of gases at various temperatures is often difficult to do. With a little chemical slight of hand we can get a formula for a much simpler answer.

$\large r_A \times \sqrt{M_A} = r_B \times \sqrt{M_B}$

Illustration : 32 ml of H2 diffuses through a fine hole in 1 minute. What volume of CO2 will diffuse in 1 minute under same condition?

Solution:

$\large \frac{V_1}{V_2} = \sqrt{\frac{M_2}{M_1}}$

$\large \frac{32}{V_2} = \sqrt{\frac{44}{2}}$

= 6.82 ml

Example: A gaseous mixture of O_{2} and X containing 20% of X diffused through a small hole in 234 secs while pure O_{2} takes 224 secs to diffuse through the same hole. Find molecular weight of X ?

Solution : We have, $\large \frac{t_{mix}}{t_{O_2}} = \sqrt{\frac{M_{mix}}{M_{O_2}}}$

$\large \frac{234}{224} = \sqrt{\frac{M_{mix}}{32}}$

$\large M_{mix} = 34.921 $

As the mixture contains 20% of X, the molar ratio of O_{2} and x may be represented as 0.8 n : 0.2 n

n being total no. of moles

$\large M_{mix} = \frac{32 \times 0.8 n + M_x \times 0.2 n}{n}$

$\large 34.921 = \frac{32 \times 0.8 n + M_x \times 0.2 n}{n}$

$\large M_x = 46.6$