[ Main ] - [ Sections ]

Background:

Graham's Law

The ratio of the rates of effusion of two gases is equal to the square root of the inverse ratio of their molecular masses or densities.

The effusion rate of a gas is inversely proportional to the square root of its molecular mass.

Mathematically, this can be represented as:

Rate1 / Rate2 = square root of (Mass2 / Mass 1)

Explanation and Discussion:

Graham's Law shows the relationship between the molar or molecular mass of a gas and the rate at which it will effuse. Effusion is the process of gas molecules escaping through tiny holes in their container. Diffusion can also be considered with Graham's Law, such as perfuming diffusing through a room.

Let us first consider why gases effuse. Containers can have small holes or pores in them. Although these openings are microscopic, they are larger than the gas molecules. Randomly, the gas molecules move around the inside of the container until they impact something. This can be another molecule or the side of the container. A gas can also, instead of hitting the side of the container, pass through one of those openings by chance. This is effusion: a random movement of a as molecule through the container's wall. A common example of this is a balloon filled with helium: first it is buoyant and floats in the air, but in a few days it hangs toward the ground or floats a few inches above the ground (if at all). The Helium has escaped through the small holes in the balloon.

With Graham's Law, you can find the effusion rates for two gases or the molecular mass of a gas. This ratio of effusion rates follows the pattern that the gas with the lesser molecular mass has a greater rate of effusion.

Calculations using Graham's Law
Let's compare the rate off effusion of two common gases, Nitrogen and Oxygen. N2, Nitrogen, has a molecular mass of 28.0 g. O2, Oxygen, has a molecular mass of 32.0 g. Therefore, to find the ratio, the equation would be:

RateN2/RateO2 = square root of 32.0 g / 28.0 g.

This works out to:

RateN2/RateO2 = 1.069

Adjusting to the appropriate accuracy, we find that the rate is 1.07. This tells us that N2 is 1.07 times as fast as O2. It is faster, but not by much.

Let's try to find a molecular mass. Let's use gas A and B. A is 0.68 times as fast as B. The mass of B is 17 g. What is the mass of A?

First, we set up the equation. Plugging in the values to our formula, we get:

0.68 = square root of 17 g/MassA

Squaring both sides gets:

0.4624 = 17 g/MassA

Then, to get the unknown alone, we exchange the extremes of our proportion:

MassA = 17 g/0.4624

Which simplifies to:

MassA = 36.7647 g

Which, when adjusted to the correct significant digits, is 37 g. Plugging this in to our formula would check the answer. (If you are interested in which gases were used, A is HCl and B is NH3).

Continued Study:
You can also test yourself. You can also learn about Thomas Graham.

Sources:
Brown, Theodore L., H. Eugene LeMay, Jr. and Bruce E. Burston, Chemistry: The Central Science, Englewood Cliffs, NJ: Prentice Hall, Inc., 1994

Dorin, Henry, Peter E. Demmin, and Dorothy L. Gabel. Prentice Hall Chemistry: The Study of Matter, Needham, Massachusetts and Englewood Cliffs, New Jersey: Prentice Hall, Inc., 1989.

Kauffman, George B., "Graham, Thomas" Groliers New Multimedia Encyclopedia, Release 6, 1993

Team # 12596