Graham's Law — Overview & Calculation

(Video) Graham's Law of Effusion

by The Science Classroom

Graham's law helps us figure out the rate of effusion for gases. This term, not to be confused with diffusion, is the process in which gas particles move through a small hole. The formula for calculating Graham's law is as follows: $$\frac{rate_A}{rate_B} = \sqrt{\frac{MM_B}{MM_A}}$$

As we've studied the physical properties of gases, we've talked about their movement, arrangement, and volume. The kinetic molecular theory taught us that gases are in constant random motion. As they move, they collide with other particles or their container. In a closed system, the collisions create pressure. But, what happens when we have an open system? Their movements and collisions cause them to diffuse or effuse. Today, we'll look at the specifics of effusion.

In the 1830s and 1840s, Thomas Graham developed a law that quantifies the effusion rates of gases. He had become curious about the work of another physical chemist. Johann Döbereiner had observed that the diffusion of hydrogen was faster than air. Graham systematically developed some experimental procedures for measuring gases. He created very tiny tubes that restricted the gas flow. So, he could make quantitative measurements of effusion. As a side note, he was also the first chemist to study diffusion in liquids. Based on his work, he developed the concepts of suspension, colloid, and mixtures.

Forms of Graham's Law

• $Rate_{effusion} \propto \sqrt{\frac{1}{MM}}$

The first form of Graham's law is a proportionality. A gas's effusion rate is proportional to the square root of its inversed molar mass. So, the heavier the gas, the slower its effuses.

As a side note, Graham initially used density instead of molar mass. Remember, in 1811 Avagadro had proposed that equal volumes of gases had the same number of molecules. That's Avagadro's law and the mole concept. But, it wasn't accepted early on. We hadn't created the notion of molar masses. So, Graham used density. Remember, you can convert between molar mass and density with a variation of the ideal gas law. So today, we use molar mass.

• $\frac{Rate_{A}}{Rate_{B}}=\sqrt{\frac{MM_{B}}{MM_{A}}}$

The second form of Graham's law is an equation. On the equation's left side, we have the effusion rate ratio. We always put the lighter gas in the ratio's numerator. We denote it as gas A. The equation's right side compares the square roots of the molar masses. Notice that A and B have flipped. Gas A is in the denominator, and B is in the numerator.

What does the equation tell us? The equation helps us to compare two different gases. If we know the molar masses of A and B, we can determine that A will effuse X times faster than B. If we have the effusion rates for A and B, we can determine that B is X times heavier than A. Let's look at some example problems to get the hang of this!

Graham's Law Example Problems

There are a few steps for completing any effusion problem.

1. Write the rates and molar masses.

2. Use Graham's law.

3. Answer the question(s). Graham's law does not always answer the question. So, you may have to do more steps. The steps of each problem are unique to that problem.

Problem 1: The noble gases helium and neon are in a porous container. Which gas will escape faster? How much more quickly will it get out of the container?

1. We can answer the first question qualitatively. Helium will escape faster because it is the lighter of the two gas.

2. To answer the second question, we use Graham's law.

• Write the rates and weights.

• $\frac{Rate_{(He)}}{Rate_{(Ne)}}=?$

• $MM_{He}=4.00$

• $MM_{Ne}=20.18$

• Use Graham's Law.

• $\frac{Rate_{A}}{Rate_{B}} = \sqrt{\frac{MM_{B}}{MM_{A}}} \rightarrow \frac{Rate_{He}}{Rate_{Ne}}= \sqrt{\frac{20.18}{4.00}}= 2.25$

• Answer the question.

• Helium gas will escape two and a quarter times faster than neon.

Problem 2: We mixed an unknown noble gas with hydrogen gas. Hydrogen gas effuses 6.45 times faster than the other gas. What is the other gas?

• Write the rates and weights.

• $Rate_{X}=1$

• $Rate_{H_{2}}=6.45$

• $MM_{H_{2}}=2.01$ Remember that hydrogen gas is diatomic!

• $MM_{X}=?$

• Use Graham's law.

• $\frac{6.45}{1}=\sqrt{\frac{X}{2.01}} \rightarrow 6.45\times \sqrt{2.01}=\sqrt{X} \rightarrow X=9.14^{2}= 83.62 \frac{g}{mol}$

• Answer the question.

• The noble gas is krypton. The difference in the decimal places is because of rounding.

Problem 3: Let's say we have two porous containers. One has helium, and the other has argon. In 4 hours, half the helium escapes. How long will it take for two-thirds of the argon to get out? Assume both containers have identical conditions.

• Write the rates and weights.

• $Rate_{He}=X$

• $Rate_{Ar}=1$

• $MM_{He}=4.00$

• $MM_{Ar}=39.9$

• Plug everything into Graham's law.

• $\frac{Rate_{A}}{Rate_{B}}=\sqrt{\frac{MM_{B}}{MM_{A}}} \rightarrow X=\sqrt{\frac{39.9}{4.00}}=3.15$

• Answer the question.

• Determine the percentage of argon that's escaped after four hours.

• $\frac{Percent_{He/4\,hrs}}{Rate_{He}}=\frac{Percent_{Ar/4\,hrs}}{Rate_{Ar}} \rightarrow \frac{0.5}{3.15}=\frac{X}{1} \rightarrow x=0.16\, Percent_{Ar}/4\,hrs$

• Calculate how long it takes for two-thirds of the argon to leave the system.

• $\frac{0.16}{4\,hrs}=\frac{.667}{X} \rightarrow X=16.08\,hrs$

An Application of Graham's Law: Isotope Separation

Are you wondering when we use Graham's law in the real world? The most monumental application of Graham's law was the separation of different isotopes via diffusion for the atomic bomb. The original atomic bombs used Uranium-235. It's incredibly radioactive and readily undergoes fission. But, it's mixed with Uranium-238 which has a relative abundance of 99 percent. U-238 does not undergo fission. So, it can't be used in bombs or nuclear power plants. To separate the two, chemists vaporized uranium samples and diffused them through a series of tubes. U-235 is the smaller isotope. So, it diffused faster. We call this process enrichment.