Science, Tech, Math › Science Graham's Law Example: Gas Diffusion-Effusion Share Flipboard Email Print Daisuke Kondo / Getty Images Science Chemistry Chemical Laws Basics Molecules Periodic Table Projects & Experiments Scientific Method Biochemistry Physical Chemistry Medical Chemistry Chemistry In Everyday Life Famous Chemists Activities for Kids Abbreviations & Acronyms Biology Physics Geology Astronomy Weather & Climate By Todd Helmenstine Todd Helmenstine is a science writer and illustrator who has taught physics and math at the college level. He holds bachelor's degrees in both physics and mathematics. our editorial process Todd Helmenstine Updated July 03, 2019 Graham's law is a gas law which relates the rate of diffusion or effusion of a gas to its molar mass. Diffusion is the process of slowly mixing two gases together. Effusion is the process that occurs when a gas is permitted to escape its container through a small opening. Graham's law states that the rate at which a gas will effuse or diffuse is inversely proportional to the square root of the molar masses of the gas. This means light gasses effuse/diffuse quickly and heavier gases effuse/diffuse slowly. This example problem uses Graham's law to find how much faster one gas effuses than another. Graham's Law Problem Gas X has a molar mass of 72 g/mol and Gas Y has a molar mass of 2 g/mol. How much faster or slower does Gas Y effuse from a small opening than Gas X at the same temperature? Solution: Graham's Law can be expressed as: rX(MMX)1/2 = rY(MMY)1/2 whererX = rate of effusion/diffusion of Gas XMMX = molar mass of Gas XrY = rate of effusion/diffusion of Gas YMMY = molar mass of Gas Y We want to know how much faster or slower Gas Y effuses compared to Gas X. To get this value, we need the ratio of the rates of Gas Y to Gas X. Solve the equation for rY/rX. rY/rX = (MMX)1/2/(MMY)1/2 rY/rX = [(MMX)/(MMY)]1/2 Use the given values for molar masses and plug them into the equation: rY/rX = [(72 g/mol)/(2)]1/2rY/rX = 1/2rY/rX = 6 Note that the answer is a pure number. In other words, the units cancel out. What you get is how many times faster or slower gas Y effuses compared to gas X. Answer: Gas Y will effuse six times faster than the heavier Gas X. If you were asked to compare how much more slowly gas X effuses compares to gas Y, just take the inverse of the rate, which in this case is 1/6 or 0.167. It doesn't matter what units you use for the rate of effusion. If gas X effuses at 1 mm/minute, then gas Y effuses at 6 mm/minute. If gas Y effuses at 6 cm/hour, then gas X effuses at 1 cm/hour. When Can You Use Grahams's Law? Graham's law may only be used to compare the rate of diffusion or effusion of gases at a constant temperature.The law breaks down, like other gas laws, when the concentration of gases becomes very high. The gas laws were written for ideal gases, which are at low temperatures and pressures. As you increase the temperature or pressure, you can expect the predicted behavior to deviate from experimental measurements.