Science, Tech, Math › Science Graham's Formula of Diffusion and Effusion Share Flipboard Email Print Thomas Graham. Wikipedia/Public Domain 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 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. Learn about our Editorial Process Updated on December 09, 2019 Graham's law expresses the relationship between the rate of effusion or diffusion of a gas and that gas's molar mass. Diffusion describes the spreading of a gas throughout a volume or second gas and effusion describes the movement of a gas through a tiny hole into an open chamber. In 1829, Scottish chemist Thomas Graham determined through experimentation that a gas's rate of effusion is inversely proportional to the square root of the gas particle's density. In 1848, he showed that the rate of effusion of a gas is also inversely proportional to the square root of its molar mass. Graham's law also shows that the kinetic energies of gases are equal at the same temperature. Graham's Law Formula Graham's law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. See this law in equation form below. r ∝ 1/(M)½ or r(M)½ = constant In these equations, r = rate of diffusion or effusion and M = molar mass. Generally, this law is used to compare the difference in diffusion and effusion rates between gases, often denoted as Gas A and Gas B. It assumes that temperature and pressure are constant and equivalent between the two gases. When Graham's law is used for such a comparison, the formula is written as follows: rGas A/rGas B = (MGas B)½/(MGas A)½ Example Problems One application of Graham's law is to determine how quickly a gas will effuse in relation to another and quantify the difference in rate. For example, if you want to compare the effusion rates of hydrogen (H2) and oxygen gas (O2), you can use their molar masses (hydrogen = 2 and oxygen = 32) and relate them inversely. Equation for comparing effusion rates: rate H2/rate O2 = 321/2 / 21/2 = 161/2 / 11/2 = 4/1 This equation shows that hydrogen molecules effuse four times faster than oxygen molecules. Another type of Graham's law problem may ask you to find the molecular weight of a gas if you know its identity and the effusion ratio between two different gases. Equation for finding molecular weight: M2 = M1Rate12 / Rate22 Uranium Enrichment Another practical application of Graham's law is uranium enrichment. Natural uranium consists of a mixture of isotopes with slightly different masses. In gaseous effusion, uranium ore is first made into uranium hexafluoride gas, then repeatedly effused through a porous substance. Through each effusion, the material passing through the pores becomes more concentrated in U-235 (the isotope used to generate nuclear energy) because this isotope diffuses at a faster rate than the heavier U-238. Cite this Article Format mla apa chicago Your Citation Helmenstine, Todd. "Graham's Formula of Diffusion and Effusion." ThoughtCo, Apr. 5, 2023, thoughtco.com/understand-grahams-law-of-diffusion-and-effusion-604283. Helmenstine, Todd. (2023, April 5). Graham's Formula of Diffusion and Effusion. Retrieved from https://www.thoughtco.com/understand-grahams-law-of-diffusion-and-effusion-604283 Helmenstine, Todd. "Graham's Formula of Diffusion and Effusion." ThoughtCo. https://www.thoughtco.com/understand-grahams-law-of-diffusion-and-effusion-604283 (accessed June 7, 2023). copy citation Featured Video