The diffusion of a gas is the process of the gas spreading out from a high concentration to a lower concentration. If you spray perfume in the corner of a room, the perfume gas molecules would spread out in the room over time. Their rate of diffusion is affected by the root mean square velocity of the gas, so that heavier molecules diffuse slower. Effusion is the process of a gas escaping through a small hole in a container into a vacuum. Here again, heavier molecules effuse slower. We can use Graham's law of effusion to mathematically relate the rate of effusion of two gases to their molar masses. According to Graham's law, the rate of effusion of gas one divided by the rate of effusion of gas two is equal to the square root of the molar mass of gas two divided by the molar mass of gas one. Since the rate of effusion and the time it takes for gas to effuse are inversely proportional, we can also say this is equal to the time it takes for gas two to effuse divided by the time for gas one to effuse. Let's practice using Graham's law of effusion. Our problem reads, "An unknown gas effuses at a rate that is 0.462 times that of nitrogen gas at the same temperature. You know the gas must have come from a gas tank in your lab: either carbon dioxide, oxygen, bromine, chlorine, or xenon. What is the identity of the unknown gas?" We'll start with Graham's law of effusion. We can make nitrogen gas one and the unknown gas, gas two. In the problem, we're told the rate of effusion for the unknown gas is 0.462 times that of nitrogen gas, so we can replace the rate of effusion for the unknown gas with 0.462 times the rate of effusion for nitrogen gas. We can also substitute in the molar mass for nitrogen. Now we can solve this equation using algebra for the molar mass of the unknown. On the left side of the equation, the rate of effusion of nitrogen gas cancels out, and we're left with 1 divided by 0.462. To get rid of the square root on the right side of the equation, we can square both sides. We end up with what's inside the square root on the right side of the equation, which is the molar mass of the unknown divided by 28.02 grams per mole. On the left side of the equation, we get 4.68 when we square 1 divided by 0.462. Rearranging the equation to solve for the molar mass of the unknown, we get 4.68 times 28.02 grams per mole. This is equal to 131 grams per mole. This is the molar mass of xenon, which should be our unknown gas.