# How to use Graham's Law to solve for time

A question from one of my GenChem practice exams:

"A 3.0 L sample of helium gas was placed in a container filled with a porous membrane. Half of the helium gas effused through the membrane in 24 hours. A 3.0 L sample of oxygen gas was placed in an identical container. How many hours will it take for half of the oxygen gas to effuse through the membrane?"

Seeing effusion here, one of the first things that came to mind was the formula from Graham's Law: $$\ce{\frac{rate of effusion of A}{rate of effusion of B} = \sqrt{\frac{m_\text{A}}{m_\text{B}}}}$$

But that, nor the original formula involving proportionality, has anything to do with time. I thought that maybe velocity could be manipulated to be related to time, but we don't know anything about the distance that has been covered by each molecule. How would you approach a problem like this?

• Any rate implicitly involves time. In your case, volumetric rates of effusion $(L/h)$ are relevant. Distance over time is not relevant here, volume over time is. Mar 1 at 20:22

Let:

A represent He

B represent $$\ce{O2}$$

$$\frac{\dot{V}_A}{\dot{V}_B}=\frac{\frac{V_A}{t_A}}{\frac{V_B}{t_B}}=\sqrt{\frac{M_B}{M_A}}$$

Solving for $$t_B$$:

$$t_B=t_A\left(\frac{V_B}{V_A}\right)\sqrt{\frac{M_B}{M_A}}$$

Substituting values:

$$t_B=24\left(\frac{1.5}{1.5}\right)\sqrt{\frac{32}{4}}$$

$$t_B=\pu{68h}$$

• Why would you use molar masses in this case? Mar 1 at 21:33
• @Mailbox Graham's law says that the rate of effusion of a gas is inversely proportional to the square root of its density. Equal volumes of gas at the same T and P contain equal numbers of gas molecules, so the rate of effusion is also inversely proportional to the square root of its molecular weight under said conditions. Mar 1 at 21:42