# Graham's Law of Effusion when Looking at Samples with Equal Mass?

So, I had a question on a chemistry test about two gas samples, one of O2 and one of N2, that had equal masses. We were given the time of effusion for the sample of N2 and then asked to calculate the time of effusion for O2. My initial reaction to the problem was to think that the nitrogen would effuse faster, as the effusion rate of nitrogen is clearly faster than that of oxygen. However, after doing the problem and adjusting for the number of mols of each sample, the oxygen had a faster time than the nitrogen. I got this question wrong (the "right answer" on the answer key had a time of effusion for oxygen that was slower than that of nitrogen). This doesn't make sense to me. When I asked my professor about it, they replied that we were supposed to use the equation t(N2)/t(O2)=sqrt(MolarMass(O2)/MolarMass(N2). I have my doubts that this is correct when looking at gas samples that have the same mass but not the same mols. It would seem that O2 should have a quicker effusion rate. Because r1/r2=sqrt(Mm2/Mm1), according to Graham's law, the ratio of Mm2/Mm1 is always going to be greater than ratio of r1/r2, so it would seem to me that the number of particles that need to diffuse should play a larger role in effusion time than how fast the individual particles are moving, at least when looking at samples of equal mass. Am I wrong here?

• Please ! Be serious ! $\ce{O2}$ and $\ce{N2}$ do not have the same mass ! – Maurice Feb 2 at 20:53
• Please read the question more carefully. I am talking about SAMPLES of N2 and O2 that have the same mass but different mols. I am not saying that O2 and N2 have the same mass. – Michael Douglas Feb 2 at 21:31
• Effusion is how isotopes of say UF$_6$ can separated. The rate of efflux of a gas with mol wt $m$ is $\rho\sqrt{RT/(2\pi m)}$ where $\rho$ is the gas density. Thus at the same density gasses undergo effusion as $1/\sqrt{m}$. (To be accurate this eqn. applies only if the mean free path is 10 times the size of the efflux pinhole.) – porphyrin Feb 3 at 9:42