A question on Graham's law of effusion...

Question

If .25 moles of a gas diffuses in 10 minutes and another gas of known molecular M1 mass diffuses by 0.5 moles in 20 minutes then the molecular mass of the unknown gas can be found by Graham's law:

r1/r2=sqrt(M2/M1). But this is only applicable at constant pressure. When some amount of gas effuses the pressure inside it decreases and so do the rate. So the ratio of the rates are not always constant . Then how can we use this formula. Please help!

Answer 1

Graham's law, which applies to both effusion and diffusion

M1r1^2 = M2r2^2

Graham's law assumes constant T and P.  It's really just a snapshot in time.  The rates are instantaneous rates, and it looks at the rate at which gas 1 flows through a pinhole into gas 2, and the rate at which gas 2 flows through the same pinhole into gas 1.

If you examine the effusion of a gas over a long period of time, you don't use Graham's law.  The rate of effusion will be greatest at the beginning of the time interval while the pressure is greatest.  The rate will decrease over time.  In fact, the pressure will follow first-order kinetics, as does radioactive decay or a decomposition reaction.

P = Po e^-kt

Where P is the pressure at some time t, Po is the pressure at time zero, t is time and k is a constant which includes the molar mass, the volume of the container, the temperature, and the area of the pinhole.  This value of k will be related to the probability of a gas molecule hitting the hole and will have units of 1/time.

A plot of the natural log of P vs t will be linear with a slope equal to -k.

At two different times we can write:

ln P1 = -kt1 + ln Po

and

ln P2 = -kt2 + ln Po

We can subtract the second from the first, eliminate Po and get:

ln (P1/P2) = k(t2 - t1)