Question

If 4g of oxygen diffuse through a very narrow hole, how much would have diffused under identical conditions?

Answer 1

Diffusion (and effusion),in fundamental terms, follow Graham's law, which can be stated as:

Rate1/Rate2 = √(M2/M1)

where:

Rate1 and Rate2 = rates of diffusion of the two gases (often in terms of moles per

unit time)

M1 and M2 = molar masses of the the two species (g/mol)

Let's call hydrogen species 1 here, and oxygen species 2.

We are told 4 g of O2 diffuse in a certain time. 4 g of O2 = 1/8 mol, since the molar mass of O2 is 32 g/mol. So the rate at which O2 is diffusing is (1/8) mol in t seconds, or ((1/8)/t) mol/s

So, for quantities to put into Graham's law, we have:

M1 = 2 g/mol (for hydrogen)

M2 = 32 g/mol (for oxygen)

Rate1 = unknown (we are trying to solve for this)

Rate2 = ((1/8)/t) mol/s

So, using Graham's law, we can now solve for the rate of hydrogen flow in mol/s.

(Rate1/Rate2) = √(M2/M1) --> (Rate1/[(1/8)/t)]) = √(32/2) = √16 = 4

Thus, hydrogen diffuses four times faster than oxygen, on a mole/second basis.

So, Rate1 = [(1/8)/t]*4 = (4/8)/t = (1/2)/t

Now, the diffusion of hydrogen is said to occur over the same period as the diffusion of oxygen, so Rate1 is measured over the same time. Thus:

Rate1 = (mol of hydrogen diffusing)/t,where t is the same t as above (for Rate2).

So, combining the last two equations, we get:

(number of moles of hydrogen diffusing)/t= (1/2)/t

Thus, the number of moles of hydrogen diffusing is 1/2 mol. 1/2 mol of molecular

hydrogen has a mass of 1/2*(2 g/mol) = 1 g.

Thus, 1 g of hydrogen would diffuse in the same time as 4 g of oxygen. This may

seem counter-intuitive, but remember that Graham's law relates molar diusion

rates, not mass diffusion rates. And, even though the hydrogen diffuses

faster, it is also 16 times lighter than oxygen. So, in mass terms, only half as much mass of hydrogen diffuses in the same time.

Answer 2

Explanation:

According to Graham's law

Rate1/Rate2 = squareroot of M2/M1

H2/O2 = squareroot 32/2 =4

4g of O2 = 1/8 moles of O2

their for H2 = 1/8/t×4 = (1/2)/t

H has a mass of 1/2×2 = 1g