 # Graham's Laws of Diffusion and Effusion Chemistry Tutorial

## Key Concepts

• Diffusion is the passage of a substance throughout another medium, eg, gases from an open bottle of perfume diffuse through the air.
• Effusion of a gas is its passage through a pinhole or orifice.
• Graham's Law of Diffusion: the rate of diffusion of one gas through another is inversely proportional to the square root of the density of the gas.

 rate of diffusion ∝ 1     √density rate of diffusion = constant × 1     √density

• Graham's Law of Effusion: the rate of effusion of a gas is inversely proportional to the square root of the density of the gas.

 rate of effusion ∝ 1     √density rate of effusion = constant × 1     √density

• Graham's Laws of Diffusion and Effusion:

rate = constant × (1 ÷ √d)
where d = density

for gases A and B under the same conditions,

 RateA = √dB RateB √dA

Since density is proportional to molecular mass (formula weight) or molar mass:

 RateA = √MB RateB √MA

where MA = molar mass (weight) of gas A
and MB = molar mass (weight) of gas B

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## Diffusion and Effusion

Have you ever been in a room when someone walks in, and, after a little while you can smell their Eau-de-Cologne, even though you have not moved closer to the source of the smell?
Or, have you been in a lift with a flatulent friend, and been unpleasantly surprised by an unwelcomed odour?
Or, have you walked pass a house and detected the succulent aroma arising from the cooking of a meal?
Clearly these smelly gases can travel a long way through the air! (and your nose is a pretty sensitive gas detector!)

I imagine this sort of experience is what prompted the English chemist Thomas Graham to study how fast gases travel through air.
Chemists refer to a gas travelling through air as diffusion in air, and the speed with which this happens is referred to as its rate.
In 1829 Thomas Graham reported the results of his observations on the rates of diffusion of various gases.

Imagine you are in the lab.
You stand at the far wall with two identical jars, one containing chlorine gas and the other containing oxygen gas.
Gas detectors are placed on the wall opposite you, 10 metres away.
If you open the two jars simultaneously, which gas will reach the gas detector quickest?

The results of the experiment might look like those in the table below:

gas time to reach detector
(seconds)
speed = distance ÷ time
oxygen gas 0.024 417
chlorine gas 0.035 286

The oxygen gas takes less time to reach the detector than chlorine gas so clearly oxygen gas has a greater rate of diffusion than chlorine gas, but why?

Graham compared the densities of gases and their rates of diffusion.1
The densities of oxygen gas and chlorine gas have been added to the table of results below:

gas speed
(m s-1)
density
(g mL-1)
oxygen gas 417 0.0013
chlorine gas 286 0.0029

The more dense chlorine gas took longer to reach the detector, that is, its rate of diffusion was slower than that of oxygen gas, but the relationship is not a simple linear relationship.
That is, the rate of diffusion of oxygen gas was 1.45 times faster than the rate of diffusion of chlorine gas, but chlorine is not 1.45 times more dense than oxygen (1.45 × 0.0013 = 0.0019 not 0.0029)
Neither is the relationship a simple inverse relationship since rate × density is not a constant
(417 × 0.0013 ≠ 286 × 0.0029)
So what is the relationship between rate and density?

What Graham found is that the rate of diffusion of a gas is inversely proportional to the square root of its density:

 rate of diffusion ∝ 1    √density

If we use a constant of proportionality we can turn this relationship into an equation:

 rate of diffusion = constant × 1    √density

Which can be rearranged to give:

rate of diffusion × √density = constant

For chlorine gas and oxygen gas at a temperature of 25°C (298 K) and a pressure of 101.3 kPa (1 atm).

gas rate density
(g mL-1)
√density rate × √density
= a constant
oxygen gas 417 0.0013 0.036 15
chlorine gas 286 0.0029 0.054 15

So now we can write:

rate(O2) × √density(O2) = 15 = rate(Cl2) × √density(Cl2)

rate(O2) × √density(O2) = rate(Cl2) × √density(Cl2)

Which we can rearrgange to give the more familiar expression:

 rate(O2) rate(Cl2) = √density(Cl2) √density(O2)

The density of a gas is related to its relative molecular weight (or its molar mass)

 density = mass   volume

Now, for an ideal gas we can use the Ideal Gas Law to calculate the volume occupied by the gas:

 volume = nRT P
where
n = moles of gas
R = gas constant
T = temperature (in Kelvin)
P = gas pressure

So the "volume" in the density equation can be replaced:

 density = mass   volume density = P × mass   nRT

Now, moles = mass ÷ molar mass
So we can replace "n" in the density equation with mass ÷ molar mass

 density = P × mass   nRT density = P × mass × molar mass mass × RT density = P × molar mass RT

Remember that R is a constant, the gas constant, so for a system at constant temperature and pressure the value of P/RT is a constant and this means that:

density ∝ molar mass

So, Graham's Law of diffusion can be rewritten as:

 rate of diffusion ∝ 1       √molar mass

And therefore, if we have two gases, A and B, the following is also true:

 rate(gas A) rate(gas B) = √molar mass(gas B) √molar mass(gas A)

Or, put another way:

rate(gas A) × √molar mass(gas A) = rate(gas B) × √molar mass(gas B)

For the oxygen gas and chlorine gas we used in the above experiment:

gas rate molar mass
(g mol-1)
√molar mass rate × √molar mass
= a constant
oxygen gas 417 32.0 5.6 2.4 × 103
chlorine gas 286 70.9 8.4 2.4 × 103

This means that the relative rate of diffusion of two gases can be used to determine their relative molecular mass (or molar mass).

Effusion is a different concept to diffusion.
Effusion refers to gas molecules under pressure escaping through a tiny orifice in the containing vessel.
The number of molecules that escape per unit time is equal to the number of molecules that hit the orifice.
The number of molecules that randomly hit the orifice is proportional to the average molecular speed (ν)

If we consider the kinetic energy (K.E.) of a molecule of mass m and speed ν then:

K.E. = ½mν2

and if we consider a system at constant temperature then:

 ν2 ∝ 1   m √ν2 ∝ √ 1   m ν ∝ √ 1   m

Remember we said above that the number of molecules that hit the orifice, and therefore escape from the vessel, was proportional to the speed of the molecules.
Then the number of particles escaping is inversely proportional to the square root of their mass (molecular mass or molar mass).

That is, the rate of effusion multiplied by the square root of the mass (relative molecular mass or molar mass) is a constant.
Which is the same expression as for the rate of diffusion!

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## Worked Example: Relative Rates of Diffusion of Gases

Question: Quantitatively compare the rates diffusion for equal moles of hydrogen gas and oxygen gas at the same temperature and pressure.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Compare rates of diffusion of hydrogen gas and oxygen gas
rate(H2(g)) = ? × rate(O2(g))

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions:
(a) constant temperature
(b) constant pressure
Names of gases:
(a) hydrogen gas ≡ H2(g)
(b) oxygen gas ≡ O2(g)
Relative quanities of gases:
moles of hydrogen gas = moles of oxygen gas
n(H2(g)) = n(O2(g))
3. What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
Then Graham's Law of Diffusion can be written as:

 rate(H2(g)) rate(O2(g)) = √molecular mass(O2(g)) √molecular mass(H2(g))

Use the Periodic Table to find the relative atomic mass (relative atomic weight) of hydrogen and oxygen atoms:
Mr(H) = 1.008
Mr(O) = 16.00

Calculate the relative molecular mass (relative molecular weight) of hydrogen gas and oxygen gas:
Mr(H2(g)) = 2 × 1.008 = 2.016
Mr(O2(g)) = 2 × 16.00 = 32.00

4. Substitute in the values for molecular mass and solve for relative rate of diffusion:

 rate(H2(g)) rate(O2(g)) = √molecular mass(O2(g)) √molecular mass(H2(g)) rate(H2(g)) rate(O2(g)) = √32.00 √2.016 rate(H2(g)) rate(O2(g)) = 5.6569 1.4199 rate(H2(g)) rate(O2(g)) = 3.984

Rearranging this equation we find that:

rate(H2(g)) = 3.984 × rate(O2(g))

A hydrogen molecule is about 2/32 or 1/16 as massive as an oxygen molecule.
We expect the less massive particle to travel faster than the more massive particle, that is, hydrogen gas should travel (diffuse) more rapidly.
This agrees with our calculated relative rate, so we confident our answer is plausible.
6. State your solution to the problem "compare rates of diffusion of hydrogen gas and oxygen gas ":

Hydrogen gas will diffuse 3.984 times faster than oxygen gas.

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## Worked Example: Using Rate of Effusion to Determine Molar Mass

Question: Gas X effuses through a pinhole at a rate of 4.73 × 10-4mol s-1.
Methane gas, CH4(g), effuses through the same pinhole at a rate of 1.43 × 10-3 mol s-1 under the same conditions of temperature and pressure.
What is the molar mass (weight) of gas X?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate molar mass of gas X
M(X(g)) = ? g mol-1

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions:
(a) constant temperature
(b) constant pressure
Rates of effusion through same pinhole:
rate(X(g)) = 4.73 × 10-4 mol s-1
rate(CH4(g)) = 1.43 × 10-3 mol s-1
3. What is the relationship between what you know and what you need to find out?
Assume Ideal Gas behaviour
Then Graham's Law of Effusion can be written as:

 rate(CH4(g)) rate(X(g)) = √molar mass(X(g)) √molar mass(CH4(g))

Which can then be rearranged to find the molar mass of gas X:

 √molar mass(CH4(g)) × rate(CH4(g)) rate(X(g)) = √molar mass(X(g))

Calculate the molar mass of CH4(g):
M(CH4(g)) = 12.01 + 4 × 1.008 = 12.01 + 4.032 = 16.042 g mol-1
4. Substitute in the values and solve for molar mass of gas X

 √molar mass(X(g)) = rate(CH4(g)) × √molar mass(CH4(g)) rate(X(g)) √molar mass(X(g)) = 1.43 × 10-3 × √16.042 4.73 × 10-4 √molar mass(X(g)) = 1.43 × 10-3 × 4.005 4.73 × 10-4 √molar mass(X(g)) = 12.11

Now, square both sides of the equation:

(√molar mass(X(g)))2 = 12.112

molar mass(X(g))= 147 g mol-1

CH4(g) effuses about 0.001/0.0001 or 10 times faster than gas X, so molecules of gas X must be more massive than molecules of CH4, therefore we expect gas X to have a greater molar mass than CH4.
147 is greater than 16.042, so our answer is plausible.
6. State your solution to the problem "molar mass of gas X ":

M(X(g)) = 147 g mol-1

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1. Remember the time, early 1800s, John Dalton had recently proposed that matter was made up of tiny invisible particles called atoms and that the atoms of one element were different to atoms of a different element. Graham needed to observe some gross property of the gases he studied, and density was an exellent choice!