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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(O_{2}) × √density(O_{2}) = 15 = rate(Cl_{2}) × √density(Cl_{2})

rate(O_{2}) × √density(O_{2}) = rate(Cl_{2}) × √density(Cl_{2})

Which we can rearrgange to give the more familiar expression:

* rate(O*_{2}) rate(Cl_{2}) |
= |
*√density(Cl*_{2}) √density(O_{2}) |

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

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

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:

Now, moles = mass ÷ molar mass

So we can replace "n" in the density equation with mass ÷ molar mass

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 × 10^{3} |

chlorine gas |
286 |
70.9 |
8.4 |
2.4 × 10^{3} |

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. = ½mν^{2}

and if we consider a system at constant temperature then:

ν^{2} |
∝ |
* 1 * m |

√ν^{2} |
∝ √ |
* 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!