Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its density.

Effusion refers to the movement of gas molecules from an enclosed space into a evacuated space.

The key concept to underlying Graham's Law is the following. The faster the speed of a molecule, the faster it will effuse. Let's compare two gases at the same temperature and pressure. Which gas will effuse faster? (Q32)

(Click here to view an animation of this process) Effusion2.html

Answer to question #32 at the end of this page.

It is important to define what we mean by "the faster it will effuse".

We can measure the time it takes the gas to effuse or we can determine the rate of effusion by measuring the number of molecules per second or the volume of gas relased per second. There is a difference between "rate" and "time".

The rate of effusion can be expressed in units of molecules/sec, cm/sec or cm3/sec.

 Because the density of a gas changes depending on the temperature, pressure and volume, it is more practical to use the moelcular weight of a gas in effusion problem calculations. The density of a gas is related to its molecular weight or molar mass.

PV = nRT

MW = mass/n

n = mass/MW

PV = mass RT/MW

P (MW/RT) = mass/V = density

The rate of effusion of a gas is inversely proportional to the square root of its molecular weight.

When comparing two gases at the the same pressure and temperature,

[ equation here ]

$$$$$$$$$$$$$$$$$$$$ Experiment $$$$$$$

To work an experimental simulation of Graham's Law of Effusion click here.




We can define the kinetic energy of a single gas moelcule as

KE = (1/2)mv2molecule

At a given temperature, different gases have the same average kinetic energy.

average KEgasA = average KEgasB

(1/2)mAuA2 = (1/2)mBuB2

Since the mass of gas A is different than the mass of Gas B, in order for this relationship to be true, the speed of the lighter molecule must be greater than the speed of the heavier gas molecule. In this case, the speed of the gas molecules is the root-mean-square speed of the molecules.

Where the root-mean-square speed, u, is the square root of the average speeds

of the molecules in a sample of gas at a specific temperature and pressure.

Does it makes sense that rate of effusion is inversly proportional to the molecular weight of the gas molecule?


Diffusion refers to the movement of gas molecules (Gas A) from a space where the gas A molecule are concentrated to a space or through a space in which there other gas molecules (Gas B) are present (a less concentrated space for Gas A).


(animation of this process) diffusionV8.html

Comparing two gases at the same pressure and temperature, lower molecular mass molecules diffuse faster than higher molecular mass molecules.


Determination of the relative rate of effusion of a gas

If we know the molecular weight of two gases, we can calculate t the relative rate of effusion of the gases (the rate of effusion of X is ____ as ast as Y).

Problem #31

Compare the relative rate of effusion of hydrogen gas to oxygen gas. At the same temperature and pressure, hydrogen gas effuses ________________as oxygen gas. Explain.


Problem #32

Hydrogen gas and oxygen gas are at the same temperature and pressure. We allow each gas to effuse through a small opening. After a very short time, approximately how many molecules of each gas would effuse?


Problem #33

A sample of O2 gas (2.0 mmol) effused through a pinhole in 5.0 sec. It will take _____ seconds for the same amont of H2 gas to effuse unde the same conditions.

Problem #34

A sample of O2 gas was found to effuse at a rate equal to two times that of an unknown gas. The molecular weight of the unknonw gas is ______________g/mol.

Determination of the molecular weight of a gas using Graham's Law of Effusion.

We can compare the rate of effusion of two gases, at the same temperature, from a container with a small hole in it into an evacuated space. If we know the molecular weight of one gas, we can calculate the molecular weight of another gas.

Problem #35

A sample of HI gas (MW = 128) effuses at 0.0962 cm/sec. A sample of butylamine gas effuses at 0.126 cm/sec. What is the molecular weight of butylamine?



Problem #36 Real World Application

One goal of the Manhattan Project (during WWII) was to create an atomic bomb using uranium. Only the pure uranium isotope 235U could be used for an atomic bomb, but 235U is the minor of the two isotopes. Chemists proposed reacting uranium with fluorne gas to form uranium hexafluoride gas. UF6 consists of a mixture of 235UF6 and 238UF6. 235U is a small fraction of the U in UF6.

235U represents 0.7% of uranium isotopes

238U represents 99.3% of uranium isotopes

To obtain pure 235U, UF6 gas was sent through a series of gas effusion tubes. Show how scientists knew that the separation of the two gases,35UF6 and 238UF6 , could be achived by sending the UF6 gas a series of long gas effusion tubes.This process actually occurred in Oak Ridge, Tennessee.


Solution to Problem #31

Hydrogen gas is lighter than oxygen gas therefore we should expect H2 to effuse at a faster rate than O2. Assuming equal temperature and pressure of the two gases, we can set up a ratio of the rates of effusion.

The rate of effusion is directly proportional to the rms speed of the molecules.


The rate of effusion of hydrogen gas is 4 times as fast as the rate of effusion for oxygen gas (compared at the same temperature and pressure).


Answer to question #32

When comparing two different gases at the same temperature, on average, a lower molecular mass gas molecule moves faster than a heigher molecular mass gas molecule. The faster the speed of a molecule, the faster it will effuse.

Molecules that are moving faster have more chances of passing through the small opening than slower moving molecules.


Solution to problem #33


Solution to Problme #36

The mass of 235UF6 < 238UF6 but the vrms of 235UF6 > 238UF6 . The 235UF6 effuses faster than 238UF6 , therefore the 235UF6 was separated from the 238UF6.