Graham's Law of Diffusion and Effusion

14th Mar 2019 @ 9 min read

Graham's law of diffusion (or Graham's law of effusion) is a law that expresses the relationship between the rate of diffusion or effusion to molar masses of particles. This empirical law was stated by Scottish chemist Thomas Graham in 1848. He established the relationship through experiments.

Difference between Effusion and Diffusion

Before we proceed further, it is necessary to understand the difference between effusion and diffusion. Many times, diffusion and effusion are interchangeably used, which is wrong; they both differ in meaning.

Diffusion

Diffusion is the movement of particles of one gas into another. The diffusion causes disorder in the system. Diffusion also occurs in liquids and solids but at a slower rate. The diffusion takes place because of the concentration difference. The particles move from higher concentration to lower concentration. A common example of diffusion is a perfume spray. When the perfume is spray in one corner of a room, we would smell it at another corner of the room.

Effusion

Effusion is the movement of particles of a gas through a tiny opening into an open container or space. Open container or space can be vacuum, atmosphere, or any other gas. In effusion, the atoms or molecules in an enclosed container are trying to escape through the aperture. When an air balloon is pricked, the gas inside the balloon starts escaping and the size of the balloon keeps shrinking. This phenomenon is the effusion of the balloon gas into the atmosphere. Another common example of the effusion is the gas leaking from a pipe.

Statement

The law states the rate of diffusion or effusion of gases is inversely proportional to the square root of their respective molar masses at a given temperature and pressure.

Expression

The rate of diffusion is given as:

$r \propto \frac{1}{\sqrt{M}}$

where r is the rate of diffusion, and M is the molar mass of a gas.

Explanation

As the law states the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This statement can be mathematically expressed as:

$r \propto \frac{1}{\sqrt{M}}$

$r = k \times \frac{1}{\sqrt{M}}$

where k is the constant of proportionality.

Graham's law is mostly used to compare the rates between different gases. For two gases: gas 1 and gas 2, the above expression can be rewritten as:

$\frac{r_1}{r_2} = \frac{\sqrt{M_2}}{\sqrt{M_1}} = \sqrt{\frac{M_2}{M_1}}$

Here, it is assumed that both the gases are at the same temperature and pressure.

The density of a gas is directly proportional to its molar mass at constant temperature and pressure. Hence, the above equation can be modified in terms of the density as:

$\frac{r_1}{r_2} = \frac{\sqrt{\rho_2}}{\sqrt{\rho_1}} = \sqrt{\frac{\rho_2}{\rho_1}}$

The graph of the rate of diffusion verses the molecular weight — Figure 2: Rate of diffusion or effusion for various gases. The rate of effusion is inversely proportional to the square root of molecular weights.

From the above figure, it can be observed that as the molecular weight increases, the rate of diffusion or effusion decreases. Thus, we can say that the diffusivity of heavier molecules is lesser than the lighter molecules.

Rate of diffusion or effusion is the volume of gas diffuse or effuse per unit time.

$r =\frac{\text{Volume diffuse or effuse}}{\text{time}}=\frac{V}{t}$

For two gases at the same temperature and pressure, we have

$\frac{r_1}{r_2} =\frac{t_2}{t_1}$

Finally, from the equations above

$\frac{r_1}{r_2} =\frac{t_2}{t_1} =\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt{\frac{M_2}{M_1}}$

Thus, we can conclude from the above equation, the rate of diffusion or effusion of a gas is inversely proportional to the square root of the density of the gas and molar mass of the molecules. And the time taken for the gas to diffuse or effuse is directly proportional to the square root of the density of the gas and molar mass of the molecules.

The rate of effusion of hydrogen is more than that of oxygen since hydrogen is lighter in comparison. So, Hydrogen effuses quickly. — Figure 3: Effusion of helium and oxygen. The helium is lighter and smaller than oxygen; so, it diffuses easily than oxygen

Kinetic Molecular Theory and Graham's Law

According to the kinetic theory, two different gases have the same kinetic energy at the same temperature.

$KE_1 =KE_2$

$\frac{1}{2}m_1v_1^2 = \frac{1}{2}m_2v_2^2$

where: v₁ and v₂ are the rms velocities of the molecule 1 and 2, m₁ and m₂ are the mass per molecule.

The mass per molecules (m) can also be expressed as molar mass (M) divided by Avogadro's constant (N_A).

$m_1 =\frac{M_1}{N_\text{A}} \qquad m_2 =\frac{M_2}{N_\text{A}}$

Substituting the above equation,

$\frac{1}{2} \frac{M_1}{N_\text{A}} v_1^2 = \frac{1}{2} \frac{M_2}{N_\text{A}} v_2^2$

$M_1 v_1^2= M_2 v_2^2$

$\frac{v_1^2}{v_2^2}= \frac{M_2}{ M_1 }$

$\frac{v_1}{v_2}= \sqrt{\frac{M_2}{ M_1 }}$

From the above expression, the rms velocities (rates) are inversely proportional to their respective molar masses. So, lighter molecules will travel faster than heavier molecules.

Finally,

$\frac{r_1}{r_2} =\frac{t_2}{t_1}=\frac{v_1}{v_2 } =\sqrt{\frac{\rho_2}{\rho_1}}= \sqrt{\frac{M_2}{ M_1 }}$

Applications of Graham's Law

Graham's law is very useful in separation processes. The gases with different densities can be separated using Graham's law.
It is also helpful in determining the molar mass of unknown gases by comparing the rate of diffusion of unknown gas to known gas.
We can separate the isotopes of an element using Graham's law. A common example is enriching uranium from its isotope. ²³⁵U is desired uranium which amounts 0.72 % of natural uranium. The separation of 235U from the rest isotope ²³⁸U is achieved by passing uranium hexafluoride gas into porous membranes. The permeate gets enriched in ²³⁵U each time it passes through the membranes. This is because the lighter isotope ²³⁵U diffuses easily compare to heavier isotope ²³⁵U.

Limitations of Graham's Law

Graham's law holds good for effusion, not for diffusion. It is an approximation for diffusion. The law is valid at constant temperature and pressure.

Examples

Example 1

Consider helium and neon gas enclosed in a container with a small aperture. Calculate the relative effusion rates?

As per Graham's law,

$\frac{r_\text{He}}{r_\text{Ne}} = \sqrt{\frac{M_\text{Ne}}{ M_\text{He }}} =\sqrt{\frac{20}{4} } =\sqrt{5} =2.2$

Therefore, the rate of effusion of helium is 2.2 times the rate of neon.

Example 2

Two flammable gases methane and hydrogen is flowing through a gas pipeline in the chemical industry. Which between the two gases will get leaked at a faster rate?

As per Graham's law,

$\frac{r_{\text{H}_2}}{r_{\text{CH}_4}} = \sqrt{\frac{M_ {\text{CH}_4} }{ M_ {\text{H}_2} }} =\sqrt{\frac{16}{2} } =\sqrt{8} =2.8$

The rate of effusion of hydrogen is 2.8 times the rate of methane. Hence, hydrogen gas will get leaked at a faster rate.

Example 3

A gas of empirical formula C_xH_y diffuses through a permeable membrane in 324 s. At the same temperature and pressure, hydrogen takes 90 s. Identify the gas?

As per Graham's law,

$\frac{t_{\text{H}_2}}{t_{\text{C}_x\text{H}_y}} =\sqrt{\frac{M_{\text{H}_2}}{M_{\text{C}_x\text{H}_y}}} =\sqrt{\frac{2}{12x+y}}$

$\frac{90}{324} =\sqrt{\frac{2}{12x+y}}$

$12x+y =25.92$

The above equation is approximately satisfying the condition: x = 2 and y = 2.

Therefore, the gas is C₂H₂ (acetylene).

Example 4

Uranium hexafluoride mixture consists of two isotopes: ²³⁵UF₆ and ²³⁸UF₆. We want to enrich ²³⁵UF₆ to 99.0 % from the present composition of 0.72 %. Calculate the relative rate of diffusion and the total number of steps to enrich ²³⁵UF₆ to the desired composition.

Let the molecular weight of ²³⁵UF₆ and ²³⁸UF₆ be M₂₃₅ and M₂₃₈.

$M_{235} &=235.043\,9+6\times 18.998\,4\\ &=349.034\,3\,\text{g}\,\text{mol}^{-1}$

$M_{238} &=238.050\,8+6\times 18.998\,4 \\ &=352.041\,2\,\text{g}\,\text{mol}^{-1}$

The relative rate of diffusion from Graham's law is

$\frac{r_{235}}{r_{238}}=\sqrt{\frac{M_{238}}{M_{235}}} =\sqrt{\frac{352.041\,2}{349.034\,3}} =1.004\,39$

The number of steps for purification be n.

$(1.004\,39)^n=\frac{0.990}{0.007\,2}=\frac{990}{7.2}$

$n=112\,4$

The number of steps is 112 4.

A large number of separators to enrich uranium-295 — Figure 4: Uranium enrichment plant. As we can see there is a large number of separation stages.
[Image source: University of Illinois, Urbana-Champaign]

Example 5

Consider a pipe as shown in the below figure. Two cotton balls are attached to both the ends of the pipe. One cotton ball is soaked in ammonia NH₃ solution and other in hydrogen chloride HCl. When ammonia and hydrogen chloride encounter one another, they react to form white ammonium chloride salt NH₄Cl. Determine the salt formed is the closest to ammonia ball or hydrogen chloride ball.

Ammonia diffuses to one end while HCl from the other end. — Figure 5: A pipe with two cotton balls attached to each end.

The molar mass of ammonia and hydrogen chloride is 17.0 g mol⁻¹ and 36.5 g mol⁻¹.

From Graham's law,

$\frac{r_{\text{NH}_3}}{r_\text{HCl}}=\sqrt{\frac{M_\text{HCl}}{M_{\text{NH}_3}}} =\sqrt{\frac{36.5}{17.0}} =1.46$

Ammonia diffuses 1.46 times faster than hydrogen chloride. Thus, ammonium chloride is the closest to hydrogen chloride.

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