Graham's Law: Diffusion And Effusion

Graham's Law: Diffusion And Effusion

Edited By Shivani Poonia | Updated on Jul 02, 2025 06:20 PM IST

Graham's law refers to the behavior of gas molecules during such processes as diffusion and effusion. The term 'diffusion' means the spread of gas molecules from an area of its higher concentration to an area of its lower concentration until uniform distribution is attained. Effusion is the process by which the gas molecules stream out from a tiny hole into a vacuum or low pressure.

This Story also Contains
  1. Graham’s Law of Diffusion
  2. Some Solved Examples
  3. Summary
Graham's Law: Diffusion And Effusion
Graham's Law: Diffusion And Effusion

Graham’s Law of Diffusion

According to it "At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density or molecular weight". It is applicable only at low pressure.

$r \propto \frac{1}{\sqrt{M}}$ or $\frac{1}{\sqrt{d}}$
Here r = rate of diffusion or effusion of a gas or liquid. M and d are the molecular weight and density respectively.

For any two gases, the ratio of the rate of diffusion at constant pressure and temperature can be shown as

$\mathrm{r}_1 / \mathrm{r}_2=\sqrt{\mathrm{M}_2 / \mathrm{M}_1}$ or $\sqrt{\mathrm{d}_2 / \mathrm{d}_1}$
Hence diffusion or effusion of a gas or gaseous mixture is directly proportional to the pressure difference of the two sides and is inversely proportional to the square root of the gas or mixture effusing or diffusing out.

Some Other Relation Based on Graham’s law
As r = V/t = Volume/time, thus:

$\begin{aligned} & \frac{V_1 t_2}{\mathrm{~V}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \text { As } \mathrm{r}=\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{\mathrm{w}}{\mathrm{t}} \\ & \text { Thus, } \frac{\mathrm{n}_1 \mathrm{t}_2}{\mathrm{n}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \frac{\mathrm{w}_1 \mathrm{t}_2}{\mathrm{w}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1} \\ & \frac{\mathrm{d}_1 \mathrm{t}_2}{\mathrm{~d}_2 \mathrm{t}_1}=\sqrt{\mathrm{M}_2 / \mathrm{M}_1}\end{aligned}$
Here n represents the number of moles, w represents weight in gram and d represents the distance traveled by a particular gas or liquid.

Differentiation Between Diffusion and Effusion

Diffusion
It is the movement of gaseous or liquid molecules without any porous bars that are, the spreading of molecules in all directions.

Effusion
It is the movement of gases molecules or liquid molecules through a porous bar that is, a small hole or orifice.

Uses of Graham’s Law

  • Detecting the presence of Marsh gas in mines.
  • Separation of isotopes by different diffusion rates. For example, U-235 and U-238
  • Detection of molecular weight and vapour density of gases using this relation.$\frac{\mathrm{r}_1}{\mathrm{r}_2}=\left(\frac{\mathrm{m}_2}{\mathrm{~m}_1}\right)^{1 / 2}$ or $\left(\frac{\mathrm{d}_2}{\mathrm{~d}_1}\right)^{1 / 2}$

Recommended topic video on (Graham's Law: Diffusion And Effusion)

Some Solved Examples

Example 1: A 4:1 molar mixture of He and CH4 is contained in a vessel at 20 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. What is the ratio of moles of He : CH4 in the mixture effusing out initially?

1)1: 8

2) 8: 1

3)1: 5

4)7: 1

Solution

We have
The molar ratio of He and CH4 is 4:1
Again, the total pressure is given = 20bar
Thus, the partial pressure ratio of He and CH4 = 16: 4

$\begin{aligned} & \frac{\mathrm{n}_{\mathrm{He}}}{\mathrm{n}_{\mathrm{CH}_4}}=\sqrt{\frac{\mathrm{M}_{\mathrm{CH} 4}}{\mathrm{M}_{\mathrm{He}}}} \times \frac{\mathrm{P}_{\mathrm{He}}}{\mathrm{P}_{\mathrm{CH}_4}} \\ & \sqrt{\frac{16}{4}} \times \frac{16}{4}=8: 1\end{aligned}$

Hence, the answer is the option (2).

Example 2: A balloon is filled with ethylene is pricked with a needle and quickly dropped in a tank of $\mathrm{H}_2$ gas under identical conditions. After a while, the balloon will:

1)Shrunk

2) Enlarge

3)Remain Unchanged.

4)Can't say

Solution

Rate of diffusion $\alpha \sqrt{\frac{1}{M}}$

This will diffuse into the balloon and its size will increase.

Hence, the answer is the option (2).

Example 3: Pure O2 diffuses through an aperture in 224 seconds, whereas a mixture of O2 and another gas containing 80% O2 diffuses from the same in 234 seconds. The molecular mass of the gas will be

1) 46.6

2)48.6

3)55

4)51.5

Solution

As we learned in

Graham’s law of diffusion -

$r_1 / r_2=\sqrt{M_1 / M_2}$ where r is the rate of diffusion of gas, and M is molar mass.

$\begin{aligned} & \frac{t_{m i x}}{t_{O_2}}=\frac{r_{O_2}}{r_{m i x}}=\sqrt{\frac{M_{m i x}}{32}} \\ & \frac{234}{224}=\sqrt{\frac{M_{\text {mix }}}{32}} \\ & M_{\text {mix }}=34.92 \\ & \therefore \chi_{O_2} \times M_{O_2}+\chi_g \times M_g=34.92 \\ & \Rightarrow 0.8 \times 32+0.2 \times M_g=34.92 \\ & \Rightarrow M_g=46.6\end{aligned}$

Hence, the answer is the option (1).

Example 4: $20 \mathrm{~cm}^3$ of $\mathrm{SO}_2$ diffuses through a porous partition in 60 seconds. Volume (in L) of $\mathrm{O}_2$ diffuse under similar conditions in 30 seconds will be :

1)12.14

2) 14.14

3)18.14

4)28.14

Solution

As we learnt in Graham’s law of diffusion -

$r_1 / r_2=\sqrt{M_1 / M_2}$ where r is the rate of diffusion of gas and M is the molar mass.

So, the volume of $\mathrm{SO}_2=20 \mathrm{~cm}^3$ and time = 60 seconds
Therefore, rate $=20 / 60$
And for Oxygen, we have time = 30 seconds. Thus,

$\begin{aligned} & \frac{20}{60} \times \frac{30}{V}=\sqrt{\frac{32}{64}} \\ & \mathrm{~V}=14.14 \mathrm{~cm}^3\end{aligned}$

Hence, the answer is the option (2).

Example 5: A 100 ml mixture containing 72% of $\mathrm{CH}_4$ by volume and the rest an unknown gas X was kept in a vessel. Due to a very fine crack, the mixture effused out. 21 ml of the mixture was lost and the remaining mixture contained 68.35% of methane by volume. Calculate the molar mass of gas X. All the measurements are made at the same temperature and pressure.

1)85.4

2) 87.1

3)77.3

4)94.2

Solution

We have an initial volume of mixture = 100ml

Thus, the mixture has 72ml of and 28ml of X.
Thus, we have:

$\frac{P_{C H_4}}{P_X}=\frac{72}{28}$

After the crack, the volume left = 100-21 = 79ml
Thus, the volume left after the crack $=(79 \times 68.35) / 100=54 \mathrm{ml}$
Thus diffused volume of $\mathrm{CH}_4=72-54=10 \mathrm{ml}$And diffused volume of $X=21-18=3 m l$
Now, we know that:

$\frac{V_{C H_4}}{V_X}=\sqrt{\frac{M_X}{M_{C H_4}}} \times \frac{72}{28}$

Thus, $M_X=87.1$

Hence, the answer is the option (2).

Summary

Graham's Law has numerous applications in various fields. Environmental scientists use it to explain the mixing of gases in the atmosphere. In medical science, it explains the process of gases during respiratory procedures and anesthetic applications. This law is applied industrially in separating isotopes and designing devices whose functions are pegged on controlled gas diffusion and effusion.

Frequently Asked Questions (FAQs)

1. What is Graham's Law and how does it relate to diffusion and effusion?
Graham's Law states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molecular mass. This means that lighter gases diffuse and effuse faster than heavier gases. In diffusion, gases spread throughout a space, while in effusion, gases pass through small openings.
2. How does the molecular mass of a gas affect its rate of diffusion?
The molecular mass of a gas is inversely related to its rate of diffusion. Gases with lower molecular masses diffuse faster than those with higher molecular masses. This is because lighter molecules have higher average velocities, allowing them to spread more quickly through space.
3. What's the difference between diffusion and effusion in gases?
Diffusion is the process by which gases spread and mix throughout a space, while effusion is the process of gases passing through small openings or pores. Both processes are governed by Graham's Law, but effusion specifically involves movement through a barrier.
4. Can Graham's Law be applied to liquids and solids?
Graham's Law specifically applies to gases. While diffusion does occur in liquids and solids, the rates and mechanisms are different due to the stronger intermolecular forces present in these states of matter. Graham's Law is not directly applicable to liquids and solids.
5. How does temperature affect the rate of diffusion according to Graham's Law?
While Graham's Law doesn't explicitly include temperature, increasing temperature does increase the rate of diffusion. Higher temperatures lead to greater kinetic energy of gas molecules, resulting in faster average velocities and more frequent collisions, which enhances the diffusion rate.
6. Why do perfumes with lower molecular mass fragrances spread faster in a room?
Perfumes with lower molecular mass fragrances spread faster because of Graham's Law. The lighter fragrance molecules have higher velocities and can diffuse more quickly through the air, allowing the scent to spread rapidly throughout the room.
7. How can Graham's Law be used to separate gases of different molecular masses?
Graham's Law can be applied in gas separation techniques like gaseous diffusion. Since lighter gases diffuse faster, they can be separated from heavier gases by allowing the mixture to diffuse through a porous barrier. This principle is used in uranium enrichment and other industrial processes.
8. What is the mathematical expression of Graham's Law?
The mathematical expression of Graham's Law is:
9. How does Graham's Law explain why helium escapes from balloons faster than air?
Helium has a lower molecular mass (4 g/mol) compared to the average molecular mass of air (about 29 g/mol). According to Graham's Law, the lighter helium molecules effuse through the tiny pores in the balloon material faster than the heavier air molecules, causing the balloon to deflate over time.
10. Can Graham's Law predict the exact time it takes for a gas to diffuse?
Graham's Law doesn't predict the exact time for diffusion, but rather the relative rates of diffusion or effusion between different gases. It allows us to compare how quickly one gas will diffuse or effuse compared to another based on their molecular masses.
11. How does the concept of mean free path relate to Graham's Law?
Mean free path is the average distance a gas molecule travels between collisions. Gases with lower molecular masses have longer mean free paths, which contributes to their faster diffusion rates as described by Graham's Law. Lighter molecules encounter fewer collisions, allowing them to travel further and diffuse more quickly.
12. Why is Graham's Law considered an empirical law?
Graham's Law is considered empirical because it was derived from experimental observations rather than theoretical principles. Thomas Graham discovered the relationship between diffusion rates and molecular masses through careful measurements, and the law was later explained by kinetic molecular theory.
13. How does pressure affect the application of Graham's Law?
Graham's Law applies equally at all pressures, as long as the gases behave ideally. However, very high pressures can cause deviations from ideal gas behavior, potentially affecting the accuracy of predictions made using Graham's Law.
14. Can Graham's Law be used to determine the molecular mass of an unknown gas?
Yes, Graham's Law can be used to estimate the molecular mass of an unknown gas. By comparing the diffusion or effusion rate of the unknown gas to that of a known gas under the same conditions, the molecular mass of the unknown gas can be calculated using the Graham's Law equation.
15. How does the kinetic molecular theory support Graham's Law?
Kinetic molecular theory supports Graham's Law by explaining that gases with lower molecular masses have higher average velocities at the same temperature. This higher velocity leads to faster diffusion and effusion rates, aligning with the predictions of Graham's Law.
16. What role does collision frequency play in Graham's Law?
Collision frequency is indirectly related to Graham's Law. Lighter gas molecules have higher velocities and thus experience more frequent collisions. However, they also have longer mean free paths. The balance between these factors results in the inverse square root relationship between diffusion rate and molecular mass described by Graham's Law.
17. How does Graham's Law apply to gas mixtures?
In gas mixtures, each gas component diffuses independently according to Graham's Law. The rate of diffusion for each gas in the mixture is inversely proportional to the square root of its own molecular mass, regardless of the presence of other gases.
18. Can Graham's Law explain why certain gases are more dangerous in enclosed spaces?
Yes, Graham's Law can help explain why some gases are more dangerous in enclosed spaces. Lighter toxic gases, like hydrogen sulfide or ammonia, can diffuse more quickly through air, potentially reaching harmful concentrations faster than heavier gases. This rapid diffusion can increase the risk in confined areas.
19. How does the size of gas molecules affect Graham's Law?
Graham's Law primarily considers molecular mass, not molecular size. However, size can indirectly affect diffusion rates through factors like intermolecular forces and collision cross-sections. In general, these effects are secondary to mass in determining diffusion rates for gases under normal conditions.
20. What is the relationship between Graham's Law and the root mean square speed of gas molecules?
The root mean square speed of gas molecules is inversely proportional to the square root of their molecular mass, which aligns with Graham's Law. This relationship arises from the kinetic energy equation and supports the observed behavior of gas diffusion and effusion rates.
21. How does Graham's Law apply to atmospheric gases?
Graham's Law explains why lighter atmospheric gases like hydrogen and helium are found in lower concentrations at lower altitudes. These gases diffuse upward more quickly than heavier gases like nitrogen and oxygen, leading to a gradual change in atmospheric composition with altitude.
22. Can Graham's Law be used to explain gas exchange in biological systems?
While Graham's Law doesn't fully explain gas exchange in biological systems, it does contribute to our understanding. For example, it helps explain why carbon dioxide (CO₂) diffuses out of blood slightly faster than oxygen (O₂) diffuses in, as CO₂ has a lower molecular mass than O₂.
23. How does Graham's Law relate to the concept of effusion half-life?
Effusion half-life is the time it takes for half of a gas to effuse through a small opening. Graham's Law can be used to compare the effusion half-lives of different gases. The effusion half-life is directly proportional to the square root of the molecular mass, aligning with Graham's Law.
24. What are the limitations of Graham's Law?
Graham's Law has several limitations: it assumes ideal gas behavior, doesn't account for molecular interactions or size effects, and doesn't consider external factors like temperature or pressure explicitly. It also becomes less accurate for gases with very high or very low molecular masses under certain conditions.
25. How does Graham's Law apply to the concept of gas chromatography?
Gas chromatography utilizes principles related to Graham's Law. While the separation in chromatography is primarily based on the interaction of gases with the stationary phase, the diffusion rates of gases through the mobile phase (carrier gas) can affect peak broadening and resolution, which are influenced by the molecular masses of the analytes.
26. Can Graham's Law explain why some gases are more easily contained than others?
Yes, Graham's Law helps explain containment difficulties for certain gases. Lighter gases like hydrogen and helium effuse through small openings or pores in containers more quickly than heavier gases. This makes them more challenging to contain and explains why special storage methods are often required for these gases.
27. How does Graham's Law relate to the concept of molecular diffusivity?
Molecular diffusivity, which quantifies a substance's ability to diffuse in a medium, is related to Graham's Law. While diffusivity depends on various factors, the molecular mass plays a significant role. Gases with lower molecular masses generally have higher diffusivities, aligning with the predictions of Graham's Law.
28. What is the significance of the square root in Graham's Law?
The square root in Graham's Law arises from the relationship between kinetic energy and velocity. Since kinetic energy is proportional to mass and velocity squared (KE = ½mv²), and all gases at the same temperature have the same average kinetic energy, the velocity must be inversely proportional to the square root of mass.
29. How does Graham's Law apply to the diffusion of gas pollutants in the atmosphere?
Graham's Law helps explain the dispersion patterns of gas pollutants in the atmosphere. Lighter pollutants will diffuse more quickly, potentially spreading over larger areas, while heavier pollutants may remain more concentrated near their source. This information is crucial for modeling air pollution and assessing environmental impacts.
30. Can Graham's Law be used to predict the behavior of gases in extreme conditions, such as in outer space?
Graham's Law can provide insights into gas behavior in space, but its application is limited. In the near-vacuum of space, gases behave more like individual molecules than continuous fluids. While lighter gases will still move faster, the concept of diffusion changes significantly due to the extremely low density and lack of a containing medium.
31. How does Graham's Law relate to the concept of vapor pressure?
While Graham's Law doesn't directly determine vapor pressure, it can help explain differences in evaporation rates. Substances with lower molecular mass tend to have higher vapor pressures and evaporate more quickly, which aligns with the faster diffusion rates predicted by Graham's Law for lighter molecules.
32. What role does Graham's Law play in understanding the composition of planetary atmospheres?
Graham's Law helps explain the composition of planetary atmospheres. Lighter gases like hydrogen and helium can escape a planet's gravity more easily due to their faster diffusion and effusion rates. This process, known as atmospheric escape, is why smaller, cooler planets struggle to retain light gases in their atmospheres.
33. How does Graham's Law apply to the concept of osmosis in gases?
While osmosis is typically associated with liquids, gaseous osmosis can occur across semipermeable membranes. Graham's Law influences this process by determining the relative rates at which different gases pass through the membrane. Lighter gases will generally osmose faster, following the same principle as effusion.
34. Can Graham's Law be used to explain the phenomenon of backdiffusion in gas systems?
Yes, Graham's Law can help explain backdiffusion. In systems where gases are flowing, lighter gases can more easily diffuse against the bulk flow direction due to their higher average molecular velocities. This phenomenon is important in various industrial processes and in understanding gas behavior in porous media.
35. How does Graham's Law relate to the concept of mean molecular speed in gases?
Graham's Law is closely related to mean molecular speed. The mean molecular speed of a gas is inversely proportional to the square root of its molecular mass, which directly aligns with Graham's Law. This relationship arises from the equipartition of energy in gases at the same temperature.
36. What is the significance of Graham's Law in understanding isotope separation?
Graham's Law is crucial in isotope separation techniques, particularly gaseous diffusion for uranium enrichment. Since isotopes have slightly different masses, they diffuse at different rates. This small difference, magnified through multiple stages, allows for the separation of isotopes, a process vital in nuclear technology and scientific research.
37. How does Graham's Law apply to the concept of gas permeation through membranes?
Graham's Law influences gas permeation through membranes, as lighter gases generally permeate faster. However, other factors like molecular size, membrane properties, and gas-membrane interactions also play significant roles. The overall permeation rate is a combination of diffusion (influenced by Graham's Law) and solubility in the membrane material.
38. Can Graham's Law explain why certain gases are more effective as heat transfer agents?
Graham's Law contributes to understanding gas effectiveness in heat transfer. Lighter gases, which diffuse faster, can more quickly distribute heat through a system. However, other properties like specific heat capacity and thermal conductivity are also crucial in determining a gas's overall effectiveness as a heat transfer agent.
39. How does Graham's Law relate to the concept of gas flux in porous materials?
Gas flux through porous materials is influenced by Graham's Law. The diffusion rates of gases through pores are inversely proportional to the square root of their molecular masses. This principle is important in understanding gas movement in materials like zeolites, activated carbon, and geological formations.
40. What role does Graham's Law play in understanding gas behavior in vacuum systems?
In vacuum systems, Graham's Law helps explain the behavior of residual gases. Lighter gases are pumped out more quickly due to their faster diffusion and effusion rates. This principle is important in designing and operating vacuum systems for various scientific and industrial applications.
41. How does Graham's Law apply to the concept of gas diffusion in soil?
Graham's Law influences gas diffusion in soil, affecting processes like soil aeration and the movement of greenhouse gases. Lighter gases like methane can diffuse through soil pores more quickly than heavier gases, impacting soil ecology and the global carbon cycle.
42. Can Graham's Law be used to explain the phenomenon of thermal diffusion in gases?
While Graham's Law primarily deals with concentration-driven diffusion, it relates to thermal diffusion (Soret effect) in gases. In a temperature gradient, lighter molecules tend to concentrate in warmer regions more than heavier ones, a behavior partially explained by the molecular speed differences described by Graham's Law.
43. How does Graham's Law relate to the concept of gas separation in industrial processes?
Graham's Law is fundamental to many industrial gas separation processes. Techniques like gaseous diffusion and effusion exploit the different diffusion rates of gases to achieve separation. This principle is applied in various industries, including petroleum refining, air separation, and hydrogen purification.
44. What is the significance of Graham's Law in understanding atmospheric mixing processes?
Graham's Law helps explain vertical mixing in the atmosphere. Lighter gases like water vapor can diffuse upward more quickly than heavier components of air. This process contributes to the distribution of gases and aerosols in different atmospheric layers, influencing weather patterns and climate.
45. How does Graham's Law apply to the concept of gas adsorption on surfaces?
While Graham's Law doesn't directly govern adsorption, it influences the rate at which gas molecules reach a surface to be adsorbed. Lighter gases, diffusing faster, may have more frequent interactions with the surface. However, the strength of adsorption depends more on the chemical properties of the gas and surface than on the gas's mass.
46. Can Graham's Law explain differences in the diffusion of isotopes in nuclear processes?
Yes, Graham's Law is crucial in understanding isotope diffusion in nuclear processes. Despite having the same chemical properties, isotopes of an element have slightly different masses. This mass difference leads to different diffusion rates, which is the basis for isotope separation techniques used in nuclear technology.
47. How does Graham's Law relate to the concept of gas permselectivity in membranes?
Graham's Law contributes to gas permselectivity in membranes, but it's not the only factor. While lighter gases generally diffuse faster through membranes, as predicted by Graham's Law, the overall selectivity also depends on factors like gas solubility in the membrane material and the membrane's pore size and structure.
48. What role does Graham's Law play in understanding gas exchange in leaves?
Graham's Law helps explain aspects of gas exchange in plant leaves. The diffusion of CO₂ into leaves and O₂ out of leaves through stomata is influenced by the molecular masses of these gases. However, other factors like concentration gradients and stomatal control mechanisms also play significant roles in this process.
49. How does Graham's Law apply to the concept of gas diffusion in porous catalysts?
In porous catalysts, Graham's Law influences the diffusion of reactant and product gases through the catalyst's pore structure. Lighter gases diffuse more quickly through the pores, potentially affecting reaction rates and selectivity. This principle is important in designing and optimizing heterogeneous catalysis processes.

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