What is Graham’s Law of Diffusion?
The Principle of Graham’s Law
Graham’s Law of Diffusion states that the rate of diffusion or effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it is expressed as:
\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]
Here:
- \( r_1 \) and \( r_2 \): Rates of diffusion or effusion of two gases
- \( M_1 \) and \( M_2 \): Molar masses of the two gases
Historical Context
Graham’s Law was formulated by Thomas Graham, a 19th-century Scottish chemist, known for his significant contributions to the study of gas behavior. His work on diffusion laid the foundation for modern physical chemistry and kinetic theory of gases.
Applications of Graham’s Law in Real Life
Graham’s Law has several important applications in real-world scenarios, especially in industries that involve gases. Some of these applications include:
- Gas Separation: In industrial processes, Graham’s Law is used to separate different gases. Lighter gases diffuse more rapidly than heavier ones, allowing for effective separation techniques, such as in air liquefaction or hydrogen separation.
- Breathable Air Composition: In the field of respiratory chemistry, understanding how gases diffuse helps in calculating the ideal mixture of gases in the air to optimize human breathing.
- Medical Gas Delivery: Graham’s Law is used in medical devices such as oxygen and anesthesia delivery systems, where understanding the rate of diffusion of gases is essential for proper dosage and safety.
- Pollution Control: In environmental science, the diffusion of gases affects the spread of pollutants. By understanding diffusion rates, scientists can predict how pollutants will spread through the atmosphere, which is critical for air quality control.
The Mathematics of Graham’s Law
Formula and Equations
The primary formula of Graham’s Law is derived from the kinetic molecular theory, which relates the rates of diffusion or effusion to the square roots of the molar masses of gases.
The equation is used to calculate any one of the four variables (\( r_1, r_2, M_1, M_2 \)) if the other three are known.
Unit Consistency in Calculations
The standard SI unit for the rate of diffusion is meters per second (\( \text{m/s} \)), while molar mass is typically expressed in grams per mole (\( \text{g/mol} \)). When performing calculations, ensure consistency in units. Common conversions include:
- \( 1 \, \text{cm/s} = 0.01 \, \text{m/s} \)
- \( 1 \, \text{mm/s} = 0.001 \, \text{m/s} \)
How to Derive Graham’s Law
Underlying Principles
The law is based on the kinetic theory of gases, which states that the average kinetic energy of gas particles is proportional to the absolute temperature. For two gases at the same temperature:
\[ \frac{\text{KE}_1}{\text{KE}_2} = 1 \quad \text{or} \quad \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2} = 1 \]
Rearranging and substituting molar masses for particle masses leads to the relationship:
\[ \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} \]
Since diffusion or effusion rates are proportional to particle velocities, the formula for Graham’s Law emerges.
Examples of Graham’s Law in Action
Example 1: Calculating \( r_1 \)
Given:
- \( r_2 = 6.0 \, \text{m/s} \)
- \( M_1 = 18 \, \text{g/mol} \) (water vapor)
- \( M_2 = 32 \, \text{g/mol} \) (oxygen)
Calculation:
\[ r_1 = r_2 \cdot \sqrt{\frac{M_2}{M_1}} = 6.0 \cdot \sqrt{\frac{32}{18}} = 6.0 \cdot 1.33 = 7.98 \, \text{m/s} \]
Example 2: Calculating \( r_2 \)
Given:
- \( r_1 = 5.0 \, \text{m/s} \)
- \( M_1 = 2 \, \text{g/mol} \) (hydrogen)
- \( M_2 = 44 \, \text{g/mol} \) (carbon dioxide)
Calculation:
\[ r_2 = \frac{r_1}{\sqrt{\frac{M_2}{M_1}}} = \frac{5.0}{\sqrt{\frac{44}{2}}} = \frac{5.0}{4.69} = 1.07 \, \text{m/s} \]
Example 3: Calculating \( M_1 \)
Given:
- \( r_1 = 2.5 \, \text{m/s} \)
- \( r_2 = 1.0 \, \text{m/s} \)
- \( M_2 = 28 \, \text{g/mol} \) (nitrogen)
Calculation:
\[ M_1 = M_2 \cdot \left(\frac{r_2}{r_1}\right)^2 = 28 \cdot \left(\frac{1.0}{2.5}\right)^2 = 28 \cdot 0.16 = 4.48 \, \text{g/mol} \]
Example 4: Calculating \( M_2 \)
Given:
- \( r_1 = 4.0 \, \text{m/s} \)
- \( r_2 = 2.0 \, \text{m/s} \)
- \( M_1 = 16 \, \text{g/mol} \) (methane)
Calculation:
\[ M_2 = M_1 \cdot \left(\frac{r_1}{r_2}\right)^2 = 16 \cdot \left(\frac{4.0}{2.0}\right)^2 = 16 \cdot 4 = 64 \, \text{g/mol} \]
FAQs About Graham’s Law
1. Can Graham’s Law apply to liquids?
No, it applies specifically to gases due to assumptions based on kinetic theory.
2. What happens at very high pressures?
At high pressures, gases deviate from ideal behavior, affecting the accuracy of Graham’s Law.
3. Can it be used for gas mixtures?
Yes, with care, the law can be extended to mixtures by considering individual component properties.
4. Is molar mass the only factor?
No, temperature and intermolecular interactions can also affect diffusion rates.
5. How accurate is Graham’s Law?
It’s precise for ideal gases but less so for real gases under extreme conditions.
Common Misconceptions
1. Graham’s Law works for liquids.
Incorrect; it only applies to gases.
2. The law is independent of temperature.
Not true; diffusion rates depend on temperature.
3. Real gases always follow Graham’s Law.
Real gases deviate at high pressures and low temperatures.
4. Diffusion and effusion are identical.
Effusion refers to gas escaping through a hole without collisions, unlike diffusion.
5. Rates depend solely on molecular weights.
Other factors like intermolecular forces can also influence the rate.
