APCHEM Gasses
Effusion Rate (AP Chemistry Rundown)
Introduction
Effusion is the process of gas particles escaping through a tiny hole into a vacuum. Think of a balloon slowly deflating – that’s effusion in action, albeit a slow one. The rate of effusion is the speed at which this happens. It’s important to understand that effusion is different from diffusion. Diffusion is the mixing of gases, while effusion is the escape of a gas through a small opening. Diffusion
Graham’s Law of Effusion
Graham’s Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In simpler terms, lighter gases effuse faster than heavier gases.
Mathematically, this can be expressed as:
$$ \frac{Rate_1}{Rate_2} = \sqrt{\frac{M_2}{M_1}} $$
Where:
- $ Rate_1 $ is the rate of effusion of gas 1
- $ Rate_2 $ is the rate of effusion of gas 2
- $ M_1 $ is the molar mass of gas 1
- $ M_2 $ is the molar mass of gas 2
This relationship makes intuitive sense. Lighter molecules, on average, move faster than heavier molecules at the same Temperature. Kinetic Molecular Theory Therefore, they are more likely to encounter and pass through the small opening.
Example Calculations
Let’s work through a couple of examples to solidify our understanding.
Example 1: Compare the effusion rates of hydrogen gas (H₂) and oxygen gas (O₂).
- Molar mass of H₂ = 2.02 g/mol
- Molar mass of O₂ = 32.00 g/mol
$$ \frac{Rate_{H_2}}{Rate_{O_2}} = \sqrt{\frac{32.00}{2.02}} \approx 3.98 $$
This means hydrogen gas effuses approximately 3.98 times faster than oxygen gas.
Example 2: An unknown gas effuses 1.42 times faster than oxygen gas. What is the molar mass of the unknown gas?
Let the unknown gas be gas 1 and oxygen gas be gas 2.
$$ \frac{Rate_1}{Rate_2} = 1.42 $$ $$ M_2 = 32.00 , g/mol $$
$$ 1.42 = \sqrt{\frac{32.00}{M_1}} $$ $$ 1.42^2 = \frac{32.00}{M_1} $$ $$ M_1 = \frac{32.00}{1.42^2} \approx 15.9 , g/mol $$
The molar mass of the unknown gas is approximately 15.9 g/mol.
Applications of Effusion
Effusion has several practical applications, including:
- Isotope separation: Different isotopes of the same element have slightly different molar masses. This difference allows for separation using effusion, although the process is often slow and requires multiple stages. Isotopes
- Determining molar mass: As demonstrated in Example 2, effusion can be used to experimentally determine the molar mass of an unknown gas.
- Leak detection: The rate of effusion can be used to detect leaks in vacuum systems or other sealed containers.