Core Concept

Graham’s Law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass. This law allows us to compare the rates at which different gases spread or escape through small openings.

Key Concepts

1. Effusion:

   • The process by which gas particles pass through a tiny opening from one container to another.

   • Example: Air leaking out of a punctured tire.

2. Diffusion:

   • The spreading of gas particles throughout a space or within another substance.

   • Example: The smell of perfume spreading in a room.

3. Relationship Between Rate and Molar Mass:

   • Graham’s Law states that lighter gases (those with lower molar mass) effuse or diffuse faster than heavier gases.

   • This is due to the higher average speed of lighter gas molecules at a given temperature.

Graham’s Law Formula

For two gases, Gas 1 and Gas 2, with molar masses $M_1$​ and $M_2$​, Graham’s Law is expressed as:

$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}$

Where:

• $\text{Rate}_1$​ and $\text{Rate}_2$​ are the rates of effusion (or diffusion) of Gas 1 and Gas 2.

• $M_1$​ and $M_2$​ are the molar masses of Gas 1 and Gas 2, respectively.

This equation shows that the rate of effusion (or diffusion) of a gas is inversely proportional to the square root of its molar mass. For example, if Gas 1 is lighter than Gas 2, Gas 1 will effuse or diffuse more quickly.

Rearranged Form for Finding Molar Mass

If the rate of effusion and the molar mass of one gas are known, Graham’s Law can also be rearranged to find the molar mass of an unknown gas:

$M_2 = M_1 \times \left( \frac{\text{Rate}_1}{\text{Rate}_2} \right)^2$

Where:

• $M_2$​ is the molar mass of the unknown gas.

• $\text{Rate}_1/\text{Rate}_2$​ is the ratio of the effusion rates.

Example 1: Comparing Effusion Rates

Problem: Hydrogen gas (H2\text{H}_2H2​) and oxygen gas (O2\text{O}_2O2​) are allowed to effuse through a small hole. If the molar mass of H2\text{H}_2H2​ is 2 g/mol2 \, \text{g/mol}2g/mol and that of O2\text{O}_2O2​ is 32 g/mol32 \, \text{g/mol}32g/mol, how many times faster will hydrogen effuse compared to oxygen?

Solution:

1. Identify Molar Masses:

   • M1=2 g/molM_1 = 2 \, \text{g/mol}M1​=2g/mol (Hydrogen)

   • M2=32 g/molM_2 = 32 \, \text{g/mol}M2​=32g/mol (Oxygen)

2. Use Graham’s Law:

   RateH2RateO2=MO2MH2=322=16=4\frac{\text{Rate}_{\text{H}_2}}{\text{Rate}_{\text{O}_2}} = \sqrt{\frac{M_{\text{O}_2}}{M_{\text{H}_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4RateO2​​RateH2​​​=MH2​​MO2​​​​=232​​=16​=4

Answer: Hydrogen gas will effuse 4 times faster than oxygen gas.

Example 2: Finding the Molar Mass of an Unknown Gas

Problem: An unknown gas effuses at half the rate of nitrogen gas (N2\text{N}_2N2​) under the same conditions. If the molar mass of nitrogen gas is 28 g/mol28 \, \text{g/mol}28g/mol, what is the molar mass of the unknown gas?

Solution:

1. Identify Given Information:

   • Rate of unknown gas = 0.5 (Rate of N2\text{N}_2N2​)

   • Molar mass of N2\text{N}_2N2​ (M1M_1M1​) = 28 g/mol28 \, \text{g/mol}28g/mol

2. Rearrange Graham’s Law:

   Munknown=MN2×(RateN2Rateunknown)2M_{\text{unknown}} = M_{\text{N}_2} \times \left( \frac{\text{Rate}_{\text{N}_2}}{\text{Rate}_{\text{unknown}}} \right)^2Munknown​=MN2​​×(Rateunknown​RateN2​​​)2

3. Substitute Values:

   Munknown=28 g/mol×(10.5)2=28 g/mol×4=112 g/molM_{\text{unknown}} = 28 \, \text{g/mol} \times \left( \frac{1}{0.5} \right)^2 = 28 \, \text{g/mol} \times 4 = 112 \, \text{g/mol}Munknown​=28g/mol×(0.51​)2=28g/mol×4=112g/mol

Answer: The molar mass of the unknown gas is 112 g/mol.

Practical Applications of Graham’s Law

1. Separating Isotopes:

   • Graham’s Law is used in separating isotopes of elements, such as uranium isotopes (235U^{235}\text{U}235U and 238U^{238}\text{U}238U) in nuclear fuel production.

2. Leak Detection:

   • Graham’s Law helps identify gas leaks by comparing effusion rates of different gases.

3. Breath Analysis:

   • Understanding effusion rates allows for the design of breath analyzers, which detect gases like alcohol vapor.