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Graham's Laws of Effusion and Diffusion

Key Concepts

Diffusion is the passage of a substance throughout another medium, eg, gases from an open bottle of perfume diffuse through the air.
Effusion of a gas is its passage through a pinhole or orifice.
Graham's Law of Diffusion: the rate of diffusion of one gas through another is inversely proportional to the square root of the density of the gas.
Graham's Law of Effusion: the rate of effusion of a gas is inversely proportional to the square root of the density of the gas.
Graham's Laws of Diffusion and Effusion:
rate = constant x (1 ÷ √d)
where d = density
for gases A and B under the same conditions,
Rate_A
= √d_B
Rate_B √d_A

Since density is proportional to molecular mass (formula weight) or molar mass:
Rate_A
= √M_B
Rate_B √M_A

where M = molar mass (weight)

Examples

Quantitatively compare the rates diffusion for equal moles of hydrogen gas and oxygen gas at the same temperature and pressure.
Organize the data:
M_H₂ = 2 g/mol
M_O₂= 32 g/mol
Write the equation:
Rate_H₂
= √M_O₂
Rate_O₂ √M_H₂

Substitute the known values:
Rate_H₂
= √32
Rate_O₂ √2

Rearrange the equation and solve:
Rate_H₂ = Rate_O₂ √ 16 = 4 x Rate_O₂
Hydrogen gas will diffuse 4 times faster than oxygen gas.
Gas X effuses through a pinhole at a rate of 4.73 x 10^-4mol s^-1.
Methane gas, CH_4(g), effuses through the same pinhole at a rate of 1.43 x 10^-3 mol s^-1 under the same conditions of temperature and pressure.
What is the molar mass (weight) of gas X?
Organize the data:
M_X = ? M_CH₄= 16 g/mol

rate_X = 4.73 x 10^-4 mol/s rate_CH₄ = 1.43 x 10^-3 mol/s

Write the equation:
Rate_X
= √M_CH₄
Rate_CH₄ √M_X

Rre-arrange this equation to solve for molecular mass of gas X:
√M_X
= rate_CH₄ √M_CH₄
Rate_X

Substitute in the known values:
√M_X
= 1.43 x 10^-3 √16
4.73 x 10^-4

√M_X
= 1.43 x 10^-3 x 4
4.73 x 10^-4

Solve for the square root of M_X:
√M_X = 12.093
Solve for the molar mass of X by squaring both sides of the equation:
M_X = 12.093² = 146.24 g/mol

Practice Questions

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