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Difficulty: MediumGraham's Law of Diffusion and Effusion

Under specified laboratory conditions, 120 cm3120\text{ cm}^3 of hydrogen gas (H2H_2) diffuses through a porous membrane in a given time interval. What volume of oxygen gas (O2O_2) will diffuse through the same membrane under identical conditions during the same time interval? [H=1,O=16H = 1, O = 16]

  1. A
    7.5 cm37.5\text{ cm}^3
  2. 30 cm330\text{ cm}^3Answer
  3. C
    60 cm360\text{ cm}^3
  4. D
    480 cm3480\text{ cm}^3

Answer

The volume of oxygen gas that will diffuse in the same time interval is 30 cm330\text{ cm}^3.
According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since oxygen (O2O_2, 32 g/mol32\text{ g/mol}) is 1616 times heavier than hydrogen (H2H_2, 2 g/mol2\text{ g/mol}), its rate of diffusion is 16=4\sqrt{16} = 4 times slower. In the same time interval, the volume of oxygen that diffuses is 120 cm34=30 cm3\frac{120\text{ cm}^3}{4} = 30\text{ cm}^3.

Step-by-Step Solution

1
Calculate the molar masses of hydrogen gas (H2H_2) and oxygen gas (O2O_2).
M(H2)=2×1=2 g/molM(H_2) = 2 \times 1 = 2\text{ g/mol}, and M(O2)=2×16=32 g/molM(O_2) = 2 \times 16 = 32\text{ g/mol}.
Graham's Law relates the rate of diffusion to the inverse square root of molecular masses.
2
Apply Graham's Law of Diffusion to find the ratio of rates of diffusion.
rH2rO2=M(O2)M(H2)=322=16=4\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M(O_2)}{M(H_2)}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4
The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
3
Relate the rate of diffusion to the volume of gas effused over a constant time interval (tt).
Since r=Vtr = \frac{V}{t} and time tt is constant, VH2VO2=4  ⟹  VO2=VH24=120 cm34=30 cm3\frac{V_{H_2}}{V_{O_2}} = 4 \implies V_{O_2} = \frac{V_{H_2}}{4} = \frac{120\text{ cm}^3}{4} = 30\text{ cm}^3.
The lighter gas diffuses 4 times faster than the heavier gas, so only a quarter of the volume of oxygen diffuses in the same time.

Key Concept

Graham's Law of Diffusion
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Topics

Graham's Law of Diffusion and Effusion

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