Under identical conditions of temperature and pressure, oxygen gas (O_2) diffuses through … — JAMB UTME

Under identical conditions of temperature and pressure, oxygen gas ( $O_2$ ) diffuses through a fine orifice at a rate of $15\text{ cm}^3/\text{s}$ . What is the rate of diffusion of hydrogen gas ( $H_2$ ) in $\text{cm}^3/\text{s}$ under the same conditions? (Relative atomic masses: $H = 1$ , $O = 16$ )

Answer: 60 / 60 cm^3/s / 60 cm³/s / 60 cm3/s

Answer

60 cm³/s

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 (

r \propto \frac{1}{\sqrt{M}}

). Molar mass of

O_2 = 32\text{ g/mol}

and molar mass of

H_2 = 2\text{ g/mol}

. The ratio

\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

. Therefore,

r_{H_2} = 4 \times 15\text{ cm}^3/\text{s} = 60\text{ cm}^3/\text{s}

Step-by-Step Solution

Calculate the molar masses of oxygen (

O_2

) and hydrogen (

H_2

M(O_2) = 2 \times 16 = 32\text{ g/mol}

and

M(H_2) = 2 \times 1 = 2\text{ g/mol}

Graham's law relates rate of diffusion directly to the inverse square root of molar mass.

Apply Graham's Law of Diffusion ratio formula:

\frac{r_{H_2}}{r_{O_2}} = \sqrt{\frac{M(O_2)}{M(H_2)}}

\frac{r_{H_2}}{15} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4

The lighter gas diffuses faster by a factor equal to the square root of the ratio of their molar masses.

Solve for the rate of diffusion of hydrogen (

r_{H_2}

r_{H_2} = 4 \times 15\text{ cm}^3/\text{s} = 60\text{ cm}^3/\text{s}

Multiply the rate of diffusion of oxygen by the calculated relative rate ratio.

Key Concept

Graham's Law of Diffusion

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