An experimental effusion cell measures gas diffusion rates through a micro-porous membrane… — JAMB UTME

An experimental effusion cell measures gas diffusion rates through a micro-porous membrane at fixed temperature and pressure. In a calibration run, $120\text{ cm}^3$ of neon gas ( $\text{Ne}$ , atomic mass $= 20\text{ g/mol}$ ) effuses through the membrane in $30\text{ seconds}$ . Calculate the volume (in $\text{cm}^3$ ) of sulfur trioxide gas ( $\text{SO}_3$ , atomic masses: $\text{S} = 32\text{ g/mol}$ , $\text{O} = 16\text{ g/mol}$ ) that will effuse through the exact same membrane in $50\text{ seconds}$ .

Answer: 100 / 100 cm3 / 100cm3 / 100 cm^3 / 100 cm³

Answer

100 cm³

The effusion rate of neon is

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

. Given

M_{\text{Ne}} = 20\text{ g/mol}

and

M_{\text{SO}_3} = 80\text{ g/mol}

, Graham's law dictates

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = 2

. Thus,

r_{\text{SO}_3} = \frac{4}{2} = 2\text{ cm}^3/\text{s}

. Over a period of

50\text{ seconds}

, the volume of

\text{SO}_3

effused is

2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

Step-by-Step Solution

Calculate the rate of effusion of neon gas (

r_{\text{Ne}}

)

r_{\text{Ne}} = \frac{120\text{ cm}^3}{30\text{ s}} = 4\text{ cm}^3/\text{s}

Effusion rate is defined as the volume of gas effusing per unit time.

Determine the molar mass of sulfur trioxide (

\text{SO}_3

)

M_{\text{SO}_3} = 32 + 3(16) = 80\text{ g/mol}

The molar mass is calculated from the constituent relative atomic masses of sulfur and oxygen.

Apply Graham's Law of Effusion to determine the effusion rate of sulfur trioxide (

r_{\text{SO}_3}

)

\frac{r_{\text{Ne}}}{r_{\text{SO}_3}} = \sqrt{\frac{M_{\text{SO}_3}}{M_{\text{Ne}}}} \implies \frac{4}{r_{\text{SO}_3}} = \sqrt{\frac{80}{20}} = \sqrt{4} = 2 \implies r_{\text{SO}_3} = 2\text{ cm}^3/\text{s}

Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

Compute the total volume of sulfur trioxide effused in

50\text{ seconds}

V_{\text{SO}_3} = r_{\text{SO}_3} \times t = 2\text{ cm}^3/\text{s} \times 50\text{ s} = 100\text{ cm}^3

Multiplying the calculated rate of effusion by the given time yields the total volume.

Key Concept

Graham's Law of Diffusion and Effusion

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