Question

At \(27^{\circ} \mathrm{C}, 500 \mathrm{~mL}\) of helium diffuses in 30 minutes. What is the time (in hours) taken for \(1000 \mathrm{~mL}\) of \(\mathrm{SO}_{2}\) to diffuse under same experimental conditions? (a) 240 (b) 340 (c) 200 (d) 440

Short answer

The time taken is 4 hours (240 minutes); option (a) 240.

Step-by-step solution

Understand the relationship using Graham's Law

Graham's law of diffusion states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Mathematically, it is given by: \[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \] where \(r_1\) and \(r_2\) are the rates of diffusion and \(M_1\) and \(M_2\) are the molar masses of the gases.

Calculate rate of diffusion for helium

The problem states that 500 mL of helium diffuses in 30 minutes. The rate of diffusion \(r_{\text{He}}\) for helium can be computed as \(\frac{500}{30}\) mL/min.

Calculate rate of diffusion for SO2

We want to find the time for the diffusion of 1000 mL of \(\text{SO}_2\). The rate of diffusion \(r_{\text{SO}_2}\) is \(\frac{1000}{t}\) where \(t\) is the time in minutes.

Find molar mass of the gases

The molar mass of helium \(M_{\text{He}}\) is 4 g/mol and for \(\text{SO}_2\), \(M_{\text{SO}_2}\) is calculated as \(32 + 2 \times 16 = 64\) g/mol.

Apply Graham's Law

Substitute values into Graham's Law: \[ \frac{500/30}{1000/t} = \sqrt{\frac{64}{4}} \]This simplifies to: \[ \frac{5}{30} \cdot \frac{t}{10} = \sqrt{16} \]Which further simplifies to: \[ \frac{t}{60} = 4 \]

Solve for time

Multiply both sides by 60 to solve for \(t\):\[ t = 60 \times 4 = 240 \text{ minutes} \]Convert minutes to hours by dividing by 60, resulting in 4 hours.

Key concepts

Rate of Diffusion

The rate of diffusion describes how fast a gas spreads out in a given space. According to Graham's Law, the rate of diffusion is inversely related to the square root of the gas's molar mass. This means lighter gases diffuse more quickly than heavier ones. When solving for a gas's rate of diffusion, you can use the formula:

• \[ r \ = \ \frac{volume}{time} \]

In simpler terms, this formula calculates how much gas changes position over a given period. For example, in the exercise, the diffusion rate for helium was computed by dividing the volume (500 mL) by the time it took to diffuse (30 minutes). This provided a rate of diffusion for helium in mL/minutes. Understanding this concept is crucial for processes like predicting how long a fragrance might spread in the air or how quickly a balloon shrinks over time.

Molar Mass Calculation

Calculating molar mass helps determine how different gases behave when they diffuse. Molar mass is the mass of one mole of a substance, usually measured in g/mol. You determine it by summing the atomic masses of all the elements in the compound.
For instance, helium has a molar mass of 4 g/mol, as it's composed of individual atoms. In contrast, sulfur dioxide (\( SO_2 \)) is composed of sulfur and oxygen atoms. Its molar mass is calculated by adding: the atomic mass of sulfur (32 g/mol) + twice the atomic mass of oxygen (16 g/mol each), resulting in:

• \[ M_{\text{SO}_2} = 32 + 2 \times 16 = 64 \, \text{g/mol} \]

Calculating molar masses allows you to effectively apply Graham's Law for comparing diffusion rates between different gases and predicting how each will behave under the same conditions.

Gas Effusion

Gas effusion involves a gas passing through a small opening into a vacuum. While diffusion describes spreading in a space, effusion is about escaping through tiny openings. Both processes are analyzed using Graham's Law.

• The formula for effusion using Graham's Law is the same:\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]

This equation shows how the rate of gas escaping through a small hole is influenced by its molar mass. Just like diffusion, lighter gases effuse more rapidly than heavier ones. Understanding this helps in scenarios like determining the speed at which a gas leaks from a container or the time it takes for a scent to escape a package.
In the provided exercise, this principle of Graham's Law underlies the calculation of how quickly helium and sulfur dioxide diffuse (or hypothetically effuse) under equivalent conditions. Knowing these properties allows chemists and engineers to predict and manipulate the behavior of various gases in practical applications.