Question

If it takes 11.2 hours for \(1.00 \mathrm{~L}\) of nitrogen, \(\mathrm{N}_{2},\) to effuse through the pores in a balloon, how long would it take for \(1.00 \mathrm{~L}\) of helium, \(\mathrm{He}\), to effuse under the same conditions?

Short answer

It would take approximately 29.65 hours for helium to effuse.

Step-by-step solution

Understand Graham's Law of Effusion

Graham's Law of effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula can be represented as: \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r_1 \) and \( r_2 \) are the rates of effusion, and \( M_1 \) and \( M_2 \) are the molar masses of the gases nitrogen and helium, respectively.

Identify the Molar Masses and Rates

The molar mass of nitrogen (\( N_2 \)) is approximately \( 28.02 \mathrm{~g/mol} \) and for helium (\( He \)) is \( 4.00 \mathrm{~g/mol} \). Since the rate of effusion is the inverse of time taken, we use \( r_1 = \frac{1}{11.2 \text{ hours}} \).

Substitute in Graham's Law Formula

Substitute the known values into Graham's Law formula: \( \frac{1/t_2}{1/11.2} = \sqrt{\frac{4.00}{28.02}} \). Simplifying gives \( t_2 = 11.2 \times \sqrt{\frac{28.02}{4.00}} \).

Calculate the Square Root Ratio

Calculate the square root of the molar mass ratio: \( \sqrt{\frac{28.02}{4.00}} \approx \sqrt{7.005} \approx 2.646 \).

Solve for Helium Effusion Time

Substitute the square root value back to find \( t_2 \): \( t_2 = 11.2 \times 2.646 \approx 29.65 \text{ hours} \).

Key concepts

Effusion

Effusion is a process where gas particles escape through small pores or openings without colliding with each other. Imagine a balloon filled with gas that has tiny holes — the slow leakage of gas out of the balloon is effusion. The rate at which this happens primarily depends on the gas's properties, such as its molar mass and temperature.
Effusion is described by Graham's Law, which allows us to compare the rates at which different gases effuse. Graham's Law tells us that lighter gases effuse faster because they move more quickly due to having less mass. This principle helps to understand why helium, a lighter gas, will behave differently from nitrogen in a controlled setting.

Molar Mass

Molar mass is a fundamental concept for understanding effusion. It is essentially the mass of one mole of a substance, typically expressed in grams per mole (g/mol).
To find the molar mass of a molecule, sum the atomic masses of its constituent atoms. For nitrogen gas (\( N_2 \)), its molar mass is approximately 28.02 g/mol (14.01 g/mol per nitrogen atom). Helium, a noble gas with a single-atom molecule (\( He \)), has a much lower molar mass of about 4.00 g/mol. Using these values in Graham's Law allows us to predict and calculate how different gases will effuse under similar conditions.

Nitrogen Gas

Nitrogen gas (\( N_2 \)) is a diatomic molecule, meaning it consists of two nitrogen atoms bonded together. It makes up around 78% of Earth's atmosphere, playing a critical role in our environment.
Nitrogen is colorless, odorless, and inert, making it essential in many industrial applications where a non-reactive atmosphere is needed. Its relatively higher molar mass compared to helium means that nitrogen will effuse more slowly, which is clearly demonstrated in our example problem where nitrogen takes 11.2 hours to effuse 1.00 L through a balloon.

Helium Gas

Helium is the second lightest element and is a noble gas with exceptional properties. As a monatomic gas, one helium atom floats alone rather than forming molecules like nitrogen. This contributes to its very low molar mass of 4.00 g/mol.
Helium is often famous for its use in balloons, diving gas tanks, and its non-reactive nature due to its full electron shell. Helium effuses faster than nitrogen due to its lower molar mass, which is evident in the exercise where helium takes longer, 29.65 hours, indicating how much quicker it effuses when conditions allow.