A sample of CO2 effuses from a container in 55 seconds. How long will it take the same amount of gaseous Xe to effuse from the same container under identical conditions?

Verified step by step guidance

1. Identify the problem as one involving effusion, which can be solved using Graham's Law of Effusion.
2. Recall Graham's Law of Effusion: \( \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of the gases.
3. Set up the equation using the given information: \( \frac{\text{Rate of effusion of CO}_2}{\text{Rate of effusion of Xe}} = \sqrt{\frac{M_{\text{Xe}}}{M_{\text{CO}_2}}} \).
4. Recognize that the rate of effusion is inversely proportional to the time taken for effusion, so \( \frac{t_{\text{Xe}}}{t_{\text{CO}_2}} = \sqrt{\frac{M_{\text{Xe}}}{M_{\text{CO}_2}}} \).
5. Substitute the known values: \( t_{\text{CO}_2} = 55 \) seconds, \( M_{\text{CO}_2} = 44.01 \) g/mol, and \( M_{\text{Xe}} = 131.29 \) g/mol, and solve for \( t_{\text{Xe}} \).

Key Concepts

Graham's Law of Effusion

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that lighter gases effuse faster than heavier gases. In this scenario, the effusion rates of CO2 and Xe can be compared using their molar masses to determine the time it will take for Xe to effuse.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For this question, the molar masses of CO2 (approximately 44 g/mol) and Xe (approximately 131 g/mol) are crucial for applying Graham's Law. The difference in molar mass directly affects the effusion rates of the gases.

Effusion

Effusion is the process by which gas molecules escape from a container through a small opening into a vacuum or lower pressure area. The rate of effusion is influenced by factors such as temperature and the size of the gas molecules. Understanding effusion is essential for solving the problem, as it involves comparing the escape rates of CO2 and Xe.