Problem 21 - Chemistry in 30 days

Question

Oxygen (one \(\mathrm{dm}^{3}\) at 1 atm) takes 2 min to effuse through an orifice. How much time (in minutes) will be taken by \(\mathrm{N}_{2}, \mathrm{He}\) and \(\mathrm{SF}_{6}\) vapours to effuse under the same conditions? (a) \(0.7,1.87,4.27\) (b) \(4.27,0.7,1.87\) (c) \(1.87,0.7,4.27\) (d) \(4.27,1.87,0.7\)

Step-by-Step Solution

Answer

The correct answer is (c) 1.87, 0.7, 4.27.

Step 1: Understand Effusion and Graham's Law

Effusion is the process through which gas molecules escape through a tiny hole into a vacuum. 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, represented by the formula: \[ r_1 / r_2 = \sqrt{M_2 / M_1} \] where \(r_1\) and \(r_2\) are the effusion rates of two gases, and \(M_1\) and \(M_2\) are their molar masses. If a gas effuses slower than another, it will take more time, inversely related to its rate.

Step 2: Identify Molar Masses

Determine the molar masses of the gases: \(\text{O}_2\) has a molar mass of 32 g/mol, \(\text{N}_2\) is 28 g/mol, \(\text{He}\) is 4 g/mol, and \(\text{SF}_6\) is 146 g/mol. We will use these values to compute the relative times for effusion compared to oxygen.

Step 3: Calculate Effusion Time for Each Gas

Calculate the time each gas takes to effuse using the formula based on Graham's Law: \[ t_2 / t_1 = \sqrt{M_2 / M_1} \]. Thus, the time for each gas to effuse compared to oxygen (taking 2 minutes) is calculated as:- For \(\text{N}_2\): \[ t_{N_2} = 2 \times \sqrt{32/28} \approx 1.87 \text{ min} \]- For \(\text{He}\): \[ t_{He} = 2 \times \sqrt{32/4} \approx 0.7 \text{ min} \]- For \(\text{SF}_6\): \[ t_{SF_6} = 2 \times \sqrt{146/32} \approx 4.27 \text{ min} \]

Step 4: Match Calculations to Options

Comparing the calculated times with the provided options, we can match each time (1.87 for \(\text{N}_2\), 0.7 for \(\text{He}\), and 4.27 for \(\text{SF}_6\)) to Choice (c): \(1.87, 0.7, 4.27\).

Key Concepts

Effusion Rate

Effusion is an intriguing process where gas molecules escape through a very small opening into a vacuum. The rate at which this happens is called the effusion rate. Graham's Law of Effusion helps us understand how the effusion rate relates to a gas's molar mass. Essentially, the law tells us that lighter gases effuse faster than heavier gases. This is because their effusion rate is inversely proportional to the square root of their molar mass. The formula for this relationship is given by: \[ r_1 / r_2 = \sqrt{M_2 / M_1} \]Where:

• \( r_1 \) and \( r_2 \) are the effusion rates of two different gases.
• \( M_1 \) and \( M_2 \) are their respective molar masses.

Understanding this relationship helps predict how quickly a gas will effuse compared to another under equivalent conditions. For instance, in the given exercise, nitrogen (\(\text{N}_2\)), helium (\(\text{He}\)), and sulfur hexafluoride (\(\text{SF}_6\)) effuse at different rates than oxygen (\(\text{O}_2\)) due to differences in their molar masses.

Molar Mass

The concept of molar mass is fundamental in studying gas behaviors, such as effusion. Molar mass refers to the mass of one mole of any substance, typically expressed in grams per mole. It becomes crucial when applying Graham's Law of Effusion because it dictates the rate at which a gas will escape through a small opening. In our exercise, each gas has a distinct molar mass that affects its effusion time:

• Oxygen (\(\text{O}_2\)): 32 g/mol
• Nitrogen (\(\text{N}_2\)): 28 g/mol
• Helium (\(\text{He}\)): 4 g/mol
• Sulfur hexafluoride (\(\text{SF}_6\)): 146 g/mol

By understanding and calculating these molar masses, we can predict how quickly or slowly each gas will effuse. For example, helium, having the smallest molar mass among the gases considered, effuses quite rapidly compared to the others.

Gas Laws

Gas laws are a set of rules that describe how gases behave under various conditions of pressure, volume, and temperature. Graham's Law of Effusion is one of these gas laws and is particularly focused on the rate of effusion of gases. However, other gas laws provide a more comprehensive understanding of how gases typically behave. Some of these include:

• Boyle's Law: It states that the pressure of a gas is inversely proportional to its volume when temperature is constant.
• Charles's Law: This tells us that the volume of a gas is directly proportional to its temperature when pressure is constant.
• Avogadro's Law: It states that the volume of a gas is directly proportional to the number of moles of gas when pressure and temperature remain constant.

By integrating these principles, students can better understand how different gases behave not just regarding effusion but in various scenarios, like compression or expansion. Even in effusion, considering these broader gas laws can offer a richer understanding of gas behaviors beyond simply applying Graham's Law. Among these laws, Graham's Law specifically assists in solving tasks like the given exercise by linking the effusion rate to molar mass. This connection is a key aspect that helps to assess and predict gas behavior under certain conditions.