Problem 18 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(X \mathrm{~mL}\) of \(\mathrm{H}_{2}\) gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is : (a) 10 seconds : He (b) 20 seconds : \(\mathrm{O}_{2}\) (c) 25 seconds : \(\mathrm{CO}\) (d) 55 seconds : \(\mathrm{CO}_{2}\)

Step-by-Step Solution

Answer

The answer is (b) 20 seconds for \( \mathrm{O}_{2} \).

Step 1: 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. Mathematically, this can be expressed as \( \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \), where \( r \) is the rate of effusion and \( M \) is the molar mass. The rate of effusion is the volume of gas effusing per unit time, so \( r = \frac{V}{t} \).

Step 2: Set up the Equation for Effusion Time Comparison

We are given that \( X \) mL of \( \mathrm{H}_2 \) gas effuses in 5 seconds. If we compare this to another gas, let's call it gas 2, we need to find when the same volume \( X \) mL effuses as described in each option. The times are given, so we can equate \( r_1 t_1 = r_2 t_2 \), leading to \( t_1 \sqrt{M_1} = t_2 \sqrt{M_2} \). Rearranging gives \( t_2 = t_1 \frac{\sqrt{M_2}}{\sqrt{M_1}} \).

Step 3: Calculate the Time for \\mathrm{H}_2 vs \\mathrm{He}

First, let's compare hydrogen to helium: the molar mass of \( \mathrm{H}_2 \) is 2 g/mol and \( \mathrm{He} \) is 4 g/mol. Using our rearranged formula, \( t_2 = 5 \frac{\sqrt{4}}{\sqrt{2}} = 5 \times \frac{2}{\sqrt{2}} = 5 \times \sqrt{2} \approx 7.07 \) seconds. This doesn't match the option time of 10 seconds.

Step 4: Calculate the Time for \\mathrm{H}_2 vs \\mathrm{O}_2

Next, for oxygen: the molar mass of \( \mathrm{O}_2 \) is 32 g/mol. Applying the formula, \( t_2 = 5 \frac{\sqrt{32}}{\sqrt{2}} = 5 \times \sqrt{16} = 5 \times 4 = 20 \) seconds, which perfectly matches the given option.

Step 5: Calculate the Time for \\mathrm{H}_2 vs \\mathrm{CO}

Now, for carbon monoxide: the molar mass of \( \mathrm{CO} \) is 28 g/mol. Using the formula, \( t_2 = 5 \frac{\sqrt{28}}{\sqrt{2}} = 5 \times \sqrt{14} \approx 18.7 \) seconds, which does not match the given option of 25 seconds.

Step 6: Calculate the Time for \\mathrm{H}_2 vs \\mathrm{CO}_2

Finally, for carbon dioxide: the molar mass of \( \mathrm{CO}_2 \) is 44 g/mol. We substitute to get \( t_2 = 5 \frac{\sqrt{44}}{\sqrt{2}} = 5 \times \sqrt{22} \approx 23.4 \) seconds, which does not match 55 seconds.

Step 7: Determine the Correct Answer

From the calculations, the only accurate match with the given options is oxygen, with an effusion time of 20 seconds, which corresponds with option (b).

Key Concepts

kinetic molecular theory

The kinetic molecular theory is a fundamental concept that describes the behavior of gases under various conditions. It is centered around the idea that gas particles are in constant motion and their speed is influenced by temperature. The theory states that gas molecules move randomly, collide with each other and the walls of their container, leading to pressure. The motion and energy of these molecules help us to understand phenomena such as diffusion and effusion.

To visualize this, imagine a room full of balloons, each representing a gas molecule. When the room heats up, the balloons move faster, bouncing off each other more frequently. The kinetic energy, associated with the motion of these molecules, is directly tied to the temperature of the gas. This concept is crucial when we consider the transition of gas particles through a small opening, or effusion.

• Gas particles are always in motion
• Collisions lead to gas pressure
• Temperature influences molecular speed and energy

effusion rate

Effusion is the process in which gas particles pass through a tiny hole from one compartment to another. According to Graham's law of effusion, the rate at which a gas effuses is inversely proportional to the square root of its molar mass. This means, lighter gases effuse faster than heavier gases if all other conditions are the same.

Picture this: your class has two books - one light and one heavy. If you throw them both, the lighter book will travel farther because it's easier to move. Similarly, hydrogen, being the lightest gas, effuses much faster than a heavier gas like carbon dioxide. This rate of effusion (\( r \)) can be mathematically expressed as:\[ r = \frac{V}{t} \]where (\( V \)) is the volume of the gas and (\( t \)) is the time. Understanding the effusion rate helps in predicting how fast different gases will pass through an opening.

• Lighter gases effuse more rapidly
• Rate is dependent on molar mass
• Involves gas transferring through small openings

molar mass comparison

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). In the context of Graham's Law, it serves as a crucial determinant of effusion rates for different gases. The law can be applied to understand how gases with different molar masses effuse through the same opening, broadening our comprehension of gas behavior.

For instance, consider hydrogen (\( \text{H}_2 \)) with a molar mass of 2 g/mol and oxygen (\( \text{O}_2 \)) with 32 g/mol. Under similar conditions, hydrogen will effuse faster because it has a smaller molar mass. This is depicted by the formula:\[ \frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} \]where (\( r \)) are the rates of effusion and (\( M \)) are the molar masses of the gases. By comparing molar masses, students can predict how long it will take for gases to effuse under identical conditions.

• Molar mass inversely affects effusion speed
• Predicts relative movement of different gases
• Helps resolve practical problems in gas behavior