Problem 81 - A Complete Resource Book in Chemistry for JEE Main

Question

If the rate of effusion of helium gas at a pressure of 1000 torr is 10 torr \(\mathrm{min}^{-1}\). Find the rate of effusion of hydrogen gas at a pressure of 2000 torr at the same temperature. (a) 20 torr \(\mathrm{min}^{-1}\) (b) 10 torr \(\mathrm{min}^{-1}\) (c) \(30 \sqrt{2}\) torr \(\mathrm{min}^{-1}\) (d) \(20 \sqrt{2}\) torr \(\mathrm{min}^{-1}\)

Step-by-Step Solution

Answer

The rate of effusion of hydrogen is \(20 \sqrt{2}\) torr \(\mathrm{min}^{-1}\).

Step 1: Understand Graham's Law of Effusion

Graham's Law relates the rates of effusion of two gases to their molar masses. It states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass: \[ \frac{\text{Rate of effusion of Gas 1}}{\text{Rate of effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}} \] where \(M_1\) and \(M_2\) are the molar masses of the gases.

Step 2: Identify Given Information for Gases

We know the rate of effusion of helium is 10 torr \(\text{min}^{-1}\) at 1000 torr, and we aim to find the rate for hydrogen at 2000 torr. The molar mass of helium is 4 g/mol, and hydrogen is 2 g/mol.

Step 3: Adjust for Pressure Difference

The rate of effusion also depends on the pressure, as the number of molecules escaping is directly proportional to the pressure. We need to adjust the rates for the pressure difference. If Rate_1 is the adjusted rate for helium at 2000 torr, then: \[ \text{Rate}_1 = 10 \text{ torr min}^{-1} \times \frac{2000}{1000} = 20 \text{ torr min}^{-1} \]

Step 4: Apply Graham's Law to Find Rate of Effusion of Hydrogen

Using the adjusted rate for helium and applying Graham's Law: \[ \frac{\text{Rate of Hydrogen}}{20} = \sqrt{\frac{4}{2}} = \sqrt{2} \] Hence, the rate of effusion of hydrogen (\( \text{Rate of Hydrogen} \)) will be: \[ \text{Rate of Hydrogen} = 20 \times \sqrt{2} \]

Step 5: Final Answer

Based on the calculation, the rate of effusion of hydrogen at 2000 torr is \(20 \sqrt{2}\) torr \(\text{min}^{-1}\). This corresponds to option (d).

Key Concepts

Rate of Effusion

The rate of effusion is a crucial concept in gas behavior. It describes how quickly a gas can escape through a tiny hole into a vacuum. This rate fundamentally depends on several key factors, and Graham's Law provides the mathematical basis for understanding these relationships. According to Graham's Law, the rate at which a gas effuses is inversely proportional to the square root of its molar mass, meaning lighter gases effuse faster than heavier ones.

To exemplify, consider two gases: helium and hydrogen. Given that helium has a higher molar mass compared to hydrogen, hydrogen will effuse at a faster rate under the same conditions due to its lower molar mass. In application, if you want to compare the rates of effusion for these gases, you would use the formula: \(\frac{\text{Rate of effusion of Gas 1}}{\text{Rate of effusion of Gas 2}} = \sqrt{\frac{M_2}{M_1}}\) where \(M\) represents molar mass. This equation highlights that as the molar mass decreases, the rate of effusion increases.

Molar Mass

Molar mass of a gas plays a significant role in its effusion rate, as highlighted by Graham's Law. A smaller molar mass means a gas particle is lighter, which directly affects its velocity. Lighter gas particles move faster and collide with the walls of their container more frequently, leading to a higher rate of effusion.

When dealing with gases like helium and hydrogen in the exercise, it's essential to know their molar masses: helium's molar mass is 4 g/mol, and hydrogen's is 2 g/mol. This difference may appear small but has a considerable impact when calculating effusion rates. Since hydrogen has a molar mass half that of helium, it effuses faster by a factor equal to the square root of the molar mass ratio \(\sqrt{\frac{4}{2}} = \sqrt{2}\). Therefore, understanding and applying the concept of molar mass helps predict how one gas will behave relative to another during effusion scenarios.

Pressure Adjustment

Pressure is another critical factor determining the rate of effusion for a gas. This is because gas pressure is directly related to the frequency of molecule-wall collisions. When the pressure increases, more molecules are present to escape through any available opening, leading to a higher effusion rate.

In our exercise, pressure adjustments are necessary for accurately predicting the effusion rate of hydrogen. Initially, helium's effusion rate is measured at 1000 torr, but the rate of hydrogen must be determined at 2000 torr. To make a fair comparison, we adjust helium's rate to the new pressure condition. By multiplying the original helium rate (10 torr min⁻¹) by the pressure ratio \(\frac{2000}{1000}\), we get an adjusted rate of 20 torr min⁻¹.

This adjustment ensures that comparisons and calculations reflect the actual conditions both gases experience, ensuring our predictions are as accurate as possible when using Graham's Law under varying pressure conditions.