Problem 13 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The ratio between the root mean square speed of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is, (a) 4 (b) 2 (c) 1 (d) \(1 / 4\)

Step-by-Step Solution

Answer

The ratio is 1; option (c).

Step 1: Understanding the Root Mean Square Speed Formula

The root mean square speed \( v_{rms} \) of a gas is given by the formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas. We will use this formula to find the ratio of the speeds of hydrogen and oxygen.

Step 2: Calculate \( v_{rms} \) of \( \mathrm{H}_{2} \) at 50 K

The molar mass \( M \) of \( \mathrm{H}_{2} \) is 2 g/mol (or 0.002 kg/mol for SI units). Substitute \( T = 50 \) K and \( M = 0.002 \) kg/mol into the formula: \[ v_{rms} = \sqrt{\frac{3 \times R \times 50}{0.002}} \]

Step 3: Calculate \( v_{rms} \) of \( \mathrm{O}_{2} \) at 800 K

The molar mass \( M \) of \( \mathrm{O}_{2} \) is 32 g/mol (or 0.032 kg/mol for SI units). Substitute \( T = 800 \) K and \( M = 0.032 \) kg/mol into the formula: \[ v_{rms} = \sqrt{\frac{3 \times R \times 800}{0.032}} \]

Step 4: Formula for the Ratio of Speeds

Since \( R \) appears in both expressions and cancels out, the ratio \( \frac{{v_{rms}^{( ext{H}_2)}}}{{v_{rms}^{( ext{O}_2)}}} \) can be expressed as follows: \[ \frac{\sqrt{\frac{3RT_{H_2}}{M_{H_2}}}}{\sqrt{\frac{3RT_{O_2}}{M_{O_2}}}} = \sqrt{\frac{T_{H_2}}{T_{O_2}} \times \frac{M_{O_2}}{M_{H_2}}} \]

Step 5: Substitute the Known Values in the Ratio Formula

Substitute \( T_{H_2} = 50 \) K, \( T_{O_2} = 800 \) K, \( M_{H_2} = 0.002 \) kg/mol, and \( M_{O_2} = 0.032 \) kg/mol into the ratio formula: \[ \begin{align*} \text{Ratio} = \sqrt{\frac{50}{800} \times \frac{0.032}{0.002}} \end{align*} \]

Step 6: Simplify the Expression

Perform the arithmetic in the expression: \[ \frac{50}{800} = \frac{1}{16} \quad \text{and} \quad \frac{0.032}{0.002} = 16. \quad \text{Thus, the overall expression becomes} \quad \sqrt{\frac{1}{16} \times 16} = \sqrt{1} = 1. \]

Step 7: Conclusion

The ratio of the root mean square speeds of \( \mathrm{H}_{2} \) at 50 K and \( \mathrm{O}_{2} \) at 800 K is 1. Therefore, the answer is option (c) 1.

Key Concepts

Kinetic Theory of Gases

The kinetic theory of gases is a simple yet powerful model that helps us understand the properties and behaviors of gases. It is based on the idea that a gas is composed of a large number of small particles, specifically molecules or atoms, that are in constant, random motion.
The core assumptions of this theory include:

• There are a large number of molecules, and they move with a range of speeds.
• Gas molecules are considered point masses, meaning their size is negligible compared to the distance they travel between collisions.
• Collisions between molecules and with the walls of the container are perfectly elastic, meaning there is no loss of energy during the collisions.
• The only forces between gas molecules are those during collisions.

These assumptions allow us to derive various gas properties, including the root mean square speed, which is a measure of the speed of gas particles, crucial for understanding the kinetic energy of gases.

Temperature Dependence of Gas Speed

The speed of gas molecules is strongly influenced by temperature. As per the kinetic theory, temperature is a measure of the average kinetic energy of gas molecules. A higher temperature corresponds to a higher average speed of the molecules.
The root mean square speed, denoted as \( v_{rms} \), is particularly useful in quantifying this speed. It is calculated using the formula:\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]where \( R \) is the universal gas constant, \( T \) is the absolute temperature in Kelvin, and \( M \) is the molar mass of the gas.
As temperature increases, \( T \) increases, resulting in an increase in \( v_{rms} \). This relationship highlights why gases expand when heated; the increased kinetic energy causes molecules to move faster, colliding more frequently and exerting greater pressure on the walls of their container.

Molar Mass of Gases

The molar mass of a gas, which is the mass of one mole of its molecules, plays a critical role in determining the speed of gas molecules. It directly impacts the root mean square speed, \( v_{rms} \), as seen in the formula:\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]where \( M \) is the molar mass.
From this equation, it's clear that gases with a smaller molar mass will have a higher \( v_{rms} \), assuming temperature remains constant. This is because lighter molecules, such as hydrogen, require less energy to move quickly compared to heavier gases, like oxygen.
In the example discussed, hydrogen (\( \mathrm{H}_2 \)) has a much smaller molar mass than oxygen (\( \mathrm{O}_2 \)), which affects the comparison of their speeds under different temperatures. Understanding the relationship between molar mass and gas speed is essential for predicting how gases will behave under various conditions.