Problem 79 - A Complete Resource Book in Chemistry for JEE Main

Question

If Vrms of \(\mathrm{H}_{2}\) at \(300 \mathrm{~K}\) is \(1.9 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the value of Vrms of \(\mathrm{O}_{2}\) at \(1200 \mathrm{~K} ?\) (a) \(1.9 \times 10^{3}\) (b) \(3.8 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (c) \(0.475 \times 10^{3} \mathrm{~m} / \mathrm{s}\) (d) \(0.95 \times 10^{3} \mathrm{~m} / \mathrm{s}\)

Step-by-Step Solution

Answer

The Vrms of \( \mathrm{O}_{2} \) at \( 1200 \mathrm{~K} \) is \( 0.95 \times 10^{3} \mathrm{~m/s} \). (d)

Step 1: Understand the Vrms Formula

The root mean square velocity (Vrms) of a gas is given by the formula: \( v_{rms} = \sqrt{\frac{3RT}{M}} \), where \( R \) is the gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

Step 2: Identify Given Values for Hydrogen

We are provided with the \( v_{rms} \) for \( \mathrm{H}_{2} \) at \( 300 \mathrm{~K} \), which is \( 1.9 \times 10^{3} \mathrm{~m/s} \). The molar mass of \( \mathrm{H}_{2} \) is approximately \( 2 \times 10^{-3} \mathrm{~kg/mol} \).

Step 3: Set Up Ratios for Comparison

For two gases under different conditions, the VRMS can be compared using the equation: \( \frac{v_{rms,1}}{v_{rms,2}} = \sqrt{\frac{T_1M_2}{T_2M_1}} \). Here, \( v_{rms,1} \) and \( T_1 \) refer to \( \mathrm{H}_{2} \), while \( v_{rms,2} \) and \( T_2 \) refer to \( \mathrm{O}_{2} \).

Step 4: Find Molar Mass of Oxygen

Oxygen's molar mass \( M_2 \) is approximately \( 32 \times 10^{-3} \mathrm{~kg/mol} \). The temperature \( T_2 \) for \( \mathrm{O}_{2} \) is given as \( 1200 \mathrm{~K} \).

Step 5: Calculate Vrms of Oxygen

Using the formula: \( v_{rms,2} = v_{rms,1} \times \sqrt{\frac{T_2M_1}{T_1M_2}} \). Substituting the known values gives: \[ v_{rms,2} = 1.9 \times 10^{3} \times \sqrt{\frac{1200 \times 2}{300 \times 32}} \]. Calculate this to get \( v_{rms,2} \).

Step 6: Finalize Calculation

Perform the calculation from the step above: \[ v_{rms,2} = 1.9 \times 10^{3} \times \sqrt{\frac{2400}{9600}} = 1.9 \times 10^{3} \times \frac{1}{2} = 0.95 \times 10^{3} \mathrm{~m/s} \].

Key Concepts

Gas Laws

Gas laws describe how gases behave under different conditions of pressure, volume, and temperature. They are fundamental to understanding various properties of gases, including their speed and energy. The root mean square velocity ( Vrms) is a core concept derived from these laws. It explains the average speed of gas particles.
The ideal gas law, expressed as \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is temperature, highlights how pressure, volume, and temperature are interrelated. Changes in temperature, as per this law, affect how fast the gas molecules move, thus influencing Vrms.

• As temperature increases, gas molecules move faster, thus increasing their energy and Vrms.
• Conversely, a decrease in temperature results in slower motion and reduced Vrms.

By understanding gas laws, we realize why varying temperatures in the problem lead to different Vrms values for hydrogen and oxygen.

Kinetic Molecular Theory

The Kinetic Molecular Theory (KMT) is crucial in explaining the behavior of gases at a microscopic level. This theory postulates that gases are composed of numerous small particles in constant, random motion. It also correlates the temperature of a gas to its molecular kinetic energy.
KMT helps us comprehend why Vrms changes with different conditions. According to this theory:

• Gas particles are in perpetual motion and exhibit random speeds.
• Average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin).
• Collisions between gas molecules are perfectly elastic, meaning there is no net loss of kinetic energy during collisions.

This explains why, in the original problem, oxygen gas at a higher temperature moves faster than hydrogen gas at a lower temperature. KMT provides the theoretical background by connecting temperature with molecular speed and kinetic energy, further clarifying the Vrms concept.

Temperature and Kinetic Energy

Temperature is a fundamental concept affecting the kinetic energy of molecules in gases. Kinetic energy, for a gas, depends on both the mass of the molecules and their speed. It is measured by the temperature of the gas.
In this context, root mean square velocity (Vrms) directly links the temperature of a gas to the average speed of its molecules through the equation:

• \( v_{rms} = \sqrt{\frac{3RT}{M}} \)
• Here, \( T \) symbolizes the temperature while \( M \) represents the molar mass of the gas.

An increase in temperature, as demonstrated in the initial exercise, results in heightened kinetic energy, leading to an increase in Vrms.
Understanding the relationship between temperature and kinetic energy allows us to predict and calculate how quickly gas molecules will be moving. It provides clarity on why oxygen in the problem, with a higher given temperature, demonstrates a different Vrms compared to hydrogen.