Problem 20 - A Complete Resource Book in Chemistry for JEE Main

Question

The rms velocity of \(\mathrm{CO}_{2}\) at a temperature \(\mathrm{T}\) (in Kelvin) is \(\times \mathrm{cm} \mathrm{sec}^{-1}\), At what temperature (in Kelvin), the \(\mathrm{rms}\) velocity of nitrous oxide would be \(4 \times \mathrm{cm} \sec ^{-1}\) ? (Atomic weights of \(\mathrm{C}, \mathrm{N}\) and \(\mathrm{O}\) are respectively 12,14 and 16 ) (a) \(16 \mathrm{~T}\) (b) \(2 \mathrm{~T}\) (c) \(4 \mathrm{~T}\) (d) \(32 \mathrm{~T}\)

Step-by-Step Solution

Answer

The temperature is \(16T\); choice (a) is correct.

Step 1: Understanding RMS Velocity Formula

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

Step 2: Calculate Molar Masses

For \( \mathrm{CO}_2 \), the molar mass \( M = 12 + 2 \times 16 = 44 \text{ g/mol} \). For \( \mathrm{N}_2\mathrm{O} \), the molar mass \( M = 2 \times 14 + 16 = 44 \text{ g/mol} \).

Step 3: Set Up Equation for Nitrous Oxide

We desire an RMS velocity of \( 4 \times \text{ cm/sec} \) for \( \mathrm{N}_2\mathrm{O} \). Using the formula, we set: \[ 4 = \sqrt{\frac{3R T_2}{44}} \].

Step 4: Solve for Temperature Ratio

Since \( v_{rms, CO_2} = \times \text{ cm/sec} \) at temperature \( T_1 \) and the molar mass of both gases is the same, the velocities are proportional to the square roots of the temperatures. Thus, \( 4 = \ \left( \times \right) \sqrt{\frac{T_2}{T_1}} \).

Step 5: Find \( T_2 \) in terms of \( T_1 \)

Simplifying \( 4 = \sqrt{\frac{T_2}{T_1}} \) gives \( 16 = \frac{T_2}{T_1} \), thus \( T_2 = 16T_1 \).

Step 6: Choose Correct Answer

Among the options, \( T_2 = 16T \) corresponds to option (a).

Key Concepts

Understanding Root Mean Square (RMS) Velocity

In thermodynamics, especially when dealing with gases, the root mean square velocity is an important concept. The RMS velocity represents an average value of the speed of gas molecules, providing insight into the kinetic energy of the gas particles.
This velocity is calculated using the formula:

• \( v_{rms} = \sqrt{\frac{3RT}{M}} \)

Here:

• \( v_{rms} \) is the root mean square velocity.
• \( R \) is the universal gas constant \( (8.314 \text{ J/mol K}) \).
• \( T \) is the absolute temperature in Kelvin.
• \( M \) is the molar mass of the gas in g/mol.

The formula reflects how the velocity increases with temperature and decreases with heavier gas molecules. A fundamental principle here is that the energy of gas particles depends on their mass and temperature. The RMS velocity provides a bridge to calculate, estimate, and understand these relations based on observable quantities like temperature.

Molar Mass Calculation

The molar mass of a substance is the mass of one mole of its entities (such as atoms, molecules, or ions), usually expressed in grams per mole (g/mol). Calculating molar mass is crucial when using the RMS velocity formula since the larger the molar mass, the lower the average speed of the gas molecules.
To calculate molar masses, you need the atomic weights of the constituent elements. For molecules, the molar mass is simply the sum of the atomic weights:

• For \( \text{CO}_2 \), the molar mass is calculated as: \( 12 + 2 \times 16 = 44 \text{ g/mol} \). Carbon has an atomic weight of 12, and oxygen has an atomic weight of 16.
• Similarly, for \( \text{N}_2\text{O} \), the molar mass is: \( 2 \times 14 + 16 = 44 \text{ g/mol} \). Nitrogen has an atomic weight of 14, and oxygen remains 16.

With these calculations, we can work within the RMS velocity formula. Whether for academic exercises or real-world applications, molar mass provides a foundation for understanding a gas's behavior at certain temperatures.

The Gas Constant (R)

The gas constant \( R \) is a vital element in the formula for RMS velocity. It acts as a bridge connecting the physical properties of gases with temperature. In the context of many gas laws, the universal gas constant is a fundamental component, defined in terms of energy per temperature increment per mole.
With a value of \( 8.314 \text{ J/mol K} \), the gas constant appears in various equations governing behavior of gases. Key attributes include:

• Consistency: \( R \) is a constant value that is universally applicable for ideal gases.
• Energy Conversion: It helps express energy changes when gas temperatures vary.
• Utility: Plays a crucial role in the equation \( PV = nRT \), where \( n \) is the number of moles, \( P \) is pressure, \( V \) is volume, and \( T \) is temperature.

In the RMS velocity equation, \( R \) helps calculate molecular speeds, interrelating temperature, molar mass, and energy. It epitomizes the link between macroscopic measurements and microscopic movements (motions of molecules). Understanding \( R \) thus enhances predictions and calculations concerning gaseous systems.