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 rms velocity of nitrous oxide would be \(4 \times \mathrm{cm} \mathrm{sec}^{-1}\) ? (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

At temperature \(16T\), the RMS velocity of nitrous oxide will be \(4x\). The answer is (a).

Step 1: Understanding Root Mean Square (RMS) Velocity

The RMS velocity for a gas is given by the formula: \[ v_{rms} = \sqrt{\frac{3RT}{M}} \] where \(v_{rms}\) is the root mean square velocity, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kg/mol.

Step 2: Calculate Molar Mass of Gases

First, we calculate the molar mass of carbon dioxide (CO\(_2\)). The atomic weights are C = 12 and O = 16, so the molar mass of CO\(_2\) is: \(M_{CO_2} = 12 + 2 \times 16 = 44 \) g/mol = 0.044 kg/mol. For nitrous oxide (N\(_2\)O), the atomic weights are N = 14 and O = 16, so the molar mass of N\(_2\)O is: \(M_{N_2O} = 2 \times 14 + 16 = 44 \) g/mol = 0.044 kg/mol.

Step 3: Set Up RMS Equation for CO2

For CO\(_2\), we are given the RMS velocity as \(x\). Therefore, using the formula: \[ x = \sqrt{\frac{3RT}{0.044}} \] (Equation 1)

Step 4: Set Up RMS Equation for N2O

For N\(_2\)O, we want the RMS velocity to be \(4x\). Thus, the equation becomes: \[ 4x = \sqrt{\frac{3RT_{N_2O}}{0.044}} \] (Equation 2)

Step 5: Relate RMS Velocity Equations

Divide Equation 2 by Equation 1: \[ \frac{4x}{x} = \frac{\sqrt{\frac{3RT_{N_2O}}{0.044}}}{\sqrt{\frac{3RT}{0.044}}} \] This simplifies to: \[ 4 = \sqrt{\frac{T_{N_2O}}{T}} \]

Step 6: Solve for TN2O

Square both sides to remove the square root: \[ 16 = \frac{T_{N_2O}}{T} \] This means: \[ T_{N_2O} = 16T \]

Step 7: Conclusion

The temperature for N\(_2\)O at which its RMS velocity is \(4x\) is \(16T\). Therefore, the correct answer is (a) \(16T\).

Key Concepts

Gas Laws

Gas laws provide a mathematical relationship between the pressure, volume, temperature, and number of moles of a gas. When it comes to root mean square (RMS) velocity, these laws help to establish how gas particles behave under various conditions.

The ideal gas law, expressed as \( PV = nRT \), where \( P \) represents pressure, \( V \) is volume, \( n \) stands for the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin, plays an important role. This law implies that if we know any three of these properties, the fourth can be calculated. Understanding these relationships allows us to work with RMS velocity since it depends on temperature and molar mass, which are interconnected through these gas properties.

In RMS velocity problems, the Ideal Gas Law helps us comprehend why changes in temperature or molar mass affect the speed of gas molecules.

Molar Mass Calculation

Molar mass is a crucial element in computing the RMS velocity of gases. It is the mass of a given substance divided by the amount of substance, usually expressed in grams per mole (g/mol).

To calculate the molar mass of a compound like carbon dioxide (\( CO_{2} \)), we sum the atomic masses of its constituent atoms. For \( CO_{2} \), the molar mass is calculated as follows:

• Carbon (C): Atomic mass = 12 g/mol.
• Oxygen (O): Atomic mass = 16 g/mol.

Adding these up: \( M_{CO_{2}} = 12 + 2 \times 16 = 44 \) g/mol. This calculation provides the necessary molar mass value for further analysis.

Similarly, for nitrous oxide (\( N_{2}O \)), the molar mass is calculated using:

• Nitrogen (N): Atomic mass = 14 g/mol.
• Oxygen (O): Atomic mass = 16 g/mol.

Resulting in: \( M_{N_{2}O} = 2 \times 14 + 16 = 44 \) g/mol. Understanding the calculation of molar mass is fundamental for any study involving gas behavior under various conditions.

Root Mean Square Formula

The root mean square (RMS) velocity is a measure of the speed of particles in a gas, calculated using the formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
Here, \( v_{rms} \) is the RMS velocity, \( R \) is the ideal gas constant, \( T \) signifies the temperature in Kelvin, and \( M \) refers to the molar mass of the gas in kg/mol.

This formula emerges from the kinetic theory of gases, which states that the average kinetic energy of gas particles is proportional to the absolute temperature of the gas. The RMS velocity helps in determining how quickly particles are moving within a gas and is essential for predicting the behavior of gases under different conditions.

It is pivotal to understand this formula because it connects various thermodynamic quantities and provides insight into the microscopic behavior of gases.

Temperature and Velocity Relationship

Temperature has a direct relationship with the velocity of gas particles. As temperature increases, the kinetic energy and hence the velocity of gas particles also increases. This relationship is captured well in the RMS velocity formula.

Examining the formula \( v_{rms} = \sqrt{\frac{3RT}{M}} \), we see how the velocity \( v_{rms} \) is directly proportional to the square root of the temperature \( T \). As such, when the temperature doubles, the RMS velocity increases square root times of the temperature. This fundamental concept is utilized when solving problems related to the movement of gas molecules, as we can compute or predict changes in particle velocity with varying temperatures.

Thus, a firm grasp on how temperature influences the velocity of gas particles is essential for predicting their behavior in various thermodynamic processes.