Problem 70 - A Complete Resource Book in Chemistry for JEE Main

Question

If the rms velocity of a gas at \(100 \mathrm{~K}\) is \(10^{4} \mathrm{~cm} \mathrm{sec}^{-1}\), what is the temperature (in \({ }^{\circ} \mathrm{C}\) ) at which the rms velocity will be \(3 \times 10^{4} \mathrm{~cm} \mathrm{sec}^{-1} ?\) (a) 900 (b) 627 (c) 327 (d) 1217

Step-by-Step Solution

Answer

The temperature is 627°C.

Step 1: Understand the rms velocity formula

The equation for the root mean square (rms) velocity of a gas is given by \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( v_{rms} \) is the rms velocity, \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the mass of one molecule of the gas.

Step 2: Relate velocities to temperatures

Using the formula for rms velocity, the ratio of rms velocities at two different temperatures can be represented as \( \frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} \). Here, \( v_{rms1} \) is the initial rms velocity \(10^{4} \mathrm{~cm} \mathrm{sec}^{-1} \) at temperature \( T_1 = 100 \mathrm{~K} \), and \( v_{rms2} \) is the required rms velocity \(3 \times 10^{4} \mathrm{~cm} \mathrm{sec}^{-1} \) at temperature \( T_2 \).

Step 3: Substitute and solve for the temperature ratio

Substitute the given values into the relation: \( \frac{3 \times 10^{4}}{10^{4}} = \sqrt{\frac{T_2}{100}} \). Simplifying gives \( 3 = \sqrt{\frac{T_2}{100}} \), which can be squared on both sides to get \( 9 = \frac{T_2}{100} \).

Step 4: Solve for \( T_2 \)

Multiply both sides by 100 to isolate \( T_2 \): \( T_2 = 900 \mathrm{~K} \).

Step 5: Convert temperature to Celsius

Convert the final temperature from Kelvin to Celsius using the formula \( \text{Celsius} = \text{Kelvin} - 273.15 \). For \( T_2 = 900 \mathrm{~K} \), the corresponding temperature in Celsius is \( 900 - 273.15 = 626.85 \approx 627 \) since it's a multiple-choice question.

Key Concepts

Temperature Conversion

Temperature conversion is an essential concept when it comes to understanding scientific measurements. The Kelvin scale is often used in scientific calculations because it starts at absolute zero, where all molecular motion stops. However, in daily life, we commonly use Celsius or Fahrenheit scales. Conversions between these scales are crucial to bridge the gap between laboratory calculations and real-world understanding.

To convert a temperature from Kelvin to Celsius, use the simple formula:

• Celsius = Kelvin - 273.15

This formula stems from the fact that 0°C corresponds to 273.15 K. Therefore, subtracting 273.15 from any Kelvin temperature provides the temperature in Celsius.

In the exercise, the temperature was calculated to be 900 K. To convert this to Celsius, subtract 273.15 from 900, resulting in approximately 627°C. This conversion helps understand how temperature values in high sensitivity scales (like Kelvin) relate to more intuitive measures (like Celsius).

Root Mean Square Velocity

The root mean square (rms) velocity is an important concept in kinetic theory. It provides an average velocity of gas particles in a system and is used to quantify the speed of molecules at a given temperature. The rms velocity is given by the formula:

• \( v_{rms} = \sqrt{\frac{3kT}{m}} \)

Here:

• \( v_{rms} \) is the root mean square velocity,
• \( k \) is the Boltzmann constant,
• \( T \) is the temperature in Kelvin,
• \( m \) is the mass of one molecule of the gas.

The rms velocity highlights how the speed of gas particles varies with temperature; a higher temperature results in a higher rms velocity.

In the context of the exercise, initial and final rms velocities were given at temperatures of 100 K and 900 K, respectively. By understanding the relationship \( \frac{v_{rms2}}{v_{rms1}} = \sqrt{\frac{T_2}{T_1}} \), we can determine how changes in temperature affect molecular velocity, providing essential insights into gas behavior.

Boltzmann Constant

The Boltzmann constant \( ( k ) \) is a fundamental physical constant that relates the average kinetic energy of particles in a gas to the temperature. It is a crucial part of the rms velocity formula:

• \( k = 1.38 \times 10^{-23} \text{ J/K} \)

This constant enables the conversion of temperature (in Kelvin) into energy terms. In the formula for rms velocity, it helps express how temperature impacts the kinetic energy and, subsequently, the velocity of gas particles.

The Boltzmann constant is significant because it bridges macroscopic and microscopic physics, showing how individual gas particles' behavior corresponds to measurable phenomena like temperature. Understanding this constant allows us to predict and rationalize the physical behavior of gases, linking molecular speed with the thermal energy present in the system. By utilizing \( k \), the calculations within the problem reinforce the relationship between temperature and molecular activity, a cornerstone of thermodynamics and statistical mechanics.