Problem 69 - University Physics with Modern Physics

Question

You have two identical containers, one containing gas \(A\) and the other gas \(B .\) The masses of these molecules are \(m_{A}=3.34 \times 10^{-27} \mathrm{kg}\) and \(m_{B}=5.34 \times 10^{-26} \mathrm{kg} .\) Both gases are under the same pressure and are at \(10.0^{\circ} \mathrm{C}\) . (a) Which molecules \((A\) or \(B)\) have greater translational kinetic energy per molecule and rms speeds? (b) Now you want to raise the temperature of only one of these containers so that both gases will have the same only one of these containers so that both gases will have the same rms speed. For which gas should you raise the temperature? (c) At what temperature will you accomplish your goal? (d) Once you have accomplished your goal, which molecules \((A\) or \(B)\) now have greater average translational kinetic energy per molecule?

Step-by-Step Solution

Answer

(a) Same kinetic energy, but A has greater rms speed. (b) Raise temperature of gas B. (c) Calculate \( T_B = 753.94 \) K. (d) Gas B has greater kinetic energy.

Step 1: Understanding Translational Kinetic Energy

The average translational kinetic energy per molecule for any gas is given by the equation: \( KE = \frac{3}{2}kT \), where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin. Since both gases are at the same temperature, the average translational kinetic energy per molecule for both gases will be the same.

Step 2: Calculating Root Mean Square Speed

The root mean square speed (rms speed) of a gas is given by the equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \), where \( k \) is the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molar mass of the gas. Since the mass of gas \( A \) is smaller than that of gas \( B \), \( A \) will have the greater rms speed at the same temperature.

Step 3: Determining Which Gas to Heat

To make the rms speeds of both gases the same, we should equalize their \( v_{rms} \). Since \( v_{rms} \) depends inversely on the square root of the mass, we need to increase the temperature of the heavier gas, \( B \), to match the rms speed of the lighter gas, \( A \).

Step 4: Calculating New Temperature for Gas B

Using the rms speed equation \( v_{rms} = \sqrt{\frac{3kT}{m}} \) and setting the equations for \( A \) and \( B \) equal: \( \sqrt{\frac{3kT_A}{m_A}} = \sqrt{\frac{3kT_B}{m_B}} \). Solve for \( T_B \): \( T_B = T_A \times \frac{m_B}{m_A} = (10+273.15) \times \frac{5.34 \times 10^{-26}}{3.34 \times 10^{-27}} \). Calculate \( T_B \).

Step 5: Comparing Translational Kinetic Energy After Heating

Once gas \( B \) is heated to the calculated temperature, its kinetic energy per molecule becomes \( KE_B = \frac{3}{2}kT_B \). Since \( T_B > T_A \), gas \( B \) will now have a greater average translational kinetic energy per molecule than gas \( A \).

Key Concepts

Translational Kinetic Energy

Translational kinetic energy refers to the motion an individual molecule experiences as it moves from one point to another in space. For gases, it's fundamentally linked to the absolute temperature of the system. The equation for the average translational kinetic energy per molecule in a gas is given by:\[ KE = \frac{3}{2}kT \]where \( k \) is the Boltzmann constant (\( 1.38 \times 10^{-23} \) J/K) and \( T \) is the temperature in Kelvin. Interestingly, this expression reveals that at a given temperature, all gases have the same average translational kinetic energy per molecule, regardless of mass. It's essential to consider the temperature in Kelvin because it's an absolute measure that begins at absolute zero, providing a true scale for kinetic energy measurements in gases.

It's enlightening to realize that even if two gases have vastly different mass values, like gas \( A \) and \( B \), their energy at the molecular level can be identical if their temperatures are the same. Hence, understanding this concept helps explain why changes in temperature are crucial for controlling molecular motion.

Root Mean Square Speed

Root mean square speed is a statistical measure of the speed of particles in a gas. It is critical for understanding how individual gas particles move at a microscopic level. The formula for rms speed \( v_{rms} \) is expressed as:\[ v_{rms} = \sqrt{\frac{3kT}{m}} \]Here, \( k \) stands for the Boltzmann constant, \( T \) is the temperature in Kelvin, and \( m \) is the molecular mass of the gas. One key takeaway from this is that rms speed has an inverse relationship with molecular mass: the lighter the particle, the faster it moves on average.

• For lighter gases like gas \( A \), the rms speed will be higher than heavier gases like gas \( B \) when at the same temperature.
• This relationship helps explain phenomena such as the diffusion rate of gases, where lighter gases spread out more rapidly.

If both gases need to have the same rms speed, adjustments in temperature become necessary, particularly heating the heavier gas.

Boltzmann Constant

The Boltzmann constant, denoted \( k \), is a fundamental constant that connects the macroscopic and microscopic worlds by relating the average kinetic energy of particles to the temperature of a gas. It is a pivotal element in statistical mechanics and thermodynamics.The constant is valued at \( 1.38 \times 10^{-23} \) J/K and serves as a bridge between the energy per molecule and the temperature expressed in Kelvin. By using \( k \) in the expression for translational kinetic energy and rms speed, we can accurately predict the behavior of gases on a molecular level.

• In the equation for translational kinetic energy, \( k \) highlights that energy scales with temperature.
• For root mean square speed, \( k \) provides a conversion from absolute temperature to velocity, demonstrating how thermal energy translates into particle motion.

In essence, the Boltzmann constant is key to understanding how energy at the atomic level influences observable properties like temperature and pressure.

Temperature Conversion

Temperature conversion is a crucial step in many physics and chemistry calculations, particularly when working with gas laws. The absolute temperature scale, Kelvin, is often used due to its convenience in scientific calculations. Here's why converting Celsius to Kelvin is necessary:

• Kelvin is an absolute scale that starts at absolute zero, providing a true reflection of energy levels as opposed to relative scales like Celsius.
• It avoids negative temperatures which can complicate kinetic energy calculations, since energy is always positive.

To convert from Celsius to Kelvin, simply add 273.15:\[ T_{K} = T_{C} + 273.15 \]By ensuring temperatures are in Kelvin, scientists and students alike can apply equations directly, maintaining accuracy and consistency. When calculating changes in kinetic energy or matching rms speeds as outlined in the original exercise, temperature conversion is a fundamental preliminary step that ensures accurate results.