Problem 67 - 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 \((A \text { or } B)\) have greater translational kinetic energy per inolecule and \(\mathrm{ms}\) speeds? Now you want to raise the temperature of only one of these containers so that both gases will have the same rms speed. (b) 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, gas A has greater \(v_{rms}\). (b) Heat gas B. (c) Raise temperature of B to 300K. (d) Gas B has greater kinetic energy.

Step 1: Identify the Known Variables

We know that the masses of gases \(A\) and \(B\) are \(m_A = 3.34 \times 10^{-27} \text{ kg}\) and \(m_B = 5.34 \times 10^{-26} \text{ kg}\) respectively. Both are at a temperature of \(10.0^\circ \text{C} = 283.15\, \text{K} \), under the same pressure.

Step 2: Understanding Kinetic Energy

The translational kinetic energy per molecule for an ideal gas is given by \(E_k = \frac{3}{2} k T\), where \(k\) is the Boltzmann constant and \(T\) is the temperature. At the same temperature, molecules of both gases will have the same translational kinetic energy per molecule, as it only depends on temperature and not on molecular mass.

Step 3: Calculate RMS Speed

The root-mean-square (rms) speed is given by \(v_{rms} = \sqrt{\frac{3kT}{m}}\). Because \(T\) is the same for both gases, \(v_{rms}\) will depend on the mass of their molecules. Gas with smaller mass will have greater \(v_{rms}\). Comparing \(m_A\) and \(m_B\), gas \(A\) will have a greater \(v_{rms}\).

Step 4: Decide which Gas to Heat

To equalize the \(v_{rms}\) of both gases, we need to raise the temperature of gas \(B\), since gas \(B\) has a higher molecular mass and thus a lower \(v_{rms}\), it needs a higher temperature to match the \(v_{rms}\) of \(A\).

Step 5: Find the Temperature to Achieve Same RMS Speed

Set the \(v_{rms}\) equations equal: \(\sqrt{\frac{3kT_A}{m_A}} = \sqrt{\frac{3kT_B}{m_B}}\). Solving for \(T_B\), we get \(T_B = T_A \times \frac{m_A}{m_B}\). Substituting known values, \(T_B = 283.15 \times \frac{3.34 \times 10^{-27}}{5.34 \times 10^{-26}} \approx 17.7\, \text{K}\).

Step 6: Evaluate Kinetic Energy After Increasing Temperature

After raising the temperature of \(B\), \(T_B > T_A\). Since kinetic energy per molecule is proportional to temperature \(E_k = \frac{3}{2} k T\), gas \(B\) will then have greater average translational kinetic energy per molecule than gas \(A\).

Key Concepts

Translational Kinetic Energy

Translational kinetic energy is a fundamental concept in the kinetic theory of gases that helps to describe the motion of gas molecules. Each molecule in an ideal gas possesses translational kinetic energy, which is the energy due to its motion from one place to another.
For an ideal gas, translational kinetic energy per molecule is given by the formula \[ E_k = \frac{3}{2}kT \]where:

• \( E_k \) is the translational kinetic energy,
• \( k \) is the Boltzmann constant \((1.38 \times 10^{-23} \text{ J/K})\),
• \( T \) is the absolute temperature in kelvins.

This equation shows that translational kinetic energy depends solely on the temperature of the gas. Because of this, at a given temperature, all molecules in an ideal gas have the same average translational kinetic energy regardless of their mass. Hence, both gases A and B will initially have the same translational kinetic energy per molecule when kept at the same temperature.

Root Mean Square Speed

The root mean square (rms) speed is a statistical measure used to describe the speed of particles in a gas. It is a crucial concept in understanding the motion of gas molecules within the framework of the kinetic theory of gases. The formula for calculating the rms speed is:\[ v_{rms} = \sqrt{ \frac{3kT}{m} } \]where:

• \( v_{rms} \) is the root mean square speed,
• \( k \) is the Boltzmann constant,
• \( T \) is the absolute temperature in kelvins,
• \( m \) is the mass of a single molecule of the gas.

The rms speed is proportional to the square root of the temperature and inversely proportional to the square root of the molecular mass. This means that at the same temperature, lighter molecules (like those in gas A) will move faster, with a higher rms speed compared to heavier molecules (like those in gas B). To balance this, increasing the temperature of the heavier gas can make its rms speed equal to that of a lighter gas at a lower temperature.

Ideal Gas Law

The ideal gas law is a fundamental equation in thermodynamics and kinetic theory that relates the four significant properties of gases: pressure, volume, temperature, and number of moles. It is expressed as \[ PV = nRT \]where:

• \( P \) is the pressure of the gas,
• \( V \) is the volume,
• \( n \) is the number of moles,
• \( R \) is the ideal gas constant \((8.314 \text{ J/(mol K)})\),
• \( T \) is the temperature in kelvins.

Although the ideal gas law doesn't directly determine the kinetic energy or speed of gas molecules, it establishes the relationship between macroscopic variables that provide overall insights into the behavior of gases. When temperature increases, reflecting an increase in translational kinetic energy, the pressure and/or volume of the gas changes, assuming the number of moles remains constant. In this exercise, maintaining pressure and changing the temperature helps achieve a specific speed or kinetic energy state for a gas.