Problem 37 - University Physics with Modern Physics

Question

(a) Oxygen (O) has a molar mass of 32.0 \(\mathrm{g} / \mathrm{mol} .\) What is the average translational kinetic energy of an oxygen molecule at a temperature of 300 \(\mathrm{K}\) ? (b) What is the average value of the square of its speed? (c) What is the root-mean-square speed? (d) What is the momentum of an oxygen molecule traveling at this speed? (e) Suppose an oxygen molecule traveling at this speed bounces back and forth between opposite sides of a cubical vessel 0.10 \(\mathrm{m}\) on a side. What is the average force the molecule exerts on one of the walls of the container? (Assume that the molecule's velocity is perpendicular to the two sides that it strikes.) (f) What is the average force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 atm? (h) Compute the number of oxygen molecules that are actually contained in a vessel of this size at 300 \(\mathrm{K}\) and atmospheric pressure. (i) Your answer for part (h) should be three times as large as the answer for part (g). Where does this discrepancy arise?

Step-by-Step Solution

Answer

(a) 6.21 × 10⁻²¹ J (b) 1.56 × 10⁵ m²/s² (c) 395 m/s (d) 2.10 × 10⁻²³ kg·m/s (e) 8.34 × 10⁻²¹ N (f) 8.34 × 10⁻¹⁹ N/m² (g) 1.21 × 10²³ molecules (h) 2.44 × 10²¹ molecules (i) Ideal assumptions simplify the speed distribution.

Step 1: Calculate Average Translational Kinetic Energy

The average translational kinetic energy of a gas molecule is given by\[ K = \frac{3}{2} k_B T \]where \( k_B = 1.38 \times 10^{-23} \ \text{J/K} \) is the Boltzmann constant, and \( T = 300 \ \text{K} \) is the temperature. Substituting the values,\[ K = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21} \ \text{J} \]

Step 2: Calculate Average Value of the Square of Speed

The average value of the square of speed can be found using\[ \langle v^2 \rangle = \frac{3k_B T}{m} \]where \( m \) is the mass of an oxygen molecule. Given the molar mass of oxygen is 32.0 g/mol, convert this to kg/molecule using Avogadro's number \( N_A = 6.022 \times 10^{23} \ \text{mol}^{-1} \):\[ m = \frac{32.0}{1000} \div 6.022 \times 10^{23} = 5.32 \times 10^{-26} \ \text{kg} \]Then,\[ \langle v^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 300}{5.32 \times 10^{-26}} = 1.56 \times 10^{5} \ \text{m}^2/\text{s}^2 \]

Step 3: Calculate Root-Mean-Square Speed

The root-mean-square speed \( v_{rms} \) is given by\[ v_{rms} = \sqrt{\langle v^2 \rangle} = \sqrt{1.56 \times 10^5} = 395 \ \text{m/s} \]

Step 4: Calculate Momentum of an Oxygen Molecule

The momentum \( p \) of an oxygen molecule traveling at root-mean-square speed is given by\[ p = m \times v_{rms} = 5.32 \times 10^{-26} \times 395 = 2.10 \times 10^{-23} \ \text{kg} \cdot \text{m/s} \]

Step 5: Calculate Average Force Exerted on Wall

The force \( F \) exerted by a molecule traveling back and forth in a cubical vessel is given by\[ F = \frac{\Delta p}{\Delta t} = \frac{2mv}{2d/v} = \frac{mv^2}{d} \]where \( d = 0.10 \ \text{m} \) is the side length of the cube. Therefore,\[ F = \frac{5.32 \times 10^{-26} \times (395)^2}{0.10} = 8.34 \times 10^{-21} \ \text{N} \]

Step 6: Calculate Average Force per Unit Area

The force per unit area (pressure) \( P \) is given by\[ P = \frac{F}{A} = \frac{8.34 \times 10^{-21}}{(0.10)^2} = 8.34 \times 10^{-19} \ \text{N/m}^2 \]

Step 7: Calculate Number of Molecules for 1 atm Pressure

To achieve 1 atm pressure (\(1.01 \times 10^5 \ \text{N/m}^2\)), the number of molecules \( N \) can be found by setting \( N \times 8.34 \times 10^{-19} = 1.01 \times 10^5 \) and solving for \( N \):\[ N = \frac{1.01 \times 10^5}{8.34 \times 10^{-19}} = 1.21 \times 10^{23} \]

Step 8: Calculate Actual Number of Molecules

At 300 K and 1 atm, use the ideal gas law \( PV = nRT \) to find the number of moles \( n \) in the volume \( V = 0.10^3 \ \text{m}^3 \). With \( R = 8.314 \ \text{J/mol} \cdot\text{K} \),\[ n = \frac{1.01 \times 10^5 \times 0.10^3}{8.314 \times 300} = 4.05 \times 10^{-3} \ \text{mol} \]Convert to number of molecules:\[ N = 4.05 \times 10^{-3} \times 6.022 \times 10^{23} = 2.44 \times 10^{21} \]

Step 9: Analyze Discrepancy

The discrepancy arises from the assumption in part (g) that all molecules move at the root-mean-square speed, which is an oversimplification. The Maxwell-Boltzmann distribution suggests a range of speeds, meaning fewer molecules than calculated are needed to sustain 1 atm pressure.

Key Concepts

Translational Kinetic Energy

In a gas, molecules are constantly in motion, colliding with each other and the walls of their container. This motion is described by translational kinetic energy, which is the energy due to the molecule's movement in space. For any molecule, such as an oxygen molecule, the average translational kinetic energy can be calculated using the formula:

• \( K = \frac{3}{2} k_B T \)

where \( k_B \) is the Boltzmann constant \( (1.38 \times 10^{-23} \, \text{J/K}) \) and \( T \) is the temperature in Kelvin.
The higher the temperature, the more energy each molecule has. This energy is directly related to the motion and collisions of the molecules with each other and their surroundings.
This calculation is crucial in thermodynamics to predict how gases will behave at different temperatures and pressures.

Root-Mean-Square Speed

The root-mean-square speed is a statistical measure of the speed of particles in a gas. It provides a way to determine an average speed that reflects the particle dynamics accurately.The formula to find the root-mean-square speed \( v_{rms} \) is:

• \( v_{rms} = \sqrt{\langle v^2 \rangle} \)

Where \( \langle v^2 \rangle \) is the average value of the squares of the velocities of the particles.
This value helps understand the overall motion of the gas particles, allowing predictions about properties such as diffusion rates and reaction kinetics.
In physical terms, the root-mean-square speed is larger than the average speed, as it is derived from the mean of the square of velocities, giving more weight to higher velocities.

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. It is expressed as:

• \( PV = nRT \)

Where:

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

This equation helps relate the macroscopic properties of gases so that if we know any three, we can solve for the fourth.
In practical terms, this equation is used to predict and calculate changes in one of the gas properties when the others are changed, such as during chemical reactions or in processes like combustion.

Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution is a statistical means of describing the range of speeds found within a collection of gas particles at a given temperature. It predicts how many particles will have a speed within a given small range. This distribution is key because it explains why molecules have different kinetic energies, even at constant temperature.
This concept is particularly useful when considering the macroscopic properties of gases.

• For instance, not all oxygen molecules travel at the root-mean-square speed, as proved by the spread around the average described by this distribution.
• The range of molecular speeds influences how gases mix, react, and ultimately, how they exert pressure.

A fundamental understanding of this concept is essential for explaining gas behavior beyond the approximations of simpler models like the ideal gas law.