Problem 37 - University Physics with Modern Physics

Question

(a) Oxygen \(\left(\mathrm{O}_{2}\right)\) 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 at a 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 aver- age force per unit area? (g) How many oxygen molecules traveling at this speed are necessary to produce an average pressure of 1 \(\mathrm{atm} ?\) (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

The average translational kinetic energy is \(6.21 \times 10^{-21}\) J, \( v_{rms} = 774\) m/s, and \( 3.16 \times 10^{24} \) molecules are needed for 1 atm. Discrepancies arise from calculation assumptions.

Step 1: Determine average translational kinetic energy

The average translational kinetic energy for a single molecule is given by the formula \( \frac{3}{2} k_B T \), where \( k_B \) is the Boltzmann constant \( (1.38 \times 10^{-23} \, \text{J/K}) \) and \( T \) is the temperature in Kelvin. For \( T = 300 \, \text{K} \), calculate:\[\text{Average kinetic energy} = \frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K} = 6.21 \times 10^{-21} \, \text{J}\]

Step 2: Calculate average square of speed

The relation between average translational kinetic energy and the square of speed \( v^2 \) is \( \frac{1}{2}mv^2 = \frac{3}{2}k_BT \). Rearrange to find \( \langle v^2 \rangle \):\[\langle v^2 \rangle = \frac{3 k_B T}{m}\]Converting the molar mass into kg (\( \frac{32.0 \, \text{g/mol}}{1000} = 0.032 \, \text{kg/mol} \)), the mass of one molecule \( m = \frac{0.032}{6.022 \times 10^{23}} \, \text{kg} \).Then,\[\langle v^2 \rangle = \frac{3 \times 1.38 \times 10^{-23} \times 300}{0.032/6.022 \times 10^{23}}\]Evaluate this to find \( \langle v^2 \rangle \approx 5.99 \times 10^5 \, \text{m}^2/\text{s}^2 \).

Step 3: Find the root-mean-square speed

The root-mean-square speed \( v_{rms} \) is the square root of \( \langle v^2 \rangle \):\[v_{rms} = \sqrt{5.99 \times 10^5 \, \text{m}^2/\text{s}^2} \approx 774 \, \text{m/s}\]

Step 4: Calculate the momentum of an oxygen molecule

The momentum \( p \) of an oxygen molecule traveling at \( v_{rms} \) is given by \( p = mv_{rms} \):\[p = \left( \frac{0.032}{6.022 \times 10^{23}} \right) \times 774 \, \text{m/s} = 4.12 \times 10^{-26} \, \text{kg m/s}.\]

Step 5: Determine average force exerted on the wall

If a molecule bounces between two walls, changing its velocity each time, the average force \( F \) it exerts is \( F = \frac{2p}{\Delta t} \), where \( \Delta t \) is the time for one round trip.\[\Delta t = \frac{0.2}{774} \, \text{s} \approx 2.58 \times 10^{-4} \, \text{s}\]\[F = \frac{2 \times 4.12 \times 10^{-26} \, \text{kg m/s}}{2.58 \times 10^{-4} \, \text{s}} \approx 3.20 \times 10^{-22} \, \text{N}\]

Step 6: Calculate average force per unit area

Force per unit area (pressure) is given by \( P = \frac{F}{A} \), where \( A = 0.1 \times 0.1 = 0.01 \, \text{m}^2 \):\[P = \frac{3.20 \times 10^{-22} \, \text{N}}{0.01 \, \text{m}^2} = 3.20 \times 10^{-20} \, \text{N/m}^2\]

Step 7: Find number of molecules needed for pressure of 1 atm

Pressure \( P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{N/m}^2 \). Using \( F = n \times \text{Force of one molecule} \), solve:\[1.013 \times 10^5 = n \times 3.20 \times 10^{-20}\]\[n = \frac{1.013 \times 10^5}{3.20 \times 10^{-20}} \approx 3.16 \times 10^{24} \text{ molecules.}\]

Step 8: Calculate actual number of molecules in the vessel

Using the ideal gas law \( PV = nRT \) with \( V = 0.10^3 \, \text{m}^3 = 0.001 \, \text{m}^3 \):\[n = \frac{1.013 \times 10^5 \times 0.001}{8.31 \times 300} \approx 3.23 \times 10^{22} \text{ molecules.}\]

Step 9: Analyze discrepancy in results

From parts (g) and (h), it is observed that the number obtained in part (h) is not exactly three times the number obtained in part (g). The likely source of this discrepancy comes from assumptions made regarding the constant speed of the gases and the idealizations in calculations. The gas molecules’ actual motion is randomized and spread out over a Maxwell-Boltzmann distribution in reality.

Key Concepts

Translational Kinetic Energy

In the realm of gases, translational kinetic energy is linked to the motion of molecules. The average translational kinetic energy of a single molecule in a gas is given by \[ \frac{3}{2} k_B T \]. This formula states that the kinetic energy is dependent solely on the temperature \( T \) (measured in Kelvin) and the Boltzmann constant \( k_B \) (which is \( 1.38 \times 10^{-23} \, \text{J/K} \)).
This means that at a higher temperature, each molecule will have a higher kinetic energy. For example, to find the average kinetic energy of an oxygen molecule at \( 300 \, \text{K} \), we calculate:
\[ \text{Average kinetic energy} = \frac{3}{2} \times 1.38 \times 10^{-23} \, \text{J/K} \times 300 \, \text{K} = 6.21 \times 10^{-21} \, \text{J} \].
Understanding this can help you predict how 'energetic' a gas's molecules are, impacting their speed and how they interact with other molecules and their container walls.

Root-Mean-Square Speed

Root-mean-square speed \( v_{rms} \) is an important concept when evaluating how gas molecules move. It gives us a measure of the average speed of the molecules in a gas.
To find the root-mean-square speed, we initially calculate the average of the square of the velocity, \( \langle v^2 \rangle \), using :
\[ \langle v^2 \rangle = \frac{3 k_B T}{m} \].
In this calculation, \( m \) is the mass of one molecule of the gas. For oxygen, converting the molar mass (32.0 g/mol) into kilograms, and using Avogadro's number, gives us the mass of a single molecule.
Plugging in the numbers, \( \langle v^2 \rangle \approx 5.99 \times 10^5 \, \text{m}^2/\text{s}^2 \).
The \( v_{rms} \) is then:
\[ v_{rms} = \sqrt{\langle v^2 \rangle} \approx 774 \, \text{m/s} \].
This speed increases with increased temperature, indicating more rapid motion as heat rises.

Ideal Gas Law

The ideal gas law is a fundamental equation in the study of gases: \[ PV = nRT \].
This equation links pressure \( P \), volume \( V \), and number of moles \( n \) of a gas to its temperature \( T \), with \( R \) being the universal gas constant, \( 8.31 \, \text{J/(mol} \cdot \text{K)} \).
The ideal gas law helps in estimating the number of gas molecules present in any given volume under specified conditions. For instance, with a cubic vessel at \( 0.001 \, \text{m}^3 \) volume and assessing the situation at atmospheric pressure (\( 1.013 \times 10^5 \, \text{N/m}^2 \)) and a temperature of \( 300 \, \text{K} \), it computes approximately \( 3.23 \times 10^{22} \) molecules.
We assume that these molecules have random, constant speeds and lack interactions, conditions that the ideal gas law utilizes to make real-world gas behavior predictable despite these simplifications.

Molecular Momentum

Momentum in gases offers a glimpse into the dynamics of collisions within and against the walls of a container. For a molecule with a certain mass \( m \) moving at speed \( v \), its momentum \( p \) is simply \( p = mv \).
The oxygen molecule at root-mean-square speed (\( 774 \, \text{m/s} \)) and calculated mass gives \( p \approx 4.12 \times 10^{-26} \, \text{kg} \cdot \text{m/s} \).
When these molecules collide with a container wall, they exert a force dependent on their momentum change. Each bounce involves altering the momentum, causing a force that's twice the impulse over the period taken for a molecule to travel to the opposite wall and back.
This fundamental concept illustrates how gases apply force, translating their energetic, kinetic interactions into measurable quantities like pressure.

Pressure Calculation

Calculating pressure involves considering how frequently and forcefully gas molecules hit the container walls. Pressure \( P \) is force per unit area, \( P = \frac{F}{A} \).
Given the recurring impact of molecules, their momentum exchange establishes an average force, which when divided by the wall area (\( 0.01 \, \text{m}^2 \) for a side of the cubic vessel), provides the pressure.
In the example provided, a single molecule's cycle of striking the walls results in an average force determined by its momentum: \( F \approx 3.20 \times 10^{-22} \, \text{N}\).
This results in pressure on the order of \( 3.20 \times 10^{-20} \, \text{N/m}^2 \), much less than standard atmospheric pressure.
To achieve higher pressures akin to 1 atm, many more molecules or higher velocities are necessary, illuminating the aggregate effect of countless molecular collisions in creating significant pressure.