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Oxygen \(\left(\mathrm{O}_{2}\right)\) has a molar mass of \(32.0 \mathrm{~g} / \mathrm{mol}\). (a) What is the root-mean-square speed of an oxygen molecule at a temperature of \(300 \mathrm{~K} ?\) (b) What is its average translational kinetic energy at that speed?

Short Answer

Expert verified
The root-mean-square speed is approximately 484 m/s, and the average translational kinetic energy is about 6.21 x 10^-21 J.

Step by step solution

01

Understanding Given Data and Formula

We are given the molar mass of oxygen (\( \mathrm{O}_2 \)) as \( 32.0 \mathrm{~g/mol} \) and the temperature \( T = 300 \mathrm{~K} \). To find the root-mean-square speed, we use the formula:\[v_\text{rms} = \sqrt{\frac{3kT}{m}}.\]Where \( k \) is the Boltzmann constant \( 1.38 \times 10^{-23} \mathrm{~J/K} \), and \( m \) is the mass of one molecule of oxygen.
02

Convert Molar Mass to Mass of a Single Molecule

Since molar mass is given in \( \mathrm{g/mol} \), convert this to kilograms and divide by Avogadro's number \( 6.022 \times 10^{23} \) to find the mass of one molecule:\[m = \frac{32.0 \times 10^{-3} \mathrm{~kg/mol}}{6.022 \times 10^{23} \mathrm{~molecules/mol}}.\]
03

Calculate Root-Mean-Square Speed

Substitute the values of \( k \), \( T \), and \( m \) into the root-mean-square speed formula:\[v_\text{rms} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \mathrm{~J/K} \times 300 \mathrm{~K}}{m}}.\]Calculate to obtain \( v_\text{rms} \).
04

Formula for Average Translational Kinetic Energy

The average translational kinetic energy \( KE \) for a molecule is given by:\[KE = \frac{3}{2} kT.\]
05

Calculate Average Translational Kinetic Energy

Using the temperature \( T = 300 \mathrm{~K} \) and the Boltzmann constant \( k = 1.38 \times 10^{-23} \mathrm{~J/K} \), substitute into the kinetic energy formula:\[KE = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300.\]Calculate to find the kinetic energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Root-Mean-Square Speed
The root-mean-square speed (commonly referred to as \(v_{\text{rms}}\)) of a gas molecule is a measure that helps us understand how fast a molecule in a gas is moving on average. If you imagine all the gas molecules, they don't travel at the same speed. Instead, some are faster and some are slower. The root-mean-square speed gives us a way to represent an average molecular speed in a mathematical manner. To calculate \(v_{\text{rms}}\), you use the formula:\[v_{\text{rms}} = \sqrt{\frac{3kT}{m}}.\]This equation includes:
  • \(k\): Boltzmann constant, which is \(1.38 \times 10^{-23} \ \text{J/K}\).
  • \(T\): Temperature in Kelvin.
  • \(m\): Mass of a single molecule in kilograms.
First, you need to find the mass of a single molecule, which involves converting the molar mass from grams per mole to kilograms, and then dividing by Avogadro's number. The root-mean-square speed is particularly useful as it accounts for the random nature of molecular motion.
Translational Kinetic Energy
Translational kinetic energy (K.E.) of a molecule relates to its motion through space. When we talk about the translational kinetic energy of gas molecules, we are looking at the energy due to their movement from one place to another. A very useful formula to determine this energy is:\[KE = \frac{3}{2} kT.\]In this formula:
  • \(k\) is the Boltzmann constant, \(1.38 \times 10^{-23} \ \text{J/K}\).
  • \(T\) is the temperature in Kelvin.
Thus, the translational kinetic energy is directly proportional to the temperature. When the temperature increases, the kinetic energy of the molecules also increases, resulting in faster molecular motion. This formula tells us that temperature is a measure of the average kinetic energy of the molecules in a system.
Boltzmann Constant
The Boltzmann constant is a fundamental constant in physics, denoted by \(k\), with a value of \(1.38 \times 10^{-23} \ \text{J/K}\). It serves as a bridge between the macroscopic and microscopic worlds. By using the Boltzmann constant, we can connect the average kinetic energy of particles (like molecules in a gas) with macroscopic quantities like temperature. Some Important Aspects:
  • Role: It plays a significant role in the formulas for both root-mean-square speed and translational kinetic energy.
  • Utility: It is fundamental in statistical mechanics and thermodynamics.
Without the Boltzmann constant, it would be challenging to translate the microscopic state of individual molecules to the broader, observable extent of a gas's behavior at different temperatures.
Molar Mass Conversion
The concept of molar mass conversion is essential when calculating properties such as root-mean-square speed. Molar mass is the mass of one mole of a substance (like oxygen, \(\mathrm{O}_2\)), usually given in grams per mole (\(\mathrm{g/mol}\)). However, to use it in calculations involving molecules, we often need to convert it to kilograms per molecule.Here's how you can perform this conversion:
  • First, convert grams to kilograms by dividing by 1000.
  • Then, divide by Avogadro's number \(6.022 \times 10^{23}\), which gives you the mass of a single molecule in kilograms.
This conversion is vital because it allows for accurate calculations of molecular speeds and energies in physics problems. By getting the mass of a single molecule, you can accurately substitute it into formulas that require the mass to be in appropriate units.

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Most popular questions from this chapter

(a) How much heat does it take to increase the temperature of 2.50 moles of an ideal monatomic gas from \(25.0^{\circ} \mathrm{C}\) to \(55.0^{\circ} \mathrm{C}\) if the gas is held at constant volume? (b) How much heat is needed if the gas is diatomic rather than monatomic? (c) Sketch a \(p V\) diagram for these processes.

Starting with \(2.50 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) gas (assumed to be ideal) in a cylinder at 1.00 atm and \(20.0^{\circ} \mathrm{C}\), a chemist first heats the gas at constant volume, adding \(1.52 \times 10^{4} \mathrm{~J}\) of heat, then continues heating and allows the gas to expand at constant pressure to twice its original volume. (a) Calculate the final temperature of the gas. (b) Calculate the amount of work done by the gas. (c) Calculate the amount of heat added to the gas while it was expanding. (d) Calculate the change in internal energy of the gas for the whole process.

If you have ever hiked or climbed to high altitudes in the mountains, you surely have noticed how short of breath you get. This occurs because the air is thinner, so each breath contains fewer \(\mathrm{O}_{2}\) molecules than at sea level. At the top of Mount Everest, the pressure is only \(\frac{1}{3}\) atm. Air contains \(21 \%\) O \(_{2}\) and \(78 \%\) N \(_{2}\), and an average human breath is \(0.50 \mathrm{~L}\) of air. At the top of Mount Everest, (a) how many \(\mathrm{O}_{2}\) molecules does each breath contain when the temperature is \(-15^{\circ} \mathrm{F},\) and \((\mathrm{b})\) what percent is this of the number of \(\mathrm{O}_{2}\) molecules you would get from a breath at sea level at \(-15^{\circ} \mathrm{F}\) ?

The surface of the sun. The surface of the sun has a temperature of about \(5800 \mathrm{~K}\) and consists largely of hydrogen atoms. (a) Find the rms speed of a hydrogen atom at this temperature. (The mass of a single hydrogen atom is \(1.67 \times 10^{-27} \mathrm{~kg} .\) ) (b) What would be the mass of an atom that had half the rms speed of hydrogen?

The average temperature of the atmosphere near the surface of the earth is about \(20^{\circ} \mathrm{C}\). (a) What is the root-mean-square speed of hydrogen molecules, \(\mathrm{H}_{2}\), at this temperature? (b) The escape speed from the earth is about \(11 \mathrm{~km} / \mathrm{s}\). Is the average \(\mathrm{H}_{2}\) molecule moving fast enough to escape? (c) Compare the rms speeds of oxygen \(\left(\mathrm{O}_{2}\right)\) and nitrogen \(\left(\mathrm{N}_{2}\right)\) with that of \(\mathrm{H}_{2}\). (d) So why has the hydrogen been able to escape the earth's gravity, but the heavier gases (such as \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) ) have not, even though none of these gases has an rms speed equal to the escape speed of the earth? (Hint: Do all the molecules have the same rms speed, or are some moving faster? since the rms speed for \(\mathrm{H}_{2}\) is greater than that of \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\), which gas would have a higher percentage of its molecules moving fast enough to escape?)

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