Problem 56
Question
What is the rms speed of the molecules in low-density oxygen gas at \(0^{\circ} \mathrm{C} ?\) (The mass of an oxygen molecule, \(\mathrm{O}_{2},\) is \(\left.5.31 \times 10^{-26} \mathrm{~kg}\right)\).
Step-by-Step Solution
Verified Answer
The rms speed of the oxygen molecules is approximately 462 m/s.
1Step 1: Convert Temperature to Kelvin
The given temperature is \(0^{\circ} \mathrm{C}\). To convert this to Kelvin, we add 273.15. Thus, the temperature in Kelvin is\(T = 273.15 \, \mathrm{K}\).
2Step 2: Write the RMS Speed Formula
The formula for the root mean square (rms) speed \(v_{\text{rms}}\) of gas molecules is given by \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\), where \(k\) is the Boltzmann constant \(1.38 \times 10^{-23}\, \mathrm{J/K}\), and \(m\) is the mass of one molecule of the gas.
3Step 3: Substitute the Values into the Formula
Using the formula from the previous step: \(v_{\text{rms}} = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \, \mathrm{J/K} \times 273.15 \, \mathrm{K}}{5.31 \times 10^{-26} \, \mathrm{kg}}}\).
4Step 4: Calculate the RMS Speed
First calculate the expression inside the square root: \(v_{\text{rms}} = \sqrt{\frac{1.1298 \times 10^{-20} \, \mathrm{J}}{5.31 \times 10^{-26} \, \mathrm{kg}}}\). \(v_{\text{rms}} \approx \sqrt{2.127 \times 10^5} \, \mathrm{m/s}\), which results in \(v_{\text{rms}} \approx 462.1 \, \mathrm{m/s}\).
Key Concepts
Kinetic Theory of GasesOxygen MoleculeTemperature ConversionBoltzmann Constant
Kinetic Theory of Gases
The kinetic theory of gases is a fundamental concept that explains the behavior of gases at the molecular level. It describes how gas molecules move and interact, leading to macroscopic properties such as pressure, volume, and temperature. According to this theory:
- Gas molecules are in constant random motion, moving in straight lines until they collide with either another molecule or the walls of a container.
- These collisions are perfectly elastic, meaning there is no loss of kinetic energy in the system.
- The average kinetic energy of the gas molecules is directly related to the temperature of the system. As temperature increases, so does the average kinetic energy.
Oxygen Molecule
An oxygen molecule, denoted as \(O_2\), consists of two oxygen atoms bonded together. Being a diatomic molecule, it is one of the most abundant elements found in nature. It is essential for life on Earth as it is a key component of the air we breathe.
In physics and chemistry, understanding the properties of an oxygen molecule is vital for calculations involving gases, especially in kinetic energy and speed computations. The mass of an oxygen molecule is particularly important when calculating the RMS speed. In this case, the mass is given as \(5.31 \times 10^{-26} \text{ kg}\).
Recognizing the molecular mass allows us to plug this value into various equations that predict the behavior of gases under different conditions, such as temperature changes.
In physics and chemistry, understanding the properties of an oxygen molecule is vital for calculations involving gases, especially in kinetic energy and speed computations. The mass of an oxygen molecule is particularly important when calculating the RMS speed. In this case, the mass is given as \(5.31 \times 10^{-26} \text{ kg}\).
Recognizing the molecular mass allows us to plug this value into various equations that predict the behavior of gases under different conditions, such as temperature changes.
Temperature Conversion
Temperature conversion is a common task in physics, especially in thermodynamics and kinetic theory calculations. To perform accurate computations, temperatures must be in Kelvin.
- The Kelvin scale is an absolute temperature scale that begins at absolute zero, the point at which all molecular motion ceases.
- To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. This shifts the zero point to absolute zero, allowing more precise mathematical handling of thermal equations.
Boltzmann Constant
The Boltzmann constant is a fundamental constant in physics that relates the average kinetic energy of particles in a gas with the temperature of the gas. It has the value \(1.38 \times 10^{-23} \text{ J/K}\) and plays a crucial role in the kinetic theory of gases, as it connects microscopic motion with macroscopic properties like temperature.
When calculating the RMS speed, the Boltzmann constant appears in the formula \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\). Here, it ensures that the units are consistent and provides a bridge between the energy content of gas molecules and their speed, given a specific temperature in Kelvin.
Understanding this constant allows us to calculate how fast molecules move in different gases and under varying thermal conditions, making it indispensable for predictions in gases' behavior and thermodynamics.
When calculating the RMS speed, the Boltzmann constant appears in the formula \(v_{\text{rms}} = \sqrt{\frac{3kT}{m}}\). Here, it ensures that the units are consistent and provides a bridge between the energy content of gas molecules and their speed, given a specific temperature in Kelvin.
Understanding this constant allows us to calculate how fast molecules move in different gases and under varying thermal conditions, making it indispensable for predictions in gases' behavior and thermodynamics.
Other exercises in this chapter
Problem 53
If the average kinetic energy per molecule of a monatomic gas is \(7.0 \times 10^{-21} \mathrm{~J},\) what is the Celsius temperature of the gas?
View solution Problem 54
What is the average kinetic energy per molecule in a monatomic gas at (a) \(10^{\circ} \mathrm{C}\) and (b) \(90^{\circ} \mathrm{C} ?\)
View solution Problem 57
(a) What is the average kinetic energy per molecule of a monatomic gas at a temperature of \(25^{\circ} \mathrm{C} ?\) (b) What is the rms speed of the molecule
View solution Problem 59
A quantity of an ideal gas is at \(0^{\circ} \mathrm{C}\). An equal quantity of another ideal gas is at twice the absolute temperature. What is its Celsius temp
View solution