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Calculate the average kinetic energies of \(\mathrm{CH}_{4}(g)\) and \(\mathrm{N}_{2}(g)\) molecules at \(273 \mathrm{~K}\) and \(546 \mathrm{~K}\).

Short Answer

Expert verified
The average kinetic energies for CH4(g) and N2(g) at 273K and 546K are as follows: - For CH4 molecule at 273K: \(KE_{avg} = 5.65 \times 10^{-21} J\) - For CH4 molecule at 546K: \(KE_{avg} = 1.13 \times 10^{-20} J\) - For N2 molecule at 273K: \(KE_{avg} = 5.65 \times 10^{-21} J\) - For N2 molecule at 546K: \(KE_{avg} = 1.13 \times 10^{-20} J\)

Step by step solution

01

Identify the Temperature Values

We are given two temperature values, 273K and 546K. We will calculate the average kinetic energies for CH4 and N2 at these temperatures.
02

Calculate the Average Kinetic Energy for CH4 at 273K

Using the formula for calculating average kinetic energy, we can determine the average kinetic energy for CH4 at 273K: \(KE_{avg} = \dfrac{3}{2} k_B T \) \(KE_{avg(CH4, 273K)} = \dfrac{3}{2} (1.38 \times 10^{-23}\,\cancel{J\,K^{-1}}) (273\,\cancel{K}) \) \(KE_{avg(CH4, 273K)} = 5.65 \times 10^{-21} J\)
03

Calculate the Average Kinetic Energy for CH4 at 546K

Using the same formula, we can determine the average kinetic energy for CH4 at 546K: \(KE_{avg} = \dfrac{3}{2} k_B T \) \(KE_{avg(CH4, 546K)} = \dfrac{3}{2} (1.38 \times 10^{-23}\,\cancel{J\,K^{-1}}) (546\,\cancel{K}) \) \(KE_{avg(CH4, 546K)} = 1.13 \times 10^{-20} J\)
04

Calculate the Average Kinetic Energy for N2 at 273K

Using the formula for calculating average kinetic energy, we can determine the average kinetic energy for N2 at 273K: \(KE_{avg} = \dfrac{3}{2} k_B T \) \(KE_{avg(N2, 273K)} = \dfrac{3}{2} (1.38 \times 10^{-23}\,\cancel{J\,K^{-1}}) (273\,\cancel{K}) \) \(KE_{avg(N2, 273K)} = 5.65 \times 10^{-21} J\)
05

Calculate the Average Kinetic Energy for N2 at 546K

Using the same formula, we can determine the average kinetic energy for N2 at 546K: \(KE_{avg} = \dfrac{3}{2} k_B T \) \(KE_{avg(N2, 546K)} = \dfrac{3}{2} (1.38 \times 10^{-23}\,\cancel{J\,K^{-1}}) (546\,\cancel{K}) \) \(KE_{avg(N2, 546K)} = 1.13 \times 10^{-20} J\) Now we have calculated the average kinetic energies for both CH4(g) and N2(g) at 273K and 546K: - For CH4 molecule at 273K: \(KE_{avg} = 5.65 \times 10^{-21} J\) - For CH4 molecule at 546K: \(KE_{avg} = 1.13 \times 10^{-20} J\) - For N2 molecule at 273K: \(KE_{avg} = 5.65 \times 10^{-21} J\) - For N2 molecule at 546K: \(KE_{avg} = 1.13 \times 10^{-20} J\)

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Key Concepts

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

Kinetic Molecular Theory
The kinetic molecular theory is a cornerstone in the understanding of gas behaviors. It states that gases are composed of a large number of particles that are in constant, random motion. These gas molecules collide with each other and the walls of their container without losing energy in elastic collisions. The theory helps us determine the relationship between the macroscopic properties of gases, such as pressure and temperature, and the microscopic behaviors of the gas molecules themselves.

According to this theory, the temperature of a gas is proportional to the average kinetic energy of its molecules. This implies that if we increase the temperature of a gas, its molecules will move faster and, as a result, the average kinetic energy will also increase. The converse is true; a lower temperature means slower molecular motion and lower average kinetic energy.
Gas Molecules Kinetic Energy
The kinetic energy of gas molecules is the energy associated with their motion. Each individual molecule possesses translational kinetic energy as it moves in three-dimensional space. While these molecules have varying speeds, what we measure in thermodynamics is the average kinetic energy of the gas molecules. The faster they move, the greater their kinetic energy.

When calculating the average kinetic energy of gas molecules, we use the formula expressed during the exercise:\[\begin{equation}KE_{avg} = \frac{3}{2} k_B T \end{equation}\]This equation highlights that average kinetic energy is directly dependent on the absolute temperature of the gas. Thus, kinetic energy serves as a microscopic indicator of the gas's temperature.
Temperature and Kinetic Energy
There is a direct and proportional relationship between temperature and the kinetic energy of gas molecules. As temperature increases, the kinetic energy of the gas molecules also increases. This is due to the fact that temperature is a measure of the average kinetic energy of the particles in a substance.

It is crucial to note that temperature is measured on an absolute scale, such as Kelvin, where 0 K is absolute zero, the point at which particles theoretically stop moving. The step-by-step solution given for different temperatures, such as 273 K and 546 K, shows how the kinetic energy of molecules changes with the temperature. The results reveal that the average kinetic energy doubles when the temperature is doubled, demonstrating the direct proportionality between these two variables.
Boltzmann Constant
The Boltzmann constant (\[\begin{equation}k_B\end{equation}\]) is a fundamental physical constant that serves as a bridge between macroscopic and microscopic physics. It appears in the equation that relates the average kinetic energy of particles to temperature, reflecting the individual components' energies on the molecular scale.

The value of the Boltzmann constant is approximately \[\begin{equation}1.38 \times 10^{-23} J/K\end{equation}\], where J represents joules and K represents Kelvin. As indicated in the exercise, when we use the Boltzmann constant in our calculations, we can quantify the average kinetic energy per molecule in a gas. It is important to realize that this constant is a significant part of the ideal gas law and the kinetic molecular theory, providing an essential link between the temperature of a gas and the average kinetic energy of its molecules.

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Most popular questions from this chapter

If you put a drinking straw in water, place your finger over the opening, and lift the straw out of the water, some water stays in the straw. Explain.

Explain the following seeming contradiction: You have two gases, \(A\) and \(B\), in two separate containers of equal volume and at equal pressure and temperature. Therefore, you must have the same number of moles of each gas. Because the two temperatures are equal, the average kinetic energies of the two samples are equal. Therefore, since the energy given such a system will be converted to translational motion (that is, move the molecules), the root mean square velocities of the two are equal, and thus the particles in each sample move, on average, with the same relative speed. Since \(A\) and \(B\) are different gases, they each must have a different molar mass. If \(A\) has a higher molar mass than \(B\), the particles of \(A\) must be hitting the sides of the container with more force. Thus the pressure in the container of gas \(A\) must be higher than that in the container with gas \(B\). However, one of our initial assumptions was that the pressures were equal.

An 11.2-L sample of gas is determined to contain \(0.50\) mole of \(\mathrm{N}_{2}\). At the same temperature and pressure, how many moles of gas would there be in a 20.-L sample?

At the same conditions of pressure and temperature, ammonia gas is less dense than air. Why is this true?

For each of the quantities listed below, explain which of the following properties (mass of the molecule, density of the gas sample, temperature of the gas sample, size of the molecule, and number of moles of gas) must be known to calculate the quantity. a. average kinetic energy b. average number of collisions per second with other gas molecules c. average force of each impact with the wall of the container d. root mean square velocity e. average number of collisions with a given area of the container f. distance between collisions

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