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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
At 273K, the average kinetic energies of both CH4(g) and N2(g) are \(KE_{avg(CH_4, 273K)} = KE_{avg(N_2, 273K)} = 5.60 \times 10^{-21}\) J. At 546K, their average kinetic energies are \(KE_{avg(CH_4, 546K)} = KE_{avg(N_2, 546K)} = 1.12 \times 10^{-20}\) J.

Step by step solution

01

Identify the given information

We are given the following information: - The temperatures: 273K and 546K - The Boltzmann constant: \(1.38 \times 10^{-23}~ J/mol~K\) We will use this information to find the average kinetic energies of CH4 and N2 molecules.
02

Calculate the average kinetic energy at 273K

Using the formula for the average kinetic energy, we will find the avg. kinetic energy for both CH4 and N2 molecules at 273K: \(KE_{avg}=\dfrac{3}{2}kT\) For CH4(g) at 273K: \(KE_{avg(CH_4, 273K)}=\dfrac{3}{2}(1.38 \times 10^{-23}~ J/mol~K)(273~K) = \) For N2(g) at 273K: \(KE_{avg(N_2, 273K)}=\dfrac{3}{2}(1.38\times10^{-23}~ J/mol~K)(273~K) = \) Since both molecules are at the same temperature (273K), their average kinetic energies will be the same.
03

Calculate the average kinetic energy at 546K

Using the same formula, we will find the average kinetic energy for CH4 and N2 molecules at 546K: \(KE_{avg}=\dfrac{3}{2}kT\) For CH4(g) at 546K: \(KE_{avg(CH_4, 546K)}=\dfrac{3}{2}(1.38\times10^{-23}~ J/mol~K)(546~K) = \) For N2(g) at 546K: \(KE_{avg(N_2, 546K)}=\dfrac{3}{2}(1.38\times10^{-23}~ J/mol~K)(546~K) = \) As in step 2, since both molecules are at the same temperature (546K), their average kinetic energies will be the same.
04

Finalize the answers

We have calculated the average kinetic energies of CH4 and N2 gas molecules at given temperatures. At 273K: - CH4(g): \(KE_{avg(CH_4, 273K)} = \) J - N2(g): \(KE_{avg(N_2, 273K)} = \) J At 546K: - CH4(g): \(KE_{avg(CH_4, 546K)} = \) J - N2(g): \(KE_{avg(N_2, 546K)} = \) J

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Key Concepts

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

Boltzmann Constant
The Boltzmann constant is a fundamental physical constant that connects the macroscopic and microscopic physical worlds. It is denoted by the symbol \(k\) and has a value of approximately \(1.38 \times 10^{-23} \, J/K\). This constant plays a crucial role in statistical mechanics and thermodynamics.

In the context of gases, the Boltzmann constant helps in determining the average kinetic energy of particles. It's particularly useful for calculating properties of individual molecules within gases. When computing the average kinetic energy of gas molecules, we use the formula:
  • \(KE_{avg} = \dfrac{3}{2}kT\)
where \(T\) is the temperature in Kelvin.

Understanding the Boltzmann constant enables us to bridge the gap between bulk thermodynamic quantities, such as temperature, and the behavior of individual particles, giving us insight into molecular motion.
Temperature
Temperature is a measure of the average kinetic energy of the particles in a substance. It's a fundamental concept in thermodynamics that dictates how energy spreads within and between systems. Measured in Kelvin (K), it is directly proportional to the average energy per degree of freedom of the particles.

In gases, increasing the temperature increases the average kinetic energy of the gas molecules. This is why the average kinetic energy formula \(KE_{avg} = \dfrac{3}{2}kT\) includes temperature as a direct multiplier. By knowing the temperature, we can determine how much energy gas molecules have on average.

Understanding temperature helps us comprehend how changes in thermal energy affect molecular speed and, consequently, properties like pressure and volume in gases under the ideal gas law.
Gas Molecules
Gas molecules are in constant, random motion, and their behavior is essential for understanding the properties of gases. These small particles, like those of
  • Methane (\(CH_4\)), and
  • Nitrogen (\(N_2\))
obey the kinetic molecular theory. This theory asserts that gas molecules are in perpetual motion, colliding elastically with each other and the walls of their container, which creates pressure.

Given temperatures, such as 273 K and 546 K, can significantly influence gas molecule behavior. At these conditions, the average kinetic energy of gas molecules can be calculated and demonstrates how temperature affects molecular movement. With formulas involving the Boltzmann constant, we can determine how energetic these molecules are under different thermodynamic conditions.

This understanding allows us to predict the behavior and properties of gases in diverse conditions, which is vital for both scientific and practical applications.

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Most popular questions from this chapter

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(g),\) ammonia, \(\mathrm{NH}_{3}(g),\) and oxygen, \(\mathrm{O}_{2}(g),\) at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of \(20.0 \mathrm{L} \mathrm{CH}_{4}(g), 20.0 \mathrm{L} \mathrm{NH}_{3}(g),\) and 20.0 \(\mathrm{L} \mathrm{O}_{2}(g) ?\) The volumes of all gases are measured at the same temperature and pressure.

Methane \(\left(\mathrm{CH}_{4}\right)\) gas flows into a combustion chamber at a rate of \(200 . \mathrm{L} / \mathrm{min}\) at 1.50 \(\mathrm{atm}\) and ambient temperature. Air is added to the chamber at 1.00 \(\mathrm{atm}\) and the same temperature, and the gases are ignited. a. To ensure complete combustion of \(\mathrm{CH}_{4}\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g),\) three times as much oxygen as is necessary is reacted. Assuming air is 21 mole percent \(\mathrm{O}_{2}\) and 79 \(\mathrm{mole}\) percent \(\mathrm{N}_{2},\) calculate the flow rate of air necessary to deliver the required amount of oxygen. b. Under the conditions in part a, combustion of methane was not complete as a mixture of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{CO}(g)\) was produced. It was determined that 95.0\(\%\) of the carbon in the exhaust gas was present in \(\mathrm{CO}_{2}\) . The remainder was present as carbon in CO. Calculate the composition of the exhaust gas in terms of mole fraction of \(\mathrm{CO}, \mathrm{CO}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . Assume \(\mathrm{CH}_{4}\) is completely reacted and \(\mathrm{N}_{2}\) is unreacted.

Suppose two \(200.0-\mathrm{L}\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

Silane, SiH, , is the silicon analogue of methane, \(\mathrm{CH}_{4}\) . It is prepared industrially according to the following equations: $$\begin{array}{c}{\mathrm{Si}(s)+3 \mathrm{HCl}(g) \longrightarrow \mathrm{HSiCl}_{3}(l)+\mathrm{H}_{2}(g)} \\ {4 \mathrm{HSiCl}_{3}(l) \longrightarrow \mathrm{SiH}_{4}(g)+3 \mathrm{SiCl}_{4}(l)}\end{array}$$ a. If \(156 \mathrm{mL}\) \(\mathrm{HSiCl}_{3} (d=1.34 \mathrm{g} / \mathrm{mL})\) is isolated when 15.0 \(\mathrm{L}\) \(\mathrm{HCl}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) is used, what is the percent yield of \(\mathrm{HSiCl}_{3} ?\) b. When \(156 \mathrm{HSiCl}_{3}\) is heated, what volume of \(\mathrm{SiH}_{4}\) at 10.0 \(\mathrm{atm}\) and \(35^{\circ} \mathrm{C}\) will be obtained if the percent yield of the reaction is 93.1\(\% ?\)

Consider the following chemical equation. $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ If 25.0 \(\mathrm{mL}\) \(\mathrm{NO}_{2}\) gas is completely converted to \(\mathrm{N}_{2} \mathrm{O}_{4}\) gas under the same conditions, what volume will the \(\mathrm{N}_{2} \mathrm{O}_{4}\) occupy?

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