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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 methane (CH4) and nitrogen (N2) gas molecules at the given temperatures are: For CH4: - At 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \mathrm{J}\) - At 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \mathrm{J}\) For N2: - At 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \mathrm{J}\) - At 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \mathrm{J}\)

Step by step solution

01

Understand the formula for average kinetic energy

The formula for the average kinetic energy (\(K_{avg}\)) of a gas molecule can be expressed as: \(K_{avg} = \cfrac{3}{2} k T\) Where: - \(K_{avg}\) is the average kinetic energy - \(k\) is Boltzmann's constant (approximately \(1.38 \times 10^{-23}\, \mathrm{J / K}\)) - \(T\) is the temperature in Kelvin (K) Step 2: Calculate average kinetic energies for CH4 at 273 K and 546 K
02

Apply the formula for methane (CH4)

To find the average kinetic energy of CH4 molecules at 273 K, plug the values into the formula: \(K_{avg} = \cfrac{3}{2} (1.38 \times 10^{-23} \, \mathrm{J / K}) (273 \, \mathrm{K})\) Calculating the average kinetic energy of CH4 molecules at 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \, \mathrm{J}\) Now, calculate the average kinetic energy of CH4 molecules at 546 K: \(K_{avg} = \cfrac{3}{2} (1.38 \times 10^{-23} \, \mathrm{J / K}) (546 \, \mathrm{K})\) Calculating the average kinetic energy of CH4 molecules at 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \, \mathrm{J}\) Step 3: Calculate average kinetic energies for N2 at 273 K and 546 K
03

Apply the formula for nitrogen (N2)

To find the average kinetic energy of N2 molecules at 273 K, plug the values into the formula: \(K_{avg} = \cfrac{3}{2} (1.38 \times 10^{-23} \, \mathrm{J / K}) (273 \, \mathrm{K})\) Calculating the average kinetic energy of N2 molecules at 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \, \mathrm{J}\) Now, calculate the average kinetic energy of N2 molecules at 546 K: \(K_{avg} = \cfrac{3}{2} (1.38 \times 10^{-23} \, \mathrm{J / K}) (546 \, \mathrm{K})\) Calculating the average kinetic energy of N2 molecules at 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \, \mathrm{J}\) Step 4: Conclude the results
04

Summarize the findings

The average kinetic energies for methane (CH4) and nitrogen (N2) gas molecules at the given temperatures are: For CH4: - At 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \, \mathrm{J}\) - At 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \, \mathrm{J}\) For N2: - At 273 K: \(K_{avg} \approx 5.67 \times 10^{-21} \, \mathrm{J}\) - At 546 K: \(K_{avg} \approx 11.34 \times 10^{-21} \, \mathrm{J}\)

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Key Concepts

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

Kinetic Energy Calculation
Have you ever wondered how fast the tiny particles that make up gases are moving? It's a number we can't easily perceive, but with the power of physics, we can calculate their average kinetic energy, which tells us something about their motion.

Kinetic energy is the energy possessed by an object due to its motion. When we talk about gases, each molecule zips around, colliding with others and the walls of its container. This microscopic hustle and bustle is described by the average kinetic energy, which depends on temperature. The faster the particles are moving on average, the higher their kinetic energy will be.

The average kinetic energy (\(K_{avg}\)) of a gas molecule is directly proportional to the temperature of the gas and is given by a deceptively simple formula: \(K_{avg} = \cfrac{3}{2} k T\). Here, \(k\) stands for Boltzmann's constant, and \(T\) is the temperature measured in Kelvin. When you input the temperature into this formula, you can figure out the energy each molecule whizzes around with. Double the temperature, and you effectively double the average kinetic energy—it's this direct relationship that highlights the essence of kinetic molecular theory.
Boltzmann's Constant
At the heart of our calculations lies a fundamental constant named after the physicist Ludwig Boltzmann. Boltzmann's constant (\(k\)) is a bridge between the macroscopic and microscopic worlds, linking the temperature we can feel to the energy of particles we can't see.

Its value is approximately \(1.38 \times 10^{-23} \mathrm{J/K}\), and it appears in various equations in statistical mechanics and thermodynamics. Why is this constant so vital? It scales the temperature when calculating microscopic energies. Without it, we wouldn't be able to directly relate the ambient temperature of a room to the kinetic energy of a single air molecule zipping around in it.

When we use Boltzmann's constant in our kinetic energy formula, we are essentially using a value that encapsulates the behavior of particles at different temperatures. It comes from deep theoretical grounds but has immense practical use: it lets us scale up from the tiny jiggles of atoms to the temperatures we deal with in everyday life.
Temperature in Kelvin
Temperature is perhaps the most familiar concept when talking about heat and energy. However, in the realm of physics, particularly when computing average kinetic energy, we use the Kelvin scale. Why not Celsius or Fahrenheit? Well, Kelvin is the SI unit of thermodynamic temperature, and it starts at absolute zero—the point where particles theoretically stop moving.

Measuring temperature in Kelvin allows us to have a direct comparison of thermal energy between different systems. For example, a temperature increase from \(273 K\) to \(546 K\) exactly doubles the average kinetic energy of a gas molecule, as we saw in our exercise. If we made the same comparison using Celsius, the math wouldn't be as straightforward because its scale doesn't start at absolute zero.

Understanding Kelvin is crucial because it tells us about the intrinsic thermal activity of particles. It's not just a number on a thermometer; it's a way to quantify the motion that constitutes temperature on the microscopic scale. From chilling liquid nitrogen to the warmth of a living room, the Kelvin scale provides a consistent and scientific measure of thermal agitation.

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Most popular questions from this chapter

An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A \(150 .\)-\(\mathrm{L}\)reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned}P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\\P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa}\end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

Xenon and fluorine will react to form binary compounds when a mixture of these two gases is heated to \(400^{\circ} \mathrm{C}\) in a nickel reaction vessel. A \(100.0\) -\(\mathrm{mL}\) nickel container is filled with xenon and fluorine, giving partial pressures of \(1.24\) atm and \(10.10\) atm, respectively, at a temperature of \(25^{\circ} \mathrm{C}\). The reaction vessel is heated to \(400^{\circ} \mathrm{C}\) to cause a reaction to occur and then cooled to a temperature at which \(\mathrm{F}_{2}\) is a gas and the xenon fluoride compound produced is a nonvolatile solid. The remaining \(\mathrm{F}_{2}\) gas is transferred to another A \(100.0\) -\(\mathrm{mL}\) nickel container, where the pressure of \(\mathrm{F}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(7.62\) atm. Assuming all of the xenon has reacted, what is the formula of the product?

In the "Méthode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density \(=1.0 \mathrm{g} / \mathrm{cm}^{3}\) ) is allowed to take place in a bottle with a total volume of \(825 \mathrm{mL}\) until \(12 \%\) by volume is ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right) .\) Assuming that the \(\mathrm{CO}_{2}\) is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is \(0.79 \mathrm{g} / \mathrm{cm}^{3} .\) )

A hot-air balloon is filled with air to a volume of \(4.00 \times\) \(10^{3} \mathrm{m}^{3}\) at \(745\) torr and \(21^{\circ} \mathrm{C}\). The air in the balloon is then heated to \(62^{\circ} \mathrm{C},\) causing the balloon to expand to a volume of \(4.20 \times 10^{3} \mathrm{m}^{3} .\) What is the ratio of the number of moles of air in the heated balloon to the original number of moles of air in the balloon? (Hint: Openings in the balloon allow air to flow in and out. Thus the pressure in the balloon is always the same as that of the atmosphere.)

A student adds \(4.00 \mathrm{g}\) of dry ice (solid \(\mathrm{CO}_{2}\) ) to an empty balloon. What will be the volume of the balloon at STP after all the dry ice sublimes (converts to gaseous \(\mathrm{CO}_{2}\) )?

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