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: \(K_{avg} \approx 11.34 \times 10^{-21} \, \mathrm{J}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Draw a qualitative graph to show how the first property varies with the second in each of the following (assume 1 mole of an ideal gas and \(T\) in kelvin). a. \(P V\) versus \(V\) with constant \(T\) b. \(P\) versus \(T\) with constant \(V\) c. \(T\) versus \(V\) with constant \(P\) d. \(P\) versus \(V\) with constant \(T\) e. \(P\) versus \(1 / V\) with constant \(T\) f. \(P V / T\) versus \(P\)

Some very effective rocket fuels are composed of lightweight liquids. The fuel composed of dimethylhydrazine \(\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}\right]\) mixed with dinitrogen tetroxide was used to power the Lunar Lander in its missions to the moon. The two components react according to the following equation: $$\left(\mathrm{CH}_{3}\right)_{2} \mathrm{N}_{2} \mathrm{H}_{2}(l)+2 \mathrm{N}_{2} \mathrm{O}_{4}(l) \longrightarrow 3 \mathrm{N}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)$$ If 150 g dimethylhydrazine reacts with excess dinitrogen tetroxide and the product gases are collected at \(127^{\circ} \mathrm{C}\) in an evacuated \(250-\mathrm{L}\) tank, what is the partial pressure of nitrogen gas produced and what is the total pressure in the tank assuming the reaction has \(100 \%\) yield?

At elevated temperatures, sodium chlorate decomposes to produce sodium chloride and oxygen gas. A \(0.8765\)-\(\mathrm{g}\) sample of impure sodium chlorate was heated until the production of oxygen gas ceased. The oxygen gas collected over water occupied \(57.2 \mathrm{mL}\) at a temperature of \(22^{\circ} \mathrm{C}\) and a pressure of \(734\) torr. Calculate the mass percent of \(\mathrm{NaClO}_{3}\) in the original sample. (At \(22^{\circ} \mathrm{C}\) the vapor pressure of water is \(19.8\) torr.)

A balloon is filled to a volume of \(7.00 \times 10^{2} \mathrm{mL}\) at a temperature of \(20.0^{\circ} \mathrm{C}\). The balloon is then cooled at constant pressure to a temperature of \(1.00 \times 10^{2} \mathrm{K}\). What is the final volume of the balloon?

Consider separate \(1.0-\mathrm{L}\) gaseous samples of \(\mathrm{He}, \mathrm{N}_{2},\) and \(\mathrm{O}_{2}\) all at \(\mathrm{STP}\) and all acting ideally. Rank the gases in order of increasing average kinetic energy and in order of increasing average velocity.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free