Oxygen molecules \(\left(\mathrm{O}_{2}\right)\) are 16 times as massive as hydrogen molecules \(\left(\mathrm{H}_{2}\right)\). Carbon dioxide molecules \(\left(\mathrm{CO}_{2}\right)\) are 22 times as massive as \(\mathrm{H}_{2}\) a. Compare the average speed of \(\mathrm{O}_{2}\) and \(\mathrm{CO}_{2}\) molecules in a volume of air. b. Does this ratio of the speeds in part (a) depend on air temperature?

Short Answer

Expert verified
The ratio of speeds of \( \text{CO}_2 \) and \( \text{O}_2 \) is \( \frac{4}{\root{2}{22}} \) and it does not depend on air temperature.

Step by step solution

01

Understand the relationship between mass and speed

The average speed of a gas molecule is inversely proportional to the square root of its molar mass, given by the equation: \[v \backsim \frac{1}{\root{2}\text{Molar Mass}}\] where \( v \) is the average speed.
02

Set up the ratio of average speeds

Given that oxygen molecules are 16 times as massive as hydrogen molecules and carbon dioxide molecules are 22 times as massive as hydrogen molecules, use the formula: \[ \frac{v(\text{CO}_2)}{v(\text{O}_2)} = \frac{ \root{2}{\text{M}_{\text{O}_2}}}{ \root{2}\text{M}_{\text{CO}_2}} \]
03

Substitute the masses into the equation

According to the problem: \( \text{M}_{\text{O}_2} = 16 \times \text{M}_{\text{H}_2} \) and \( \text{M}_{\text{CO}_2} = 22 \times \text{M}_{\text{H}_2} \). Thus, substituting these values, the ratio becomes: \[ \frac{v(\text{CO}_2)}{v(\text{O}_2)} = \frac{ \root{2}{16\text{M}_{\text{H}_2}}}{ \root{2}{22\text{M}_{\text{H}_2}}} \] which simplifies to: \[ \frac{\root{2}{16}}{\root{2}{22}} = \frac{4}{\root{2}{22}} \]
04

Simplify the ratio

Simplify the ratio further to get: \[ \frac{v(\text{CO}_2)}{v(\text{O}_2)} = \frac{\root{2}(4 \times 4) /\root{2}(2 \times 11)} = \frac{4}{\root{2}{22}} \]
05

Consider temperature dependence

The average speed of the molecules is also proportional to the square root of temperature by formula: \( v \backsim \root{2}\text{T} \), where \( T \) is the temperature. Given that both are at the same temperature, this dependence cancels out: \[ \frac{v(\text{CO}_2)}{v(\text{O}_2)} \]in this case, does not depend on air temperature.

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.

Molecular Mass and Speed
Gas molecules move at different speeds depending on their masses. The average speed of a gas molecule is inversely proportional to the square root of its molar mass (\text{M}). This relationship is shown by the equation: \(v \backsim \frac{1}{\root{2}{\text{M}}}\). The lighter the molecule, the faster it moves on average. For instance, hydrogen molecules (\text{H}_2) are very light compared to oxygen (\text{O}_2) and carbon dioxide (\text{CO}_2). As a result, hydrogen molecules will have a higher average speed. In the exercise, oxygen is 16 times heavier than hydrogen and carbon dioxide is 22 times heavier. So, to compare the speeds, you would use the mass in this inverse relationship.
Temperature Dependence
The speed of gas molecules is also affected by temperature. As the temperature increases, molecules move faster. This relationship is given by the equation: \(v \backsim \root{2}{\text{T}}\), where \(T\) represents the temperature. When comparing gas molecules at the same temperature, the temperature factor cancels out because both gases are exposed to the same thermal energy. Therefore, while the actual speed of the molecules will change with temperature, the ratio of their speeds stays constant if the mass ratio is considered the same. In our exercise, the speed ratio between \(\text{CO}_2\) and \(\text{O}_2\) will remain unaffected by temperature differences.
Kinetic Theory of Gases
The kinetic theory of gases explains how gas particles move and interact. It assumes that gas molecules are in constant, random motion and that the pressure and temperature of a gas are a result of collisions between molecules and the walls of the container. According to this theory:
  • Molecules have negligible volume compared to the distance between them.
  • No intermolecular forces act between them, except during collisions.
  • Collisions are perfectly elastic, meaning no kinetic energy is lost.
  • The average kinetic energy of gas molecules is directly proportional to the temperature of the gas in Kelvin.
These principles explain why lighter molecules move faster and how temperature affects their speed, aligning with our observations in the given exercise. Understanding this theory helps to predict and explain behavior such as diffusion and pressure changes in gases.

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

Less massive molecules tend to escape from an atmosphere more often than more massive molecules because a. the gravitational force on them is less. b. they are moving faster. c. they are more buoyant. d. they are smaller and so experience fewer collisions on their way out.

Atmospheric pressure is caused by the weight of a column of air above you pushing down. At sea level on Earth, this pressure is equal to \(10^{5}\) newtons per square meter \(\left(\mathrm{N} / \mathrm{m}^{2}\right)\) a. Estimate the total force on the top of your head from this pressure. b. Recall that the acceleration due to gravity is \(9.8 \mathrm{m} / \mathrm{s}^{2}\). If the force in part (a) were caused by a kangaroo sitting on your head, what would the mass of the kangaroo be? c. Assume a typical kangaroo has a mass of 60 kg. How many kangaroos would have to be sitting on your head to be equal to the extremely massive kangaroo in part (b)? d. Why are you not crushed by this astonishing force on your head?

On which of these planets is the atmospheric greenhouse effect strongest? a. Venus b. Earth c. Mars d. Mercury

Mars: a. Go to http://www.planetfour.org, a Zooniverse Citizen Science Project in which people examine images of the surface of Mars. Log in or create a Zooniverse account if you don't have one. Read through "About": Where did these data come from? What are the goals of this project? Why is it useful to have many people look at the data? Read through "Classify": "Show Tutorial" and "See Examples" and "FAQs." Now classify some images. b. Go to the website for the \(M A V E N\) mission, which entered the orbit of Mars in 2014 . (http://lasp.colorado.edu/home/ maven). What are the scientific goals of the mission? Is this mission a lander, an orbiter, or a flyby? What instruments are on this mission? How will this mission contribute to the understanding of climate change on Mars? Go to the NASA Web page for \(M A V E N\) (http://www.nasa.gov/ mission_pages/maven/main/index.html). Are there any results?

In what ways does plant life affect the composition of Earth's atmosphere?

See all solutions

Recommended explanations on Physics 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