/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 101 Consider three identical flasks ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider three identical flasks filled with different gases. Flask A: \(\mathrm{CO}\) at 760 torr and \(0^{\circ} \mathrm{C}\) Flask B: \(\mathrm{N}_{2}\) at 250 torr and \(0^{\circ} \mathrm{C}\) Flask C: \(\mathrm{H}_{2}\) at 100 torr and \(0^{\circ} \mathrm{C}\) a. In which flask will the molecules have the greatest average kinetic energy? b. In which flask will the molecules have the greatest average velocity?

Short Answer

Expert verified
a. All flasks A, B, and C have equal average kinetic energy because they are at the same temperature (\(0^{\circ} \mathrm{C}\)). b. Flask C (Hydrogen gas) has the greatest average velocity among the molecules of all flasks due to its lower molecular mass as compared to flasks A and B, which contain Carbon monoxide and Nitrogen, respectively.

Step by step solution

01

Determine if temperature differs among flasks

The flasks are all at \(0^{\circ} \mathrm{C}\). Since temperature is the same for all three flasks, the molecules are sharing the same average kinetic energy. To find the difference in velocity, let's look at molecular mass.
02

Determine molecular mass

The molecular masses for the gases in each flask are as follows: Flask A: Carbon monoxide (CO) has a molecular mass of 12 + 16 = 28 g/mol. Flask B: Nitrogen (N2) has a molecular mass of 28 g/mol (14 x 2 = 28 g/mol). Flask C: Hydrogen (H2) has a molecular mass of 2 g/mol (1 x 2 = 2 g/mol).
03

Calculate the average kinetic energy

Since temperature is the same for all three flasks, the molecules in each flask have the same average kinetic energy. Therefore, the answer for part (a) is that all flasks A, B, and C have equal average kinetic energy.
04

Determine the average velocity of molecules in each flask

The relationship between molecular mass and molecular velocity is given by the equation: \(v_{rms} = \sqrt{\dfrac{3RT}{M}}\) where \(v_{rms}\) is the root mean square velocity, R is the gas constant (8.314 J/mol K), T is the temperature (0°C = 273.15 K), and M is the molar mass in kg/mol. For Flask A, the average velocity is: \[ v_{A} = \sqrt{\dfrac{3(8.314)(273.15)}{0.028}}\] For Flask B, the average velocity is: \[ v_{B} = \sqrt{\dfrac{3(8.314)(273.15)}{0.028}}\] For Flask C, the average velocity is: \[ v_{C} = \sqrt{\dfrac{3(8.314)(273.15)}{0.002}}\] Now, comparing the average velocities: Since 0.028 > 0.002, \(v_{C} > v_{A} = v_{B}\). Therefore, the answer for part (b) is that Flask C (Hydrogen gas) has the greatest average velocity among the molecules of all flasks.

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 91Ó°ÊÓ!

Key Concepts

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

Average Molecular Velocity
When we talk about gases, each molecule is in constant motion, randomly bouncing around and occasionally colliding with other molecules. The concept of average molecular velocity refers to the speed at which these molecules move on average. The velocity is not uniform because individual molecules travel at different speeds depending on collisions or the type of gas involved. However, factors such as molecular mass and temperature are key in determining the average velocity of the gas molecules. In our flask problem, even though the temperature is the same, differing molecular masses result in different average velocities for gases. This is why, among Carbon monoxide, Nitrogen, and Hydrogen gases—although they all share the same temperature—their varying molecular weights lead to differences in their average velocities.
Gas Laws
Gas laws help us understand and predict the behavior of gases under different conditions. They establish relationships among pressure, volume, temperature, and number of molecules in a gas. In the context of our exercise, we can observe these laws at work. All gases in the flasks are at the same temperature but at different pressures, which are given in torr. Although gases like CO and N extsubscript{2} have similar molecular velocities due to their identical molecular masses, differences in pressure and molecular mass influence properties such as density and velocity, as per the gas laws.
Molecular Mass
Molecular mass is a crucial factor when discussing the behavior of gases, especially concerning their velocity. Molecular mass, often measured in grams per mole (g/mol), represents the mass of a single molecule of a compound or element. To find it, we sum the atomic masses of each atom in a molecule’s structure. For instance, Carbon monoxide (CO) has a molecular mass of 28 g/mol—carbon being 12 g/mol and oxygen 16 g/mol. When temperatures are equal among gas samples, lighter molecules (like H extsubscript{2}) naturally travel faster than heavier ones. This is part of why, in our flasks, the light hydrogen molecules achieve greater velocities than those in the flasks containing heavier carbon monoxide and nitrogen.
Root Mean Square Velocity
Root mean square velocity (\(v_{rms}\)) is a specific measure of the average speed of gas molecules, calculated using the formula:\[v_{rms} = \sqrt{\dfrac{3RT}{M}}\]In this equation, \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in kg/mol. This equation provides a glimpse into how kinetic energy impacts molecular speeds and further solidifies the connection between molecular mass and velocity.In our example, since the flasks have different gases but are kept at the same temperature, the differences in \(v_{rms}\) arise due to their distinct molecular masses. Flask C, containing hydrogen with the lowest molecular mass, displays the highest root mean square velocity, demonstrating the inverse relationship between mass and speed in gas molecules.

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

A person accidentally swallows a drop of liquid oxygen, \(\mathrm{O}_{2}(l)\), which has a density of \(1.149 \mathrm{~g} / \mathrm{mL}\). Assuming the drop has a volume of \(0.050 \mathrm{~mL}\), what volume of gas will be produced in the person's stomach at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) and a pressure of \(1.0 \mathrm{~atm}\) ?

You have a sealed, flexible balloon filled with argon gas. The atmospheric pressure is \(1.00 \mathrm{~atm}\) and the temperature is \(25^{\circ} \mathrm{C}\). Assume that air has a mole fraction of nitrogen of \(0.790\), the rest being oxygen. a. Explain why the balloon would float when heated. Make sure to discuss which factors change and which remain constant, and why this matters. Be complete. b. Above what temperature would you heat the balloon so that it would float?

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 \(\mathrm{CO}_{2}(\mathrm{~g})\) 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 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.

At room temperature, water is a liquid with a molar volume of \(18 \mathrm{~mL}\). At \(105^{\circ} \mathrm{C}\) and 1 atm pressure, water is a gas and has a molar volume of over \(30 \mathrm{~L}\). Explain the large difference in molar volumes.

A 20.0-L nickel container was charged with \(0.500\) atm of xenon gas and \(1.50\) atm of fluorine gas at \(400 .{ }^{\circ} \mathrm{C}\). The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming \(100 \%\) yield?

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.