/*! 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} Q. 50 FIGURE P20.50 shows the thermal... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

FIGURE P20.50shows the thermal energy of0.14molof gas as a function of temperature. What isCv for this gas?

Short Answer

Expert verified

The Cvfor gas isCV=29J/mol·K

Step by step solution

01

Definition of thermal energy

Thermal energy is defined as the energy stored within a system that is relevant to its temperature. Heat is the term used to describe the passage of thermal energy.

Thermodynamics is a branch of physics that studies how heat is transferred across systems and how work is performed in the process.

02

Step 2:Expalation

If the temperature changes by ∆T, then the thermal energy for a gas changes by equation (20.30)in the form

ΔEth=nCVΔT---------(1)

Solve equation (1)for Cvto be in the form

CV=ΔEthnΔT------------(2)

From figure P20.50, the temperature changes from 100°Cto 200°C. So, the change in temperature is

ΔT=200°C-100°C=100°C

At this change in temperature, the thermal energy changes by

ΔEth=Ef-Ei

role="math" localid="1648440626213" =1892J-1492J

=400J

Now, we plug the values for∆Eth, nand ∆Tinto equation (2)to get Cvby

CV=ΔEthnΔT

=400J(0.14mol)(100K)

=29J/mol·K

03

The thermal energy of gas

The thermal energy of gas isCv=29J/mol.K

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!

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

The mean free path of a molecule in a gas is 300nm. What will the mean free path be if the gas temperature is doubled at (a) Constant volume and (b) Constant pressure?

6. Suppose you could suddenly increase the speed of every molecule in a gas by a factor of 2.

a. Would the RMS speed of the molecules increase by a factor of 21/2,2,or22? Explain.

b. Would the gas pressure increase by a factor of 21/2,2or 22? Explain.

At what temperature does thermsspeed of (a)a nitrogen molecule and (b)a hydrogen molecule equal the escape speed from the earth's surface? (c)You'll find that these temperatures are very high, so you might think that the earth's gravity could easily contain both gases. But not all molecules move withVrms. There is a distribution of speeds, and a small percentage of molecules have speeds several times Vrms . Bit by bit, a gas can slowly leak out of the atmosphere as its fastest molecules escape. A reasonable rule of thumb is that the earth's gravity can contain a gas only if the average translational kinetic energy per molecule is less than 1%of the kinetic energy needed to escape. Use this rule to show why the earth's atmosphere contains nitrogen but not hydrogen, even though hydrogen is the most abundant element in the universe.

Atoms can be "cooled" to incredibly low temperatures by letting them interact with a laser beam. Various novel quantum phenomena appear at these temperatures. What is the rmsspeed of 100nK?

The rms speed of the molecules in 1.0gof hydrogen gas is1800ms .
a. What is the total translational kinetic energy of the gas molecules?
b. What is the thermal energy of the gas?
c. 500Jof work are done to compress the gas while, in the same process, 1200Jof heat energy are transferred from the gas to the environment. Afterward, what in the rms speed of the molecules?

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.