/*! 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. 3.9 In solid carbon monoxide, each C... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In solid carbon monoxide, each CO molecule has two possible orientations: CO or OC. Assuming that these orientations are completely random (not quite true but close), calculate the residual entropy of a mole of carbon monoxide.

Short Answer

Expert verified

The residual entropy for a carbon monoxide is calculated to be5.76JK-1.

Step by step solution

01

Given

Carbon monoxide is the given solid. CO is its chemical formula. Each molecule can be arranged in two different ways: CO or OC.

02

Calculation

Multiplicity for Nmolecules is given as:

Ω=2N………..(1)

And Residual entropy is given as:

Sresidual=klnΩ………(2)

Where, kis Boltzmann constant.

In the given case, N=6.022×1023

By substituting this value in equation (1), we get,

Ω=26.022×1023

Now, by substituting this value of Ωand k=1.38×10-23JK-1in equation (2), we get,

Sresidual=1.38×10-23ln26.022×1023Sresidual=1.38×10-236.022×1023ln2Sresidual=5.76JK-1

03

Final answer

Hence, the residual entropy is calculated to be5.76JK-1.

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

In Problem 2.18 you showed that the multiplicity of an Einstein solid containing N oscillators and q energy units is approximately

Ω(N,q)≈q+Nqqq+NNN

(a) Starting with this formula, find an expression for the entropy of an Einstein solid as a function of N and q. Explain why the factors omitted from the formula have no effect on the entropy, when N and q are large.

(b) Use the result of part (a) to calculate the temperature of an Einstein solid as a function of its energy. (The energy is U=qϵ, where ϵis a constant.) Be sure to simplify your result as much as possible.

(c) Invert the relation you found in part (b) to find the energy as a function of temperature, then differentiate to find a formula for the heat capacity.

(d) Show that, in the limit T→∞, the heat capacity is C=Nk. (Hint: When x is very small, ex≈1+x.) Is this the result you would expect? Explain.

(e) Make a graph (possibly using a computer) of the result of part (c). To avoid awkward numerical factors, plot C/Nkvs. the dimensionless variable t=kT/ϵ, for t in the range from 0 to about 2. Discuss your prediction for the heat capacity at low temperature, comparing to the data for lead, aluminum, and diamond shown in Figure 1.14. Estimate the value of ϵ, in electron-volts, for each of those real solids.

(f) Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through x3 in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than(ϵ/kT)2 in the final answer. When the smoke clears, you should find C=Nk1-112(ϵ/kT)2.

Figure 3.3 shows graphs of entropy vs. energy for two objects, A and B. Both graphs are on the same scale. The energies of these two objects initially have the values indicated; the objects are then brought into thermal contact with each other. Explain what happens subsequently and why, without using the word "temperature."

Consider an Einstein solid for which both N and q are much greater than 1. Think of each oscillator as a separate "particle."

(a) Show that the chemical potential is

role="math" localid="1646995468663" μ=-kTlnN+qN

(b) Discuss this result in the limits N≫qand N≪q, concentrating on the question of how much Sincreases when another particle carrying no energy is added to the system. Does the formula make intuitive sense?

A cylinder contains one liter of air at room temperature ( 300K) and atmospheric pressure 105N/m2. At one end of the cylinder is a massless piston, whose surface area is 0.01m2. Suppose that you push the piston in very suddenly, exerting a force of 2000N. The piston moves only one millimeter, before it is stopped by an immovable barrier of some sort.

(a) How much work have you done on this system?

(b) How much heat has been added to the gas?

(c) Assuming that all the energy added goes into the gas (not the piston or cylinder walls), by how much does the internal energy of the gas increase?

(d) Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium).

Consider an ideal two-state electronic paramagnet such as DPPH, with μ=μB. In the experiment described above, the magnetic field strength was 2.06T and the minimum temperature was 2.2K. Calculate the energy, magnetization, and entropy of this system, expressing each quantity as a fraction of its maximum possible value. What would the experimenters have had to do to attain99% of the maximum possible magnetization?

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.