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Find an expression for the entropy of the two-dimensional ideal gas considered in Problem 2.26. Express your result in terms of U,AandN.

Short Answer

Expert verified

The Entropy of the two-dimensional ideal gaas isS=Nkln2mπUA(Nh)2+2.

Step by step solution

01

Step: 1 Definition of Entropy:

Entropy is described as a measure of the degree of unpredictability in a system, or in other words, the growth in disorder.Entropy is a measure of disarray that has an impact on many facets of our existence. In reality, it's akin to a tax imposed by nature. If problem is not addressed, it will worsen over time. Energy dissipates, and systems disintegrate. We consider something to be more entropic if it is more disordered.

02

Step: 2 Derivative part

The entropy substance as

S=kln(Ω)

where,

localid="1650262053991" Ωis the number of microstates substance accessible.

The localid="1650262058146" 2-dideal gas multipilicity is

Ω=(πA)N(N!)2h2N(2mU)N

where,localid="1650262061679" Ais the area gas.

By using Stirling's approximation,

n!2πnnnenN!2πNNNeN(N!)22πN2N+1e2N

03

Step: 3 Finding Ω value:

Substituting we get,

Ω(2mπUA)N2πN2N+1e2Nh2N

Where localid="1650262075329" Nis the large,the couple of factors away is

Ω(2mπUA)NN2Ne2Nh2NΩ(2mπUA)N(Nh)2Ne2NΩ(2mπUA)(Nh)2e2N

04

Step: 4 Finding entropy of ideal gas:

Taking logarithm on both sides,

ln(ab)=ln(a)+ln(b)andlnab=ln(a)ln(b)

is taking into account,so

ln(Ω)Nln(2mπUA)(Nh)2e2ln(Ω)Nln(2mπUA)(Nh)2+Nlne2ln(Ω)Nln2mπUA(Nh)2+2

The entrpy of ideal gas gives as,

S=Nkln2mπUA(Nh)2+2.

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Most popular questions from this chapter

Consider a two-state paramagnet with 1023elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down.

(a) How many microstates are "accessible" to this system?

(b) Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)?

(c) Is it correct to say that, if you wait long enough, a system will eventually be found in every "accessible" microstate? Explain your answer, and discuss the meaning of the word "accessible."

Consider again the system of two large, identical Einstein solids treated in Problem 2.22.

(a) For the case N=1023, compute the entropy of this system (in terms of Boltzmann's constant), assuming that all of the microstates are allowed. (This is the system's entropy over long time scales.)

(b) Compute the entropy again, assuming that the system is in its most likely macro state. (This is the system's entropy over short time scales, except when there is a large and unlikely fluctuation away from the most likely macro state.)

(c) Is the issue of time scales really relevant to the entropy of this system?

(d) Suppose that, at a moment when the system is near its most likely macro state, you suddenly insert a partition between the solids so that they can no longer exchange energy. Now, even over long time scales, the entropy is given by your answer to part (b). Since this number is less than your answer to part (a), you have, in a sense, caused a violation of the second law of thermodynamics. Is this violation significant? Should we lose any sleep over it?

Problem 2.20. Suppose you were to shrink Figure2.7until the entire horizontal scale fits on the page. How wide would the peak be?

Consider a system of two Einstein solids, with N{A} = 300, N{B} = 200 and q{total} = 100 (as discussed in Section 2.3). Compute the entropy of the most likely macrostate and of the least likely macrostate. Also compute the entropy over long time scales, assuming that all microstates are accessible. (Neglect the factor of Boltzmann's constant in the definition of entropy; for systems this small it is best to think of entropy as a pure number.) 65

Using the same method as in the text, calculate the entropy of mixing for a system of two monatomic ideal gases, Aand B, whose relative proportion is arbitrary. Let Nbe the total number of molecules and letx be the fraction of these that are of speciesB . You should find

ΔSmixing=Nk[xlnx+(1x)ln(1x)]

Check that this expression reduces to the one given in the text whenx=1/2 .

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