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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=Nkln2mUA(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()

where,

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

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

=(A)N(N!)2h2N(2mU)N

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

By using Stirling's approximation,

n!2nnnenN!2NNNeN(N!)22N2N+1e2N

03

Step: 3 Finding Ω value:

Substituting we get,

(2mUA)N2N2N+1e2Nh2N

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

(2mUA)NN2Ne2Nh2N(2mUA)N(Nh)2Ne2N(2mUA)(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()Nln(2mUA)(Nh)2e2ln()Nln(2mUA)(Nh)2+Nlne2ln()Nln2mUA(Nh)2+2

The entrpy of ideal gas gives as,

S=Nkln2mUA(Nh)2+2.

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Most popular questions from this chapter

For either a monatomic ideal gas or a high-temperature Einstein solid, the entropy is given by times some logarithm. The logarithm is never large, so if all you want is an order-of-magnitude estimate, you can neglect it and just say . That is, the entropy in fundamental units is of the order of the number of particles in the system. This conclusion turns out to be true for most systems (with some important exceptions at low temperatures where the particles are behaving in an orderly way). So just for fun, make a very rough estimate of the entropy of each of the following: this book (a kilogram of carbon compounds); a moose of water ; the sun of ionized hydrogen .

Compute the entropy of a mole of helium at room temperature and atmospheric pressure, pretending that all the atoms are distinguishable. Compare to the actual entropy, for indistinguishable atoms, computed in the text.

For an Einstein solid with four oscillators and two units of energy, represent each possible microstate as a series of dots and vertical lines, as used in the text to prove equation 2.9.

Use the Sackur-Tetrode equation to calculate the entropy of a mole of argon gas at room temperature and atmospheric pressure. Why is the entropy greater than that of a mole of helium under the same conditions?

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?

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