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

Fun with logarithms.
a Simplify the expressionealnb. (That is, write it in a way that doesn't involve logarithms.)
b Assuming that b<<a, prove that ln(a+b)(lna)+(b/a). (Hint: Factor out the afrom the argument of the logarithm, so that you can apply the approximation of part d of the previous problem.)

The mathematics of the previous problem can also be applied to a one-dimensional random walk: a journey consisting of Nsteps, all the same sic, cache chosen randomly to be cither forward or backward. (The usual mental image is that of a drunk stumbling along an alley.)

(a) Where are you most likely to find yourself, after the end of a long random walk?

(b) Suppose you take a random walk of 10,000steps (say each a yard long). About how far from your starting point would you expect to be at the end?

(c) A good example of a random walk in nature is the diffusion of a molecule through a gas; the average step length is then the mean free path, as computed in Section 1.7.Using this model, and neglecting any small numerical factors that might arise from the varying step size and the multidimensional nature of the path, estimate the expected net displacement of an air molecule (or perhaps a carbon monoxide molecule traveling through air) in one second, at room temperature and atmospheric pressure. Discuss how your estimate would differ if the clasped time or the temperature were different. Check that your estimate is consistent with the treatment of diffusion in Section1.7.

Suppose you flip1000 coins.
a What is the probability of getting exactly 500heads and 500tails? (Hint: First write down a formula for the total number of possible outcomes. Then, to determine the "multiplicity" of the 500-500"macrostate," use Stirling's approximation. If you have a fancy calculator that makes Stirling's approximation unnecessary, multiply all the numbers in this problem by 10, or 100, or1000, until Stirling's approximation becomes necessary.)
bWhat is the probability of getting exactly 600heads and400 tails?

For an Einstein solid with each of the following values of N and q , list all of the possible microstates, count them, and verify formula Ω(N,q)=q+N1q=(q+N1)!q!(N1)!

(a) N=3,q=4

(b)N=3,q=5

(c) N=3,q=6

(d) N=4,q=2

(e) N=4,q=3

(f) N=1,q=anything

(g) N= anything, q=1

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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