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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?

Short Answer

Expert verified

(a) The entropy of this method , assuming that each one of the microstates are allowed Stotal=3.826JK1

(b) The entropy again, assuming that the system in its presumably macro state Smp=3.826JK1

(c) The entropy is given by , but over while scales it's given by which is ever slightly larger than .Smp

(d) Placea superb excellent resistance between two surfaces, keeping them from transmitting energy. In effect, this shows that the model is stuck in its current condition.

Step by step solution

01

Einstein Spheres (a)

(a) Suppose two big, similar Einstein spheres, every with Nclocks and q=2Namount of energy. We had (from issue 2.22) for the state during which all business can make are able:

Ωtotal24N8πN

This complete agency's volatility is:

Stotal=klnΩtotal

If we modify Ωtotalwith , we get:

Stotal=kln24N8πN

ln(ab)=ln(a)+ln(b),lnab=ln(a)ln(b)andlnab=bln(a)

Stotal=k4Nln(2)12ln(8πN)

k=1.38×1023JK1andN=1023:

Stotal=1.38×10234×1023×ln(2)12×ln8π×1023

Stotal=3.826JK1

02

Two solids (b)

(b) Its presumably situation, that the energy is spread evenly between the 2 solids, had a redundancy (from problem 2.22):

Ωmp24N4πN

The volatility of its most frequent state was even as follows:

Smp=klnΩmp

Assuming we swap Ωmp, we get:

Smp=kln24N4πN

Stotal=k[4Nln(2)ln(4πN)]

k=1.38×1023JK1andN=1023

Smp=1.38×10234×1023×ln(2)×ln4π×1023

Smp=3.826JK1

Hence, whether object is viewed as either an unified location with all feasible microstates or as 2 independent systems for his or her most plausible shape, the energy is virtually defined solely more by 24Nvariable.

03

Entropy (c) and (d)

(c) The entity is liberal to pursue each one of the microstates included in Stotal, over longer timeframes, yet it's extremely likely to be in or near its most frequent state at any given time. Thus, we may claim that the entropy is supplied by Smp over small scales, while it's given by Stotal very long time scales, which has become so marginally bigger than Smp.

(d) Think about a situation within which we elect some extent t so when system is in its possibly efficient than one (i.e., the entropy is Smp) and so place a wonderful resistance between two surfaces, keeping them from transmitting energy. In effect, this shows that the model is stuck in its current condition.

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Most popular questions from this chapter

This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.

(a) Consider two identical Einstein solids, each with Noscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly 2N. How many different macrostates (that is, possible values for the total energy in the first solid) are there for this combined system?

(b) Use the result of Problem2.18to find an approximate expression for the total number of microstates for the combined system. (Hint: Treat the combined system as a single Einstein solid. Do not throw away factors of "large" numbers, since you will eventually be dividing two "very large" numbers that are nearly equal. Answer: 24N/8πN.)

(c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of Problem 2.18to find an approximate expression for the multiplicity of this macrostate. (Answer:24N/(4πN) .)

(d) You can get a rough idea of the "sharpness" of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the height of the peak, while part (b) tells you the total area under the entire graph. As a very crude approximation, pretend that the peak's shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case N=1023.

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

How many possible arrangements are there for a deck of 52playing cards? (For simplicity, consider only the order of the cards, not whether they are turned upside-down, etc.) Suppose you start w e in the process? Express your answer both as a pure number (neglecting the factor of k) and in SI units. Is this entropy significant compared to the entropy associated with arranging thermal energy among the molecules in the cards?

Calculate the number of possible five-card poker hands, dealt from a deck of 52 cards. (The order of cards in a hand does not matter.) A royal flush consists of the five highest-ranking cards (ace, king, queen, jack, 10) of any one of the four suits. What is the probability of being dealt a royal flush (on the first deal)?

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.

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