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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."

Short Answer

Expert verified

(a) The accessible microstates to the present system is Ω≈2N=2N23

(b) There are 3.15576×1026microstates will it explore in ten billion years.

(c) For all this to develop, we'd to require almost a vast quantity of days (at least here on scope of the planet's present era).

Step by step solution

01

Microstates (a)

(a) Assume we get a paramagnet has N=1023 magnets, and shut to half them are as in spin-up state, giving in N↑=N↓=12N,, whereby N↓ signifies the energy units..This technique provides quantity of microstates:

Ω=N↓+N↑+1N↓=N2+N2+1N2

when N is simply a good number

Ω≈NN2=N!N2!N2!

For such design variables, we will employ Stirling's approximation:

n!≈2πnnne−n

Ω≈2πNNNe−N2πN2N2N2e−N22

Ω≈2πNNNe−N2πN2N2Ne−N

localid="1650343548696" Ω≈2πNNNπNN2N

Ω≈2πN2NπNΩ≈2N+12πN

Ω≈2N=2N23

where every other line probably applies for Namounts about 1023, although this makes 2N a really vast concentration, but dividing by the denominator won't make a big difference.

02

Independent States (b) and (c)

(b) Above a ten-billion-year interval, if the machine was during a changing microstate a billion views a second, it'd examine:

109×1010×365.25×24×3600=3.15576×1026microstates

The variability of microstates allowed is:

21023≈103×1022

As both a result, the quantity of independent states accessed over 10 billion centuriesis simply alittle fraction among those possible.

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Most popular questions from this chapter

Rather than insisting that all the molecules be in the left half of a container, suppose we only require that they be in the leftmost 99%(leaving the remaining 1%completely empty). What is the probability of finding such an arrangement if there are 100molecules in the container? What if there are 10,000molecules? What if there are 1023?

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?

Describe a few of your favorite, and least favorite, irreversible processes. In each case, explain how you can tell that the entropy of the universe increases.

Write e1023in the form 10x, for somex.

Use a pocket calculator to check the accuracy of Stirling's approximation forN=50 . Also check the accuracy of equation 2.16forlnN! .

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