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 Ω2N=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

ΩNN2=N!N2!N2!

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

n!2πnnnen

Ω2πNNNeN2πN2N2N2eN22

Ω2πNNNeN2πN2N2NeN

localid="1650343548696" Ω2πNNNπNN2N

Ω2πN2NπNΩ2N+12πN

Ω2N=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:

21023103×1022

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

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.

Use Stirling's approximation to find an approximate formula for the multiplicity of a two-state paramagnet. Simplify this formula in the limit NNto obtain ΩNe/NN. This result should look very similar to your answer to Problem 2.17; explain why these two systems, in the limits considered, are essentially the same.

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?

Suppose you flip four fair coins.

(a) Make a list of all the possible outcomes, as in Table 2.1.

(b) Make a list of all the different "macrostates" and their probabilities.

(c) Compute the multiplicity of each macrostate using the combinatorial formula 2.6, and check that these results agree with what you got by bruteforce counting.

The mixing entropy formula derived in the previous problem actually applies to any ideal gas, and to some dense gases, liquids, and solids as well. For the denser systems, we have to assume that the two types of molecules are the same size and that molecules of different types interact with each other in the same way as molecules of the same type (same forces, etc.). Such a system is called an ideal mixture. Explain why, for an ideal mixture, the mixing entropy is given by

ΔSmixing=klnNNA

where Nis the total number of molecules and NAis the number of molecules of type A. Use Stirling's approximation to show that this expression is the same as the result of the previous problem when both Nand NAare large.

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