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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