Calculate the Entropy Change if 5 Previously Tossed Dice Are Turned to All "1"

Question

Relevant Equations: S = Boltzmann*ln(W) where S is entropy and W is the number of microstates.

I have thought about this two ways.

1 way. Look at each die separately. Let macrostate 1 = number of dots is 1, macrostate 2 is number of dots is not 1.

Then the die has 1 microstate for macrostate 1, and 5 microstates for macrostate 2.

Thus the entropy is S = Boltzmann * ln(5) if the die does not read 1.

Now we repeat this for all 5 die and so our entropies add to get S = n*Boltzmann*ln(5) where n is the number of die that do not show 1.

Since we are looking for the change in S, and the number of microstates for rolling a 1 is just 1, our entropy for that is 0.

Thus, our change in entropy is (-n)*boltzmann*ln(5). Our entropy decreases.

Way number 2 - examine all die at once.

macrostate 1 = all 1. macrostate 2 = not all 1.

Then there are 6^5 = 7776 total microstates, and only 1 of those is all 1. Then there are 7775 microstates that do not give one, and so S = boltzmann * ln(7775). Like above, we subtract that since our final state of all 1 has entropy 0.

Thus, our change is -Boltzmann*ln(7775).

Obviously I am doing something wrong, because I got a number for way 2 but a formula for way 1. Any help would be greatly appreciated.

Answer 1

What you've calculated in Method 1 is the change in entropy from all five dice showing not 1 to all five dice showing 1.   You haven't included the possibility that one or more die starts out showing 1.   Hence you've counted only $5^5$ microstates.

What you've calculated in Method 2 is the change in entropy from some of the five dice show not 1 to all five dice show 1.   This includes the possibility that one or more die starts out showing 1 with the exception of the possibility that all five dice show 1.   Hence you've counted $6^5-1$ microstates.

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However, you really should not exclude the possibility that all five die show 1 after being tossed randomly.   They could start out in the desired microstate as there's no bias against it.

(Note that the initial and final macrostates in problems such as this one are not necessarily composed of mutually exclusive microstates.)

Try:

• Initial Macrostate: "the dice are randomly tossed": $6$ microstates for each die; or $6^5$ microstates for the five dice.
• Final Macrostate: "the dice are all set to show 1". $1$ microstate for each die; or $1$ microstate for the five dice.

Using with this your method 1 and method 2 should give the same result: $-5 k_{\small B}\ln 6$