Question

In: Chemistry

what is the entropy of freshly shuffled deck of cards? (Physical chemistry)

what is the entropy of freshly shuffled deck of cards? (Physical chemistry)

Solutions

Expert Solution

In thermodynamics, we measure things like pressure, temperature, and volume. These are high-level descriptions. Together, these state variables form a macrostate. The low-level description is the position and momentum of each particle. This information forms a microstate.

In the card example, I assume we take the order of the cards to be a microstate. However, it is unspecified what characteristic of the cards make up the macrostate.

If the macrostate is, "the cards are in a pile", then you're right; shuffling doesn't change the entropy. The entropy would always be log52!, because any microstate would correspond to that macrostate.

A more interesting macrostate might be the number of times a higher card comes before a lower card. If we label the cards 1 - 52, we could then look at all   pairs of cards. Whenever the higher card comes before the lower card, we add one to our tally.

In the fresh deck, this number would be zero (depending on how we number the cards) because the cards are in exactly the right order and go from lowest to highest.

When we shuffle the deck, there would be some higher cards before lower ones, and as we continued shuffling the number would increase until, with a perfectly random order, it would be likely that very nearly half the pairs of cards appear in the wrong order.

If NN pairs of cards appear in the wrong order, we could then solve a combinatorics problem to figure out how many possible orderings there are for that N. The result would be Ω(N). The entropy would be

S=logΩ

and this entropy would be maximum for N


This choice of macrostate is arbitrary; I picked it as an example. If you made some other choice, such as the number of cards touching only cards of the same color, you would get a different result for the entropy. However, for most choices you would find that the original deck had low entropy, and for all choices you would find that a randomly-shuffled deck would be most likely to have nearly the highest possible entropy.


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