# Entropy problem

1. Mar 4, 2008

### Benzoate

1. The problem statement, all variables and given/known data
How many possible arrangements are there for a deck of 52 playing cards?(For simplicity, consider only the order of the cards , not whether they are turned upside down, etc.) suppose you start with a sorted deck and shuffle it repeatedly , so that all possible arrangements becomes accessible? how much entropy have you created in the process? express your answer both as a pure number(neglecting the factor k) and the SI units.Is the entropy significant compared to the entropy asociated with arranging thermal energy among the molecules in the cards?

2. Relevant equations

S=k*ln(omega), k is neglected in this problem.

omega=(q+N)!/((q)!(N)!)
3. The attempt at a solution

a)How many possible arrangements are there for a deck of 52 playing cards?

the number of arrangements is just N factorial or in my case 52!

b)suppose you start with a sorted deck and shuffle it repeatedly , so that all possible arrangements becomes accessible how much entropy have you created in the process?

so would I just calculate the total number of omega 's: In other words, would I calculate all the posible q's? for instance , omega(q=0)=(0+52)/((0!)(52!)+omega(q=1)=(1+52)/((1!)(52!))+....+omega(q=51)=(51+52)!/((51!)(52!))+omega(q=52)=(52+52)!/((52!)(52!)) and then proceed to take the natural log of all the total sums of the omega's to calculate my entropy?

2. Mar 4, 2008

### genneth

I'm not sure where your equation for omega comes from, but omega is supposed to just be the number of possible arrangements. So your increase in entropy is just:

$$\Delta S = k \left( \ln 52! - \ln 1 \right)$$

3. Mar 4, 2008

### Benzoate

Is my calculation for the total number of arrangements correct? I do not understand how you obtained ln 1.

i obtained my equation for omega on p. 63 of Daniel's V. Schroeder Thermal physics textbook.

Last edited: Mar 4, 2008
4. Mar 4, 2008

### genneth

The number of arrangements is correct. The 1 comes from the fact that there is only one way to arrange the deck in a sorted manner.

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