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 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?

• In a standard pack of playing cards there are $N=52$different cards, so they can be arranged in:

$\Omega =N!=52!=8.07\times {10}^{67}$ways

• The size of this number is why it's highly unlikely that any card game that relies on dealing cards from a shuffled deck will ever repeat itself. The entropy of a shuffled deck is therefore:

$\begin{array}{c}S=k\mathrm{ln\Omega }\end{array}$

substitute with$\begin{array}{c}\\ k=1.38\times {10}^{-23}\mathrm{J}\cdot {\mathrm{K}}^{-1}\\ \end{array}$ $S=\left(1.38\times {10}^{-23}\right)\ln \left(8.07\times {10}^{67}\right)=2.157\times {10}^{-21}\mathrm{J}·{\mathrm{K}}^{-1}$

$\begin{array}{c}S=\left(1.38\times {10}^{-23}\right)\ln \left(8.07\times {10}^{67}\right)=2.157\times {10}^{-21}\mathrm{J}\cdot {\mathrm{K}}^{-1}\end{array}$

Without Boltzmann's constant we have,

$S=\ln \Omega =\ln (8.07x{10}^{67})=156.36$

• Although playing cards aren't made of an Einstein solid, the multiplicity of the macrostate in which thermal energy is exchanged among the cards will be something of similar order. The approximate multiplicity in the high temperature case for an Einstein solid with oscillators and q >> N energy quanta is

$\Omega \approx {\left(\frac{qe}{N}\right)}^N$

• For Non the order of 10²³ ,$\Omega$is a very large number, so the thermal entropy of the cards is vastly greater than the entropy generated by shuffling the deck.