9.7 Statistical interpretation of entropy and the second law of  (Page 4/8)

Entropy increases in a coin toss

Suppose you toss 100 coins starting with 60 heads and 40 tails, and you get the most likely result, 50 heads and 50 tails. What is the change in entropy?

Strategy

Noting that the number of microstates is labeled $W$ in [link] for the 100-coin toss, we can use $\mathrm{\Delta }S=S_{\text{f}}-S_{\text{i}}=k\text{ln}{W}_{\text{f}}-k\text{ln}{W}_{\text{i}}$ to calculate the change in entropy.

Solution

The change in entropy is

where the subscript i stands for the initial 60 heads and 40 tails state, and the subscript f for the final 50 heads and 50 tails state. Substituting the values for $W$ from [link] gives

Discussion

This increase in entropy means we have moved to a less orderly situation. It is not impossible for further tosses to produce the initial state of 60 heads and 40 tails, but it is less likely. There is about a 1 in 90 chance for that decrease in entropy ( $–2\text{.}7\times {\text{10}}^{–\text{23}}\text{J/K}$ ) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get $\mathrm{\Delta }S=–\text{92}\times {\text{10}}^{–\text{23}}\text{J/K}$ . There is about a $1\text{ in }{\text{10}}^{\text{30}}$ chance of this change occurring. So while very small decreases in entropy are unlikely, slightly greater decreases are impossibly unlikely. These probabilities imply, again, that for a macroscopic system, a decrease in entropy is impossible. For example, for heat transfer to occur spontaneously from 1.00 kg of $\mathrm{0º}\text{C}$ ice to its $\mathrm{0º}\text{C}$ environment, there would be a decrease in entropy of $1\text{.}\text{22}\times {\text{10}}^3\text{J/K}$ . Given that a $\mathrm{\Delta }S{\text{ of 10}}^{–\text{21}}\text{J/K}$ corresponds to about a $1\text{ in }{\text{10}}^{\text{30}}$ chance, a decrease of this size ( ${\text{10}}^3\text{J/K}$ ) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.