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

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## 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 $\Delta S={S}_{\text{f}}-{S}_{\text{i}}=k\text{ln}{W}_{\text{f}}-k\text{ln}{W}_{\text{i}}$ to calculate the change in entropy.

Solution

The change in entropy is

$\Delta S={S}_{\text{f}}–{S}_{\text{i}}=k\text{ln}{W}_{\text{f}}–k\text{ln}{W}_{\text{i},}$

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

$\begin{array}{lll}\Delta S& =& \left(1\text{.}\text{38}×{\text{10}}^{–\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\right)\left[\text{ln}\left(1\text{.}0×{\text{10}}^{\text{29}}\right)–\text{ln}\left(1\text{.}4×{\text{10}}^{\text{28}}\right)\right]\\ & =& \text{2.7}×{\text{10}}^{–\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}\end{array}$

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 ( $–2\text{.}7×{\text{10}}^{–\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}$ ) to occur. If we calculate the decrease in entropy to move to the most orderly state, we get $\Delta S=–\text{92}×{\text{10}}^{–\text{23}}\phantom{\rule{0.25em}{0ex}}\text{J/K}$ . There is about a 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 $0º\text{C}$ ice to its $0º\text{C}$ environment, there would be a decrease in entropy of $1\text{.}\text{22}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J/K}$ . Given that a corresponds to about a chance, a decrease of this size ( ${\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{J/K}$ ) is an utter impossibility. Even for a milligram of melted ice to spontaneously refreeze is impossible.

## Problem-solving strategies for entropy

1. Examine the situation to determine if entropy is involved.
2. Identify the system of interest and draw a labeled diagram of the system showing energy flow.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). You must carefully identify the heat transfer, if any, and the temperature at which the process takes place. It is also important to identify the initial and final states.
5. Solve the appropriate equation for the quantity to be determined (the unknown). Note that the change in entropy can be determined between any states by calculating it for a reversible process.
6. Substitute the known value along with their units into the appropriate equation, and obtain numerical solutions complete with units.
7. To see if it is reasonable: Does it make sense? For example, total entropy should increase for any real process or be constant for a reversible process. Disordered states should be more probable and have greater entropy than ordered states.

## Section summary

• Disorder is far more likely than order, which can be seen statistically.
• The entropy of a system in a given state (a macrostate) can be written as
$S=klnW,$
where $k=1.38×{10}^{–23}\phantom{\rule{0.25em}{0ex}}\text{J/K}$ is Boltzmann’s constant, and $lnW$ is the natural logarithm of the number of microstates $W$ corresponding to the given macrostate.

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