In the Northern Hemisphere, summer has arrived. The time has come for us to chase the general public with water guns, jump through the neighbour’s garden sprinklers’ water rays, and either carefully place a soggy, wet sea cucumber on a human’s belly during their beach nap^{1}The author does not approve of this. Sea cucumbers should be left alone., or simply dump them (the human) in the actually-still-too-cold seawater, especially if you love them. At the end of the day, after all those wet adventures, nothing will beat hanging your clothes out to dry in a soothing breeze of fresh alpine air.

A few years ago, a friend asked what exactly causes wet clothes to dry. How does that work, exactly? I thought it was a great question because what may seem like a simple problem actually exposes one of the fundamental aspects of the way our universe works and, at the same time, forms one of the main causes for headaches among undergraduates: the second law of thermodynamics.

TL;DR? Don’t like mathematics (high school level) and prefer to read the ‘dashboard’ version? Skip to the bottom. Everyone else, please read on!

### A box of gas

Let’s first paint ourselves a simpler picture than the actual situation where we wear whatever is the latest summer catwalk beach fashion. For now, we will also ignore the Sun, we will ignore the wind, and we will ignore the humidity of the air.

Imagine, your colourful pair of swimming trunks is actually a simple box with a hundred gas molecules. The particles bounce chaotically back and forth against each other and the walls of the box itself.

Now, let there be a hole in the wall. Imagine, by pure chance, one molecule escaping the box through the hole, arriving in another container of exactly the same size. This obviously means there are now only 99 gas molecules left in the original box.

Have a look at Figure 1a. Assuming all gas molecules look exactly alike, how many ways do we have to arrange them in order to get the same result? Well, instead of this particular molecule having escaped, *any other* *one* of these hundred molecules could have escaped just as well. And so, as each one of the hundred gas particles was capable of escaping the box, exactly a hundred possibilities could have led to the same outcome (i.e. 1 escaped, 99 remain). In other words, exactly one hundred different configurations, or microstates, will entail the microstate of the box where it lost one molecule while 99 remain inside. Let’s call this number $W$. And let’s call that number for the microstate where one molecule escaped (and 99 remain inside), $W(1)$. So, $W(1)=100$.

Now, imagine not one, but two particles flew out, as is depicted in Figure 1b. Well, this means that a different number of arrangements would have led to this situation or microstate. As concluded above, for the first particle, one hundred possibilities existed as there were as many particles in the box, originally. For the second particle, however, only 99 possibilities existed since one had left the building already! Since for every 100 possibilities for the first molecule, 99 other possibilities exist for the second molecule, we calculate that the total number of possibilities leading to this particular state (i.e. 2 escaped, 98 remain), is $100 \times 99 = 9900$. However, since it doesn’t matter which of the two particles leaves the box first and which second, as they look exactly alike, we can divide that number by two, giving a total number of $4950$ possibilities. And so, $W(2) = 4950$.

More accurately, in general, to calculate the possible combinations in a situation like this, we use the formula

\begin{equation} W(k) = \frac{n!}{k!(n – k)!}, \end{equation}

where $n$ is the number of molecules in the box initially, which is 100, and $k$ is the number of molecules having escaped through the hole. $W$ is the letter we will further use to denote the number of possible arrangements of our gas molecules for each situation (e.g. 0 escaped & 100 remain, $W(0)$, or 4 escaped & 96 remain, $W(4)$, et cetera).

Here is a table with a few results. We included the situation where no single molecule has left the box. Obviously, the number of possible arrangements of the molecules leading to this situation, i.e. 0 escaped & 100 remain, is exactly one. We also included two more configurations where three and four particles have left the box. Note how quickly the possible arrangements increase.

Nr of escaped molecules / k | Nr of remaining molecules / (100 - k) | Nr of possible arrangements of all the molecules leading to the same situation / W(k) |
---|---|---|

0 | 100 | 1 |

1 | 99 | 100 |

2 | 98 | 4950 |

3 | 97 | 161 700 |

4 | 96 | 3 921 225 |

### Probabilities

How far can we take this? In this simplistic model, we can imagine the number of molecules remaining in the box becoming equal to the number having escaped into the other box: 50 escaped, 50 remain. So, let’s add $W(50)$ to the table. Also, let’s add more configurations to the table and have even more molecules escape the box until none are left, just to see what happens to the number of possible arrangements.

Nr of escaped molecules / k | Nr of remaining molecules / (100 - k) | Nr of possible arrangements of all the molecules leading to the same situation / W(k) |
---|---|---|

0 | 100 | 1 |

1 | 99 | 100 |

2 | 98 | 4950 |

3 | 97 | 161 700 |

4 | 96 | 3 921 225 |

... | ... | ... |

50 | 50 | 100 891 344 545 564 193 334 812 497 256 |

... | ... | ... |

96 | 4 | 3 921 225 |

97 | 3 | 161 700 |

98 | 2 | 4950 |

99 | 1 | 100 |

100 | 0 | 1 |

As you can see, the number of possible arrangements *decreases* again after the box has reached its natural state of equilibrium (i.e. 50 escaped, 50 remain). This is, of course, only logical as the situation ‘flips’, as it were. More and more molecules end up escaping the box rather than remaining.

If we were to calculate the probability of one or the other situations occurring, how should we go about this? Well, for instance, take the situation, or state, in which precisely zero gas molecules escaped. What is the probability of this state occurring?

We would need to know the number of possible arrangements ($W$) in the state where there are 0 which escaped and 100 remain (that number is exactly 1, so, $W(0) = 1$), divided by the total of all possible arrangements in all states (the sum of all $W$’s). Mathematically, what we are calculating is the possibility $P$ where 0 molecules have escaped, in other words, for $P(0)$, we write:

\begin{equation} P(0) = \frac{W(0)}{\text{total of all }W} = \frac{1}{\text{total of all }W}. \end{equation}

Of course, the total of all $W$ still needs calculation. To do that, we use equation (1) to write down an equation for the sum of all $W$ in the previous table:

\begin{align} \text{total of all }W &= \text{the sum of }W(0)\text{ to }W(100), \\ &= \sum_{n=0}^{100} \frac{100!}{n!(100-n)!}. \end{align}

The answer to equation (4) is 1 267 650 600 228 229 401 496 703 205 376.

That is a large number of total possible arrangements of all possible states. So, you can imagine that the probability of $P(0)$ occurring is inconceivably small: following equation (2), we get about $8 \times 10^{-29}$%. This would be 0% rounded to the nearest whole percentage.

Likewise, we can calculate the probability of 50 escaping, 50 remaining, or P(50). This turns out to have a probability of 8% (to the nearest whole percentage). In the following plot of probabilities for each state, you can see which state is most likely to occur.

So, in a way, given enough time, the arrangements of water molecules will converge to the state with the highest probability. Here, this is the situation where 50 molecules escaped, 50 remain. This is the so-called equilibrium point. Though there may be fluctuation around its equilibrium—plus or minus one or a few particles—it is perfectly fair to say that the probability of no molecules remaining, P(0), and the probability of all molecules escaping, P(100), are near zero. Intuitively, this is what you would expect: while it is theoretically possible, in practice, you will never live long enough to ever witness all the molecules randomly gathering in just one box.

### Back to our wet clothes

In real life, however, our pair of swimming trunks are not a box nor are there only one hundred water molecules. There are billions of water molecules in a liquid phase held together by the fabric of our garment. Also, there is the Sun. And there might be wind or even just a slight breeze. Besides, not looking like a box, swimmers don’t have a hole attached to a second box.

However, if the box is a metaphor for our wet clothes, then the second box is a metaphor for its environment, the open air. And now, it gets interesting.

In our example of the two boxes, no further external forces played any role^{2}A system that is thermally isolated from its surroundings is called an adiabatic system. There was no wind, no Sun, no air humidity to consider. In reality, of course, they should be taken into account. Here’s what they do: Sun heats up the water in the trunks, causing its molecules to gain energy, aiding escape from the fabric. Wind causes the air molecules to bump into water molecules, removing excess water vapour around the clothes, aiding the water molecules to evaporate even further. As long as the relative air humidity isn’t too high, drier air helps to evaporate the water even further.

So, what do these circumstances do, exactly? They shift the equilibrium point in our plot to the right. States where more and more molecules escape and less remain in the box, that is, stay in the swimming trunks, get a higher probability of occurring due to these circumstances.

In other words, if our boxes would be subjected to the elements, it would cause the equilibrium point to shift from 50 escaped, 50 remain, towards 95 escaped, 5 remain, for instance.

Moreover, since the air outside is practically infinitely large compared to our swimming trunks—and not at all like the spatially limited second box of our metaphor—the number of ways water molecules can be arranged by random motion, in a system of clothes hanging in the air outside, is significantly leaning towards the state where most escape into the air, even without wind and sunshine. Even though it may take longer, that state is practically inevitable.

It was the Austrian physicist Ludwig Boltzmann who elaborated on this very statistical nature of states of a system, specifically in terms of its possible configurations of microscopically small molecules per one and the same end state.

### Entropy and the second law of thermodynamics

Now, because the number of possible arrangements, $W(k)$, gets very big very fast, Boltzmann calculated their natural logarithm value. This is a very neat function when dealing with incredibly large numbers and exponential growth. On any respectable high school calculator, this can be done using the ‘ln’-button.

Boltzmann then proceeded to multiply these log values with a constant $k$^{3}which is a different $k$ than the one we used earlier. The value of this $k = 1.380649 \times 10^{-23}\text{JK}^{-1}$. to link the phenomenon of mechanically mixing stuff (arrangements of molecules) with the thermodynamical phenomenon of entropy of heat. We now call this constant $k$ the Boltzmann constant. He then wrote down the famous expression which we now call Boltzmann’s equation for entropy. In Vienna, in the city’s Central Cemetery, his gravestone is engraved with this very formula:

\begin{equation} S = k\log_e W. \end{equation}

So, now we get a new table with values for entropy $S$:

Nr of escaped molecules / k | Nr of remaining molecules / (100 - k) | Nr of possible arrangements of all the molecules leading to the same situation / W(k) | Probability / % to 2 sig.fig. | Entropy / S $(\times 10^{-22})$ |
---|---|---|---|---|

0 | 100 | 1 | $7.9 \times 10^{-29}$ | 0 |

1 | 99 | 100 | $7.9 \times 10^{-27}$ | 0.636... |

2 | 99 | 4950 | $3.9 \times 10^{-25}$ | 1.17... |

3 | 97 | 161 700 | $1.3 \times 10^{-23}$ | 1.66... |

4 | 96 | 3 921 225 | $3.1 \times 10^{-22}$ | 2.10... |

... | ... | ... | ... | |

50 | 50 | 100 891 344 545 564 193 334 812 497 256 | 8.0 | 9.22... |

... | ... | ... | ... | |

96 | 4 | 3 921 225 | $3.1 \times 10^{-22}$ | 2.10... |

97 | 3 | 161 700 | $1.3 \times 10^{-23}$ | 1.66... |

98 | 2 | 4950 | $3.9 \times 10^{-25}$ | 1.17... |

99 | 1 | 100 | $7.9 \times 10^{-27}$ | 0.636... |

100 | 0 | 1 | $7.9 \times 10^{-29}$ | 0 |

As you can see, entropy $S$ increases towards the equilibrium point, only to decrease beyond that, up to the point where it is zero again. Note that the highest value of entropy also has the highest probability value. This means that the state of the box and its surroundings (the other box) will tend to maximum entropy. This also means that an equilibrium point entails maximum entropy.

Going back to our system of wet clothing and their surroundings (the air) this means that, here too, the state of wet clothing will tend to maximum entropy. This value will correspond to the situation where most of the water molecules have escaped the clothes.

This is the second law of thermodynamics: the entropy of the Universe tends to a maximum.

### Why do wet clothes dry?

While external factors such as sunshine, wind, and relatively low air humidity do cause the probability distribution to shift more towards the state where most water molecules escape the clothes, based on the random motion of molecules alone, statistically, they should leave the fabric anyway (even though this is a slower process than with sunshine, wind, et cetera).

This is because the number of ways in which water molecules remain inside the clothes is simply almost infinitely small compared to the number of ways where they are *not* inside the clothes. This leads to the statistical fact that the probability of water molecules not remaining in the clothing outweighs that of them remaining in the clothing.

There are simply more places for water molecules to be in the open air than there are within the constrained spatial dimensions of someone’s tight swimming trunks. Or anyone’s, really.

Ultimately, wet clothes dry because the entropy of the Universe tends to a maximum.

*Photo of Boltzmann’s grave by **Daderot**, under **CC BY-SA 3.0**.*