Molecular Physics > Probability of a thermodynamical state > Problem 5.8.4

Statement

$5.8.4.$ Estimate the probability that the air density in the volume of 0.1 mm3 of any
part of your room will be twice as high as the usual density. What should be
the volume of this section so that this probability is large enough?

Solution

What should happen is the same number of molecules that are already in $V=1 mm^3$ should appear in $V.$
Total of molecules in the room (my bathroom after i take a shower $:)$ ):
$$N=N_{A} \cdot \frac{PV_0}{RT}\approx N_{A} \cdot \frac{10^5 \cdot 10}{8.3 \cdot 300}\approx10^{26}$$
We need:$$ n=N\cdot \frac{V}{V_0} \approx 10^{15}$$Probability of existing of $n$ molecules is $1$.
Probability of existing in $V$ for $1$ molecule in $p= \frac{V}{V_0}.$

So final answer is
$$ P= (\frac{V}{V_0})^n=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}=x^{Nx}, x=\frac{V}{V_0}$$In our case $\frac{V}{V_0}=10^{-10}$ and $P=10^{-10^{15}}$, which is close to Savchenko's estimation. The upper power depends strongly on temperature and volume.

We can plot $P(x)$ and see that $P=10^{-4}% $ if $x=\frac{V}{V_0}\approx10^{-27}$, so
$V=10^{-17} mm^3$

Intereating to notice, that with decreasing of $x$, $P$ goes to $1$.

Explanation in terms of our model is that if we consider a volume less than length of free fly of molecule, there should be all pairs of colliding molesules, so double density.

But much more truthful is the next fact:

The key postulate of thermodynamix say that we consider a system that has many enough partickles, so fluctuations of $n$ are negligible. But with decreasing of volume, we can no longer have this fact true.

Answer

$P=10^{-10^{15}}$,

$P=10^{-4}% $ if $V=10^{-17} mm^3$,

$P=(\frac{V}{V_0})^{N \cdot \frac{V}{V_0}}$