Rather than insisting that all the molecules be in the left half of a container, suppose we only require that they be in the leftmost $99\%$(leaving the remaining $1\%$completely empty). What is the probability of finding such an arrangement if there are $100$molecules in the container? What if there are $10,000$molecules? What if there are ${\mathbf{10}}^{\mathbf{23}}$?

The number of microstates available to an $N$molecule $3-d$ideal gas with energy $U$contained in volume $V$is approximately:

$\Omega =\frac{V^N}{N!h^{3N}}\frac{{\pi }^{\frac{3N}{2}}}{\left(\frac{3N}{2}\right)!}(2mU{)}^{\frac{3N}{2}}$

where, $$\begin{gathered} \Omega (U,V,N)=f\left(N\right)V^N{U}^{\frac{3N}{2}} \\ f\left(N\right)=\frac{(2\pi m{)}^{\frac{3N}{2}}}{\left(\frac{3N}{2}\right)!N!h^{3N}} \end{gathered}$$

Since one of the assumptions of statistical mechanics is that all microstates are equally probable, it should be possible, just by change, to find that even if a gas has a total volume V available to it, sometimes all the molecules will clump up in some smaller portion of the volume, leaving the remaining space empty (a vacuum). How likely is this to happen?

Effectively, what we're asking is how likely is it that the volume occupied by the gas will spontaneously reduce from $V$to $aV$, where$0<a<1$. Since the volume is all that changes (both $N$and $U$are unchanged), we can look at formula $\left(1\right)$and find that reducing the volume reduces multiplicity to:

$\Omega (U,V,N)=f\left(N\right)(aV{)}^N{U}^{\frac{3N}{2}}$

Thus the probability that this will happen spontaneously is :

substitute equation $\left(1\right)$, localid="1650265451996" $$\begin{gathered} P\left(a\right)=\frac{\Omega \left(a\right)}{\Omega } \\ P\left(a\right)=\frac{(aV{)}^N}{(V{)}^N}=a^N \end{gathered}$$

Since $N$is a large number, even a value of a close to $1$is still very unlikely. For example, if $a=0.99$we find:

For $N=100$:

$P(0.99)=0.{99}^{100}=0.366$

For $N=10000$:

$P(0.99)=0.{99}^{10000}=2.248\times {10}^{-44}$

For$N={10}^{23}$:

$$\begin{gathered} P(0.99)=0.{99}^{{10}^{23}} \\ 0.{99}^{{10}^{23}}={10}^{-x} \end{gathered}$$

take the natural logarithm for both sides:

localid="1650267033492" $$\begin{gathered} {10}^{23}\ln (0.99)=-x\ln \left(10\right) \\ -0.01\times {10}^{23}=-x\times 2.3 \\ x=4.348\times {10}^{20} \end{gathered}$$

the probability is therefore:

$P(0.99)={10}^{-4.348\times {10}^{20}}$

In fact, even for only $100$molecules, the chance of the gas crowding into a smaller volume is virtually zero for $a<0.95$as we can see from a plot: