Question

In statistical mechanics, we frequently use the approximationN! = N In N-N, where N is of the order of Avogadro’s number. Write out ln N! using Stirling’s formula, compute the approximate value of each term for N = 10²³ , and so justify this commonly used approximation.

Short answer

The approximation is justified.

Step-by-step solution

Given Information

The approximation is N! =N InN- N .

Definition of the sterling’s formula.

Sterling’s formula is used to simplify formulas involving factorial.

$\mathit{n}\mathbf{!}\mathbf{\square }{\mathit{n}}^{\mathbf{n}}{\mathit{e}}^{\mathbf{-}\mathbf{n}}\sqrt{\mathbf{2}\mathbf{\pi n}}$.

Justify the approximation.

The approximation is N! = N In N - N.

The sterling’s formula is $n!\square n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}$.

Take log on both side on the formula mentioned above.

The equation becomes as follows.

$\begin{array}{rcl}In\left(n!\right)& =& In\left(n^n{e}^{-n}\sqrt{2\mathrm{\pi n}}\right)\left(1+\frac{1}{12n}\right)\\ & =& In{n}^n+In{e}^{-n}+In\sqrt{n}+In\left(1+\frac{1}{12n}\right)\\ & =& nInn-n+\frac{1}{2}Inn+In\left(1+\frac{1}{12n}\right)\end{array}$

Substitute n = 10²³ in the above equation.

The equation becomes as follows.

$$\begin{gathered} nInn=5.526\times {10}^{26} \\ where,n={10}^{23} \\ Inn=55In\left(1+\frac{1}{12n}\right) \\ \approx 0 \end{gathered}$$

Hence, In(n!) = nIn n - n .

The approximation is justified.