91Ó°ÊÓ

• The entropy of a substance is given as:

$S=k\ln \left(\mathrm{Î\copyright }\right)$ $\left(1\right)$

• where $\mathrm{Î\copyright }$is the number of microstates accessible to the substance. For a $3-d$ideal gas, this is given by Schroeder's equation $2.40:$

$\mathrm{Î\copyright }=\frac{V^N}{N!h^{3N}}\frac{(2m\mathrm{Ï€}U{)}^{\frac{3N}{2}}}{\left(\frac{3N}{2}\right)!}$ $\left(2\right)$

• where $V$is the volume, $U$is the energy, $N$is the number of molecules, $m$is the mass of a single molecule and $h$is Planck's constant. We can further approximate this formula by using Stirling's approximation for the factorials:

$\begin{array}{c}n!≈\sqrt{2\mathrm{π}n}{n}^n{e}^{−n}\end{array}$

we get,

role="math" localid="1650298531903" $\begin{array}{c}\\ N!≈\sqrt{2\mathrm{π}N}{N}^N{e}^{−N}\\ \\ \left(\frac{3N}{2}\right)!≈\sqrt{2\mathrm{π}\left(\frac{3N}{2}\right)}{\left(\frac{3N}{2}\right)}^{\left(\frac{3N}{2}\right)}{e}^{−\left(\frac{3N}{2}\right)}\end{array}$ $\left(3\right)\&\left(4\right)$

$\begin{array}{c}\left(\frac{3N}{2}\right)!N!≈\sqrt{2\mathrm{π}\left(\frac{3N}{2}\right)}{\left(\frac{3N}{2}\right)}^{\left(\frac{3N}{2}\right)}{e}^{−\left(\frac{3N}{2}\right)\sqrt{2\mathrm{π}N}{N}^N{e}^{−N}}\\ \\ \left(\frac{3N}{2}\right)!N!≈2\mathrm{π}{\left(\frac{3N}{2}\right)}^{\left(\frac{3N+1}{2}\right)}{N}^{\left(N+\frac{1}{2}\right)}{e}^{−\frac{5N}{2}}\end{array}$

• When $N$is large, we can throw away a couple of factors:

$\begin{array}{c}\left(\frac{3N}{2}\right)!N!≈{\left(\frac{3N}{2}\right)}^{\left(\frac{3N}{2}\right)}{N}^N{e}^{−\frac{5N}{2}}\\ \\ \left(\frac{3N}{2}\right)!N!≈\left(3\right)\left(\begin{array}{c}3N\\ 2\end{array}\right)\left(2\right)\left(\begin{array}{c}3N\\ 2\end{array}\right)N\left(\begin{array}{c}5N\\ 2\end{array}\right)C_2^{5N}\end{array}$ $\left(5\right)$

substitute from $\left(5\right)$into $\left(2\right)$, we get:

$\begin{array}{c}\mathrm{Î\copyright }≈\frac{V^N}{h^{3N}}\frac{(2m\mathrm{Ï€}U{)}^{\frac{3N}{2}}}{{\left.\left(3\right)\left(\frac{3N}{2}\right)(2{)}^{−\left(\frac{3N}{2}\right)}{N}^{\left(\frac{5N}{2}\right.}\right)}^{−\frac{6N}{2}}}\\ \\ \mathrm{Î\copyright }≈\frac{V^N(m\mathrm{Ï€}U{)}^{\frac{3N}{2}}}{h^{3N}}\frac{(2{)}^{\frac{3N}{2}}}{\left(3{)}^{\left(\frac{3N}{2}\right)}\right(2{)}^{−\left(\frac{3N}{2}\right)}{N}^{\left(\frac{5N}{2}\right)}{e}^{−\frac{5N}{2}}}\\ \\ \mathrm{Î\copyright }≈\frac{V^N(m\mathrm{Ï€}U{)}^{\frac{3N}{2}}}{h^{3N}}\frac{(2{)}^{3N}{e}^{\frac{5N}{2}}}{\left.(3{)}^{\left(\frac{3N}{2}\right.}\right)N\left(\frac{5N}{2}\right)}\end{array}$

$\begin{array}{r}\mathrm{Î\copyright }≈V^N{\left(\frac{2}{h}\right)}^{3N}{\left(\frac{m\mathrm{Ï€}U}{3}\right)}^{\frac{3N}{2}}{\left(\frac{e}{N}\right)}^{\frac{5N}{2}}\\ \\ \mathrm{Î\copyright }≈{\left[\frac{V}{N}{\left(\frac{2}{h}\right)}^3{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}\right]}^N{e}^{\frac{5N}{2}}\\ \\ \mathrm{Î\copyright }≈{\left[\frac{V}{N}{\left[{\left(\frac{2}{h}\right)}^2\right]}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}\right]}^N{e}^{\frac{5N}{2}}\\ \\ \mathrm{Î\copyright }≈{\left[\frac{V}{N}{\left(\frac{4}{h^2}\right)}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}\right]}^N{e}^{\frac{5N}{2}}\\ \mathrm{Î\copyright }\\ ≈{\left[\frac{V}{N}{\left(\frac{4m\mathrm{Ï€}U}{3h^{2}N}\right)}^{\frac{3}{2}}\right]}^N{e}^{\frac{5N}{2}}\end{array}$

take the natural logarithm for both sides, and take into account role="math" localid="1650302000554" $\ln \left(ab\right)=\ln \left(a\right)+\ln \left(b\right)\text{and}\ln \left(\frac{a}{b}\right)=\ln \left(a\right)−\ln \left(b\right)$, so:

$\begin{array}{c}\ln \left(\mathrm{Î\copyright }\right)≈N\ln \left[\frac{V}{N}{\left(\frac{4}{h^2}\right)}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}\right]+\ln \left(e^{\frac{5N}{2}}\right)\\ \\ \ln \left(\mathrm{Î\copyright }\right)≈N\ln \left[\frac{V}{N}{\left(\frac{4}{h^2}\right)}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}\right]+\frac{5N}{2}\\ \\ \ln \left(\mathrm{Î\copyright }\right)≈N\ln \left[\frac{V}{N}{\left(\frac{4}{h^2}\right)}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}+\frac{5}{2}\right]\end{array}$

• This gives the entropy of an ideal gas (from equation (1)) as:

$S=Nk\ln \left[\frac{V}{N}{\left(\frac{4}{h^2}\right)}^{\frac{3}{2}}{\left(\frac{m\mathrm{Ï€}U}{3N}\right)}^{\frac{3}{2}}+\frac{5}{2}\right]$