91Ó°ÊÓ

Consider two Einstein solids with N_A=300, N_B= 200 and q = 100 .

The most likely macrostate will see the energy divided proportionately between the two solids,

so q_A=60 and q_B=40 .

The multiplicity of this macrostate is give by:

$\mathrm{Î\copyright }={\mathrm{Î\copyright }}_A{\mathrm{Î\copyright }}_B$

where,${\mathrm{Î\copyright }}_A=\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right){\mathrm{Î\copyright }}_A=\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right)$

The multiplicity of this macrostate is therefore:

$\begin{array}{r}\mathrm{Î\copyright }=\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right)\left(\begin{array}{c}{q}_A+N_A−1\\ q_A\end{array}\right)\\ \mathrm{Î\copyright }=\frac{\left(q_A+N_A−1\right)!}{q_A!\left(N_A−1\right)!}\frac{\left(q_B+N_B−1\right)!}{q_B!\left(N_B−1\right)!}\end{array}$

substitute with

$\begin{array}{r}{N}_A=300,N_B=200,q_A=60\text{and}{q}_B=40,\text{so:}\\ \end{array}$

$\begin{array}{r}\\ \mathrm{Î\copyright }=\frac{(60+300−1)!}{60!(300−1)!}\frac{(40+200−1)!}{40!(200−1)!}=6.866×{10}^{114}\\ \\ S=\mathrm{\pm ô²ÔÎ\copyright }=\ln \left(6.866×{10}^{114}\right)=264.42\end{array}$

• The least likely macrostate would find all the energy in the smaller solid, so that qA= 0 and qB=100. In that case:

$\begin{array}{c}\mathrm{Î\copyright }=\frac{\left(q_A+N_A−1\right)!}{q_A!\left(N_A−1\right)!}\frac{\left(q_B+N_B−1\right)!}{q_B!\left(N_B−1\right)!}=\frac{\left(299\right)!}{100!\left(199\right)!}\\ \mathrm{Î\copyright }=2.772×{10}^{81}\\ \\ S=\mathrm{\pm ô²ÔÎ\copyright }=\ln \left(2.772×{10}^{81}\right)=187.52\end{array}$

• Over long time scales, the interaction between the solids mean that all microstates are accessible. In this case the multiplicity is:

$\begin{array}{c}\mathrm{Î\copyright }=\left(\begin{array}{c}{q}_A+q_B+N_A+N_B−1\\ q_A+q_B\end{array}\right)=\left(\begin{array}{l}599\\ 100\end{array}\right)\\ \mathrm{Î\copyright }=9.262×{10}^{115}\\ \\ S=\mathrm{\pm ô²ÔÎ\copyright }=\ln \left(9.262×{10}^{115}\right)=267\\ \end{array}$

• Thus the most probable state with the solids divided has almost as much entropy as when the whole system is a single state. In most calculators, these binomial coefficients are non calculable, so you can use the following python code to calculate these cofficients.