Question

Consider a system of two Einstein solids, with N{A} = 300, N{B} = 200 and q{total} = 100 (as discussed in Section 2.3). Compute the entropy of the most likely macrostate and of the least likely macrostate. Also compute the entropy over long time scales, assuming that all microstates are accessible. (Neglect the factor of Boltzmann's constant in the definition of entropy; for systems this small it is best to think of entropy as a pure number.) 65

Short answer

The macrostate is

1. For the most likely macrostate: S = 264.2
2. For the least likely macrostate: S = 187.52
3. For the most likely macrostateS=267

Step-by-step solution

Step :1 Expression of Einstein solids

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:

$\mathrm{\Omega }={\mathrm{\Omega }}_A{\mathrm{\Omega }}_B$

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

The multiplicity of this macrostate is therefore:

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

substitute with

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

$\begin{array}{r}\\ \mathrm{\Omega }=\frac{(60+300-1)!}{60!(300-1)!}\frac{(40+200-1)!}{40!(200-1)!}=6.866\times {10}^{114}\\ \\ S=\mathrm{ln\Omega }=\ln \left(6.866\times {10}^{114}\right)=264.42\end{array}$

Step :2 Expression of solve 

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

$\begin{array}{c}\mathrm{\Omega }=\frac{\left(q_A+N_A-1\right)!}{q_A!\left(N_A-1\right)!}\frac{\left(q_B+N_B-1\right)!}{q_B!\left(N_B-1\right)!}=\frac{\left(299\right)!}{100!\left(199\right)!}\\ \mathrm{\Omega }=2.772\times {10}^{81}\\ \\ S=\mathrm{ln\Omega }=\ln \left(2.772\times {10}^{81}\right)=187.52\end{array}$

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

$\begin{array}{c}\mathrm{\Omega }=\left(\begin{array}{c}{q}_A+q_B+N_A+N_B-1\\ q_A+q_B\end{array}\right)=\left(\begin{array}{l}599\\ 100\end{array}\right)\\ \mathrm{\Omega }=9.262\times {10}^{115}\\ \\ S=\mathrm{ln\Omega }=\ln \left(9.262\times {10}^{115}\right)=267\\ \end{array}$

• 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.