Question

Consider a system of two Einstein solids, \(A\) and \(B\), each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed.

(a) How many different macrostates are available to this system?

(b) How many different microstates are available to this system?

(c) Assuming that this system is in thermal equilibrium, what is the probability of finding all the energy in solid \(A\) ?

(d) What is the probability of finding exactly half of the energy in solid \(A\) ?

(e) Under what circumstances would this system exhibit irreversible behavior?

Short answer

(a) 21 macrostates are available

(b) ${\Omega }_{\text{overall}}=6.892\times {10}^{10}$microstates are available

(c) the probability that all the energy is in solid $A$at thermal equilibrium,$P=1.453\times {10}^{-4}$

(d) the probability in this case is$P=0.1238$

Step-by-step solution

find macrostates and microstates 

Suppose we have two systems of Einstein solids, $A$and $B$with $N_A=N_B=10$and $q_A+q_B=20$

(a) the number of macrostates is therefore:

$q+1=20+1=21$

because we start counting 0 to 20.

(b) the number of microstates formula is given by:

${\Omega }_{\text{overall}}\left(N_{\text{overall}},q_{\text{overall}}\right)=\left(\begin{array}{c}{q}_{\text{overall}}+N_{\text{overall}}-1\\ q_{\text{overall}}\end{array}\right)=\frac{\left(q_{\text{overall}}+N_{\text{overall}}-1\right)!}{q_{\text{overall}}!\left(N_{\text{overall}}-1\right)!}$

where,

$q_{\text{overall}}=q_A+q_B=20,N_{\text{overall}}=N_A+N_B=20$

so,

width="385">Ωoverall=20+20-120=(20+20-1)!20!(20-1)!=6.892×1010