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 macro states 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) The macro states are available to the current system is $21macrostates$

b) The micro states are available to the present system is ${\mathrm{\Omega }}_{\text{overall}}=6.892\times {10}^{10}$

c) The probability of finding all the energy in solid A is $P=1.453\times {10}^{-4}$

d) The probability of finding exactly half of the energy in solid A is $P=0.1238$

Step-by-step solution

macro and micro states (a) and (b)

Assume they're two Einstein physics structures, $A$and $B$, with $N_A=N_B=10$and $q_A+q_B=20\text{.}$.

(a) As a metric, the count of macro states is:

$q+1=20+1=21$

because we began with $0$ and switch our attention resolute $20$.

(b) The formula for said length of microstates is:

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

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

$N_{\text{overall}}=N_A+N_B$

$\begin{array}{c}{\mathrm{\Omega }}_{\text{overall}}=\left(\begin{array}{c}20+20-1\\ 20\end{array}\right)\\ \end{array}$

$\begin{array}{c}=\frac{(20+20-1)!}{20!(20-1)!}\\ \end{array}$

$\begin{array}{c}=6.892\times {10}^{10}\\ \end{array}$

${\mathrm{\Omega }}_{\text{overall}}=6.892\times {10}^{10}$

Equilibrium (c)

(c)The likelihood that while in equilibrium state, most of the facility is in rigid $A$, resulted in:

$q_A=20$$q_B=0$$N_A=10$$N_B=10$

The cumulative number is calculated based:

$\begin{array}{c}{\mathrm{\Omega }}_{\text{total}}={\mathrm{\Omega }}_A{\mathrm{\Omega }}_B\\ \end{array}$

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

$\begin{array}{c}{\mathrm{\Omega }}_A=\left(\begin{array}{c}20+10-1\\ 20\end{array}\right)=\frac{(20+10-1)!}{20!(10-1)!}=10015005\\ \\ \end{array}$

$\begin{array}{c}{\mathrm{\Omega }}_B=\left(\begin{array}{c}{q}_B+N_B-1\\ q_B\end{array}\right)=\frac{\left(q_B+N_B-1\right)!}{q_{B3}!\left(N_B-1\right)!}\\ \end{array}$

${\mathrm{\Omega }}_B=\left(\begin{array}{c}0+10-1\\ 0\end{array}\right)=\frac{(0+10-1)!}{0!(10-1)!}=1$

${\mathrm{\Omega }}_{\text{total}}={\mathrm{\Omega }}_A{\mathrm{\Omega }}_B=10015005\times 1=10015005$

As an answer, this same chances inside this instance is:

localid="1650298575703" $P=\frac{{\mathrm{\Omega }}_{\text{total}}}{{\mathrm{\Omega }}_{\text{overall}}}$

localid="1650298609352" $=\frac{10015005}{6.892\times {10}^{10}}$

localid="1650298489904" $=1.453\times {10}^{-4}$

Evenly Distributed (d)

(d)The likelihood that while in equilibrium state, most of the power is in rigid , resulted in:

$q_A=10$$q_B=10$$N_A=10$$N_B=10$

The cumulative number is calculated based:

$\begin{array}{c}{\mathrm{\Omega }}_{\text{total}}={\mathrm{\Omega }}_A{\mathrm{\Omega }}_B\\ \end{array}$

localid="1650382124449" $\begin{array}{c}\\ \end{array}\begin{array}{c}\\ {\mathrm{\Omega }}_A=\left(\begin{array}{c}{q}_A+N_A-1\\ q_A\end{array}\right)=\frac{\left(q_A+N_A-1\right)!}{q_A!\left(N_A-1\right)!}\\ \end{array}$

$\begin{array}{c}{\mathrm{\Omega }}_A=\left(\begin{array}{c}20+10-1\\ 20\end{array}\right)=\frac{(10+10-1)!}{10!(10-1)!}=92378\\ \\ \end{array}$

$\begin{array}{c}{\mathrm{\Omega }}_B=\left(\begin{array}{c}{q}_B+N_B-1\\ q_B\end{array}\right)=\frac{\left(q_B+N_B-1\right)!}{q_{B3}!\left(N_B-1\right)!}\\ \end{array}$

${\mathrm{\Omega }}_B=\left(\begin{array}{c}0+10-1\\ 0\end{array}\right)=\frac{(10+10-1)!}{10!(10-1)!}=92378$

As a solution, this same chances inside this instance is:

$P=\frac{{\mathrm{\Omega }}_{\text{total}}}{{\mathrm{\Omega }}_{\text{overall}}}$

$=\frac{8.5337\times {10}^9}{6.892\times {10}^{10}}$

$=0.1238$

Equally Distributed (e)

(e) it'll be more likely that it photon energy will spread them divided between the $2$ solids, and when that does, it's rare that its position will revert about one solid having similar more quanta than others. That is, an irreversible state is that the one where the quanta are spaced equally.