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Eq. (21-298)
Mos NjMos Nj 31 Dec 2016 23:42
in discussion Chapter 21 / Solutions » Eq. (21-298)

Hello all,

Could anyone help me in proving Eq. (21-292)? It seems that is quite simple but I cannot understand why the ensemble average in the right-hand side of Eq. (21-297) reduces to Eq. (21-298).

Regards
MN

Eq. (21-298) by Mos NjMos Nj, 31 Dec 2016 23:42
Re: Q2-12
VerktajVerktaj 14 Nov 2016 04:43
in discussion Chapter 2 / Solutions » Q2-12

You have a distribution

(1)
\begin{align} W(\text{a})=\frac{\mathscr{A}!}{\prod_ja_j!}\prod_j\Omega^{a_j} \end{align}

with the constraints

(2)
\begin{align} \sum_ja_j&=\mathscr{A}\\ \sum_ja_jE_j&=\mathscr{E} \end{align}

Now, you have the logarithm of your distribution (using Stirling's approximation in the second equality)

(3)
\begin{align} \ln W(a_j)=\ln{\mathscr{A}!}+\sum_ja_j\ln{\Omega_j}-\sum_j\ln{a_j!}=\mathscr{A}\ln{\mathscr{A}}-\mathscr{A}+\sum_ja_j\ln{\Omega_j}-\left(\sum_ja_j\ln{a_j}-a_j\right) \end{align}

Then, the Lagrangian of optimization is

(4)
\begin{align} \mathscr{L}=\ln{W(a_j)}-\alpha\left(\sum_ja_j-\mathscr{A}\right)-\beta\left(\sum_ja_jE_j-\mathscr{E}\right) \end{align}

Taking the derivative with respect to $a_j$

(5)
\begin{align} \frac{\partial\mathscr{L}}{\partial a_j}&=\left[\frac{\partial\left(\mathscr{A}\ln{\mathscr{A}}-\mathscr{A}\right)}{\partial a_j}\right]+\left[\frac{\partial\left(\sum_ja_j\ln{\Omega_j}\right)}{\partial a_j}\right]-\left[\frac{\partial\left(\sum_ja_j\ln{a_j}-a_j\right)}{\partial a_j}\right]-\alpha-\beta E_j=0\\ &=\left(\ln{\mathscr{A}}+1\right)+\left(\ln{\Omega_j}\right)-\left(\ln{a_j}+1\right)-\alpha-\beta E_j=0 \end{align}

Rearranging you have

(6)
\begin{align} \ln{\frac{\mathscr{A}}{a_j}}=\alpha+\beta E_j-\ln{\Omega_j}\qquad\qquad\text{or}\qquad\qquad\ln{\frac{a_j}{\mathscr{A}}}=\ln{\Omega_j}-\alpha-\beta E_j \end{align}

Taking the exponential and summing over $j$:

(7)
\begin{align} \sum_j\frac{a_j}{\mathscr{A}}=1=\sum_j\Omega_je^{-\alpha}e^{-\beta E_j} \end{align}

where $e^{-\alpha}=\left(\sum_j\Omega_je^{-\beta E_j}\right)^{-1}$ and $\beta=1/k_BT$. Finally you have the answer:

(8)
\begin{align} a_j^*=\frac{\Omega_je^{-E_j/k_BT}}{\sum_j\Omega_je^{-E_j/k_BT}} \end{align}
Re: Q2-12 by VerktajVerktaj, 14 Nov 2016 04:43
Problem 4-10
O_omarO_omar 22 Feb 2016 23:39
in discussion Chapter 4 / Solutions » Problem 4-10

—-

Problem 4-10 by O_omarO_omar, 22 Feb 2016 23:39
problem 1.33
Junior moreiraJunior moreira 22 Sep 2015 19:05
in discussion Chapter 1 / Solutions » problem 1.33

Please help me with this problem!

problem 1.33 by Junior moreiraJunior moreira, 22 Sep 2015 19:05

All I see here is "unsupported math environment ,tabular'" can someone repost the solution?

Re: Problem 5-7
heath_henleyheath_henley 18 Sep 2015 17:51
in discussion Chapter 5 / Solutions » Problem 5-7

Results above using the one particle partition function, should be replaced with the full partition function

(1)
\begin{equation} Q = q_{trans}^N/N! \end{equation}

Will result in a factor of N multiplying P, E, and C.

Entropy is calculated using:

(2)
\begin{align} S = -(\frac{\partial A}{\partial T})_A = -(\frac{\partial ln Q}{\partial T})_A = Nk(lnq_{trans} +1) \end{align}
Re: Problem 5-7 by heath_henleyheath_henley, 18 Sep 2015 17:51
Problem 5-7
heath_henleyheath_henley 18 Sep 2015 17:28
in discussion Chapter 5 / Solutions » Problem 5-7

Given

(1)
\begin{align} \epsilon = \frac{h^2}{8ma^2} (s_x^2 + s_y^2) \end{align}

(1) Find the density of states and then use it to find the q_trans:

We can write:

(2)
\begin{align} R^2 = (s_x^2 + s_y^2) = \epsilon\frac{8ma^2}{h^2} \end{align}

We assume that the number of states ($\Phi(\epsilon)$) with energy $\epsilon$ can be approximated by 1/4 the area of a circle with radius R. Then we have

(3)
\begin{align} \Phi(\epsilon) = \frac{\pi R^2}{4} = \frac{2\pi\epsilon ma^2}{h^2} \end{align}

Then the number of states between $\epsilon$ and $\epsilon +d\epsilon$ is given by:

(4)
\begin{align} \Phi(\epsilon+d\epsilon)-\Phi(\epsilon) = \frac{2\pi ma^2}{h^2}d\epsilon = \omega(\epsilon)d\epsilon \end{align}

Now using:

(5)
\begin{align} q_{trans} = \int_0^\infty \omega(\epsilon)exp(-\beta\epsilon) d\epsilon \end{align}

gives:

(6)
\begin{align} q_{trans} = \frac{2\pi mkT}{h^2}A \end{align}

(2) Find $q_{trans}$ using another method:

This time we will sum over states instead of summing over levels, we write:

(7)
\begin{align} q_{trans} = \int_0^\infty \int_0^\infty exp(\frac{-\beta h^2}{8ma^2}s_x^2)exp(\frac{-\beta h^2}{8ma^2}s_y^2) ds_x ds_y = (\int_0^\infty exp(\frac{-\beta h^2}{8ma^2}s^2)ds)^2 \end{align}

Evaluating the integral gives:

(8)
\begin{align} q_{trans} = \frac{2\pi mkT}{h^2}A \end{align}

Which is fortunately the same we obtained by summing over levels.

(3) Find heat capacity, U, S, and EOS

To find the EOS we use:

(9)
\begin{align} P = kT (\frac{\partial ln(q)}{\partial A})_T = \frac{kT}{A} \end{align}

To find E:

(10)
\begin{align} E = kT^2 (\frac{\partial ln(q)}{\partial T})_A = kT \end{align}

Finally to find $C_A$:

(11)
\begin{align} C_A = (\frac{\partial E}{\partial T})_A = k \end{align}
Problem 5-7 by heath_henleyheath_henley, 18 Sep 2015 17:28
Re: Question1-23
DrJalaliDrJalali 10 Sep 2015 23:15
in discussion Chapter 1 / Solutions » Question1-23

Yes, What part are you having difficulty with specially? Also, I believe the answers will be posted on Monday to HW2.

Re: Question1-23 by DrJalaliDrJalali, 10 Sep 2015 23:15
Q2-12
Shanshan CuiShanshan Cui 10 Sep 2015 21:14
in discussion Chapter 2 / Solutions » Q2-12

Can anyone solve this problem?

Q2-12 by Shanshan CuiShanshan Cui, 10 Sep 2015 21:14
Question1-23
Shanshan CuiShanshan Cui 10 Sep 2015 20:54
in discussion Chapter 1 / Solutions » Question1-23

Can anyone solve this problem?Thanks a lot!

Question1-23 by Shanshan CuiShanshan Cui, 10 Sep 2015 20:54
question
sagharsaghar 02 Nov 2014 14:12
in discussion Chapter 12 / Solutions » question

can anyone please solve Q 16 of chapter 12?

question by sagharsaghar, 02 Nov 2014 14:12
Re: Problem 2-14
Mon CBarMon CBar 27 Feb 2014 04:58
in discussion Chapter 2 / Solutions » Problem 2-14

This is almost the same than 2-17. You need to use the same equations to solve it, but here it goes:

To get the pressure, we use (sorry for the notation, but I really don't know how to write the solution in a better way. You can copy-paste in LaTeX…)

\bar{P} = kT (\frac{\partial ln Q}{\partial V})_N

with Q(N,V,T) = \frac{1}{N!} (frac{2 \pi m k T}{h^2})^{3N/2} V^N.

Doing the derivatives we get

\bar{P} = \frac{kTN}{V}.

In order to get the energy, we use

\bar{E} = kT^2 (\frac{\partial ln Q}{\partial T})_N

and the same Q, of course. So the energy is

\bar{E} = \frac{3 kTN}{2}

Re: Problem 2-14 by Mon CBarMon CBar, 27 Feb 2014 04:58
Problem 2-14
Symon SajibSymon Sajib 07 Feb 2014 00:31
in discussion Chapter 2 / Solutions » Problem 2-14

I have tried problem 2-14 but cant solve it. can anyone help?

Problem 2-14 by Symon SajibSymon Sajib, 07 Feb 2014 00:31
(1)
\begin{align} \Xi(V,T,\mu)=\sum\mathit{Q(N,V,T)}e^{\beta\mu\bar{N}} \end{align}
(2)
\begin{align} \Xi(V,T,\mu)=\sum\mathit{Q(\bar{N},V,T)}e^{\beta\mu\bar{N}} \end{align}
(3)
\begin{align} \Xi(V,T,\mu)=\sum\mathit{Q(\bar{N},V,T)}e^{\beta\mu\bar{N}} \end{align}
(4)
\begin{align} \ln\Xi(V,T,\mu)=\ln\mathit{Q+\beta\mu\bar{N}} \end{align}
(5)
\begin{align} \ln\Xi(V,T,\mu)=\frac{S}{K}-\frac{\bar{E}}{kT}+\beta\mu\bar{N} \end{align}
(6)
\begin{align} \ln\Xi(V,T,\mu)=\frac{1}{kT}\mathit{pV} \end{align}
(7)
\begin{align} \triangle\mathit{(p,T,N)=\sum}\mathit{Q(N,\bar{V},T)}e^{-\beta pV} \end{align}
(8)
\begin{align} \ln\Delta(\mathit{p,T,N)=\ln}\sum\mathit{Q(N,\bar{V,T)}e^{-\mathit{\beta pV}}} \end{align}
(9)
\begin{align} =\ln\mathit{Q-\beta pV} \end{align}
(10)
\begin{align} =\frac{\mathit{S}}{K}-\frac{E}{kT}-\beta pV \end{align}
(11)
\begin{align} =-\mathit{\frac{A}{kT}}-\beta pV \end{align}
(12)
\begin{align} =-\mathit{\frac{G}{kT}} \end{align}
(13)
\begin{align} \phi(V,E,\beta\mu)=\sum\Omega(\mathit{\bar{N,V,E})e}^{\beta\mu\bar{N}} \end{align}
(14)
\begin{align} \phi(V,E,\beta\mu)=\sum\Omega(\mathit{\bar{N,V,E})e}^{\beta\mu\bar{N}} \end{align}
(15)
\begin{align} \mathit{\ln\mathit{\phi=\ln\Omega}}+\beta\mu\bar{N} \end{align}
(16)
\begin{align} \ln\phi=\frac{S}{K}-\frac{\bar{E}}{kT}+\beta\mu\bar{N} \end{align}
(17)
\begin{align} \mathit{\ln\phi=-}\frac{H}{kT} \ln\phi=\frac{S}{K}-\frac{\bar{E}}{kT}+\beta\mu\bar{N} \end{align}
(18)
\begin{align} \ln\psi(V,T,\mu_{1},\mathit{N_{1}})=\sum(N_{1},N_{2},\mathit{T},V)e^{\beta\mu N_{1}} \end{align}
(19)
\begin{align} \ln\psi(V,T,\mu_{1},\mathit{N_{1}})=\sum(\bar{N}_{1},N_{2},\mathit{T},V)e^{\beta\mu\bar{N}_{1}} \end{align}
(20)
\begin{align} \ln\psi=\ln[\mathit{Qe^{\beta\bar{\mu\bar{N_{1}}}}}] \end{align}
(21)
\begin{align} \mathit{\ln\psi=\frac{S}{K}}-\frac{\bar{E}}{kT}+\beta\mu\bar{N_{1}}=\frac{ST-\bar{E}}{kT}-\beta\mu\bar{N_{1}}=\mathit{\frac{A}{kT}}-\beta\mu\bar{N} \end{align}
(22)
\begin{align} \ln\psi=\frac{A}{kT}-\beta\mu\bar{N} \end{align}
(23)
\begin{align} W(p,\gamma,T,N)=\sum\sum Q(N,V,AT)e^{-\beta pV}e^{\beta\gamma A} \end{align}
(24)
\begin{align} W(p,\gamma,T,N)=\sum\sum Q(N,\bar{V},\bar{A,}T)e^{-\beta\bar{P}V}e^{\beta\gamma\bar{A}} \end{align}
(25)
\begin{align} \ln\mathit{W=}\ln[\mathit{Q}e^{-\beta p\bar{V}}e^{\beta\gamma\bar{A}}] \end{align}
(26)
\begin{align} \ln\mathit{W=\frac{S}{K}}-\frac{E}{kT}-\beta\bar{p}V+\beta\gamma A=\frac{ST-E}{kT}-\beta\bar{p}V+\beta\gamma A \end{align}
(27)
\begin{align} \ln W=\frac{-A}{kT}-\beta\bar{p}V+\beta\gamma A=-G \end{align}
Problem 3-15 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 04:40

State and use Euler's therom to show

(1)
\begin{align} \mathit{p=kT(\frac{\partial\ln\Xi}{\partial V}})=kT\frac{\ln\Xi}{V} \end{align}
(2)
\begin{align} 1-\mathit{p=kT}\frac{\ln\Xi}{V} \end{align}

From equation (1-56)

(3)
\begin{align} \mathit{dG=-SdT+Vdp+\sum}\mu jdN \end{align}

In The Grand Canonical Ensemble

(4)
\begin{align} \mathit{G=\mu N=}E+pV-TS \end{align}

Thus we have

(5)
\begin{align} pV=-E+TS-\mu N \end{align}
(6)
\begin{align} \mathit{p=\frac{1}{V}}(-E+TS-\mu N)\rightarrow1 \end{align}

From equation (3-12)

(7)
\begin{align} S=\frac{\bar{E}}{T}-\frac{\bar{N}\mu}{T}+\mathit{k}\ln\Xi \end{align}

We can rewrite (1) in term of Entropy in equation (3-12)

(8)
\begin{align} p=\frac{1}{V}[-\mathit{E+T}(\frac{\bar{E}}{T}-\frac{\bar{N\mu}}{T}+\mathit{k}\ln\Xi)-\mu N] \mathit{p=\frac{1}{V}}[-E+\bar{\bar{E}-N\mu+}Tk\ln\Xi-\mu N] \end{align}
(9)
\begin{align} p=\frac{1}{v}[\mathit{KT}\ln\Xi] \end{align}
(10)
\begin{align} p=kT\frac{\ln\Xi}{V} \end{align}

Apply Euler's theorem

(11)
\begin{align} nF(x_{1,}x_{2,}...,x_{N)=x_{1}}\frac{\partial F}{\partial x_{1}}+.....+\frac{\partial F}{\partial x_{N}} v\ln\Xi(V,T,\mu)=v\frac{\partial\ln\Xi}{\partial v}=v[\frac{\partial}{\partial v}(\frac{pV}{kT})]=v\frac{p}{kT} \end{align}

Therefor

(12)
\begin{align} v\ln\Xi(V,T,\mu)=v\frac{p}{kT} \end{align}

We can rewrite it as

(13)
\begin{align} p=kT\frac{\ln\Xi}{v} \end{align}

And we can use the same method to state that p=kT(\frac{\partial\ln\Xi}{\partial V})

(14)
\begin{align} (\frac{\partial\ln\Xi}{\partial v})_{\mu,T}=\frac{\ln\Xi}{v}=\mathit{\frac{p}{kT}} \end{align}
(15)
\begin{align} p=kT(\frac{\partial\ln\Xi}{\partial v}) \end{align}
Problem3-4 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 04:12

\mathit{B}
in thermal equlibrim,
the total number of states of the system is given be Eq 2-22

(1)
\begin{align} \mathit{W(a,b)=\frac{A!}{\Pi_{j}a_{j}!}} \frac{B!}{\Pi_{j}b_{j}!} \end{align}

The lagrangian with out set of three constraints is

For two systems \mathit{A}and
\mathit{B}
in thermal equlibrim,
the total number of states of the system is given be Eq 2-22

(2)
\begin{align} \mathit{W(a,b)=\frac{A!}{\Pi_{j}a_{j}!}} \frac{B!}{\Pi_{j}b_{j}!} \end{align}

The lagrangian with out set of three constraints is

(3)
\begin{align} \backslash mathcal\{L\}=\backslash ln\backslash\{W(a,b)\}-\backslash alpha\_\{A\}\backslash suma\_\{k\}-\backslash alpha\_\{B\}\backslash sumb\_\{k\}-\backslash beta\backslash sum(a\_\{k\}E\_\{kA\}+b\_\{k\}E\_\{kB\}) \end{align}
(4)
\begin{align} =A\mathit{\ln(A)-\sum a_{k}}\ln(a_{k})+\beta\ln(\beta)-\sum b_{k}\ln(b_{k)}-\sum-\alpha_{A}\sum a_{k}-\alpha_{B}\sum b_{k}-\beta\sum(a_{k}E_{kA}+b_{k}E_{kB} \end{align}
(5)
\begin{align} =A\mathit{\ln(A)-\sum a_{k}}\ln(a_{k})+\beta\ln(\beta)-\sum b_{k}\ln(b_{k)}-\sum-\alpha_{A}\sum a_{k}-\alpha_{B}\sum b_{k}-\beta\sum(a_{k}E_{kA}+b_{k}E_{kB} \end{align}

We now need to find the set of a_{j}
s and
b_{j}
s that maximize
the function

(6)
\begin{align} \frac{d\mathcal{L}}{da_{j}}=-\ln(a_{j}^{*})-1-\alpha_{A}-\beta E_{jA}=0 \end{align}
(7)
\begin{align} \frac{d\mathcal{L}}{db_{j}}=-\ln(b_{j}^{*})-1-\alpha_{B}-\beta E_{jB}=0 \end{align}

We are maximizing \mathit{B},
ignoring what state \mathit{A}is

in and vice veras. For each of theses we may sum over all js
and
apply the constrains \sum_{j}a_{j}=A
and \sum_{j}b_{j}=B

(8)
\begin{align} e^{-\alpha\prime_{A}}=\frac{\sum_{j}e^{-\beta E_{jA}}}{A} \end{align}
(9)
\begin{align} e^{-\alpha\prime B}=\frac{\sum_{j}e^{-\beta E_{j}B}}{B} \end{align}

Where \alpha\prime=\alpha+1.
This gives

(10)
\begin{align} a_{j}{}^{*}=A\frac{e^{-\beta E_{jA}}}{\sum_{j}e^{-\beta E_{jA}}}=A\frac{e^{-\beta E_{jA}}}{QA} \end{align}
(11)
\begin{align} b_{j}^{*}=B\frac{e^{-\beta EjB}}{\sum_{j}e^{-\beta E_{jB}}}=B\frac{e^{-\beta E_{jB}}}{Q_{B}} \end{align}

the probability of finding A in state i and B in the state j is
then

(12)
\begin{align} p_{ij}=\frac{a\bar{i}}{A}\frac{b\bar{i}}{B}=\frac{ai^{*}}{A}\frac{b_{i}^{*}}{B}=\frac{e^{-\beta E_{jA}}}{Q_{A}}\frac{e^{-\beta E_{jB}}}{Q_{B}}=P_{iA}P_{iB} \end{align}
(13)
\begin{align} =A\mathit{\ln(A)-\sum a_{k}}\ln(a_{k})+\beta\ln(\beta)-\sum b_{k}\ln(b_{k)}-\sum-\alpha_{A}\sum a_{k}-\alpha_{B}\sum b_{k}-\beta\sum(a_{k}E_{kA}+b_{k}E_{kB} \end{align}

We now need to find the set of a_{j}
s and
b_{j}
s that maximize
the function

(14)
\begin{align} \frac{d\mathcal{L}}{da_{j}}=-\ln(a_{j}^{*})-1-\alpha_{A}-\beta E_{jA}=0 \end{align}
(15)
\begin{align} \frac{d\mathcal{L}}{db_{j}}=-\ln(b_{j}^{*})-1-\alpha_{B}-\beta E_{jB}=0 \end{align}

We are maximizing \mathit{B},
ignoring what state \mathit{A}is

in and vice veras. For each of theses we may sum over all js
and
apply the constrains

(16)
\begin{align} \sum_{j}a_{j}=A and \sum_{j}b_{j}=B \end{align}
(17)
\begin{align} e^{-\alpha\prime_{A}}=\frac{\sum_{j}e^{-\beta E_{jA}}}{A} \end{align}
(18)
\begin{align} e^{-\alpha\prime B}=\frac{\sum_{j}e^{-\beta E_{j}B}}{B} \end{align}

Where \alpha\prime=\alpha+1.
This gives

(19)
\begin{align} a_{j}{}^{*}=A\frac{e^{-\beta E_{jA}}}{\sum_{j}e^{-\beta E_{jA}}}=A\frac{e^{-\beta E_{jA}}}{QA} \end{align}
(20)
\begin{align} b_{j}^{*}=B\frac{e^{-\beta EjB}}{\sum_{j}e^{-\beta E_{jB}}}=B\frac{e^{-\beta E_{jB}}}{Q_{B}} \end{align}

the probability of finding A in state i and B in the state j is
then

(21)
\begin{align} p_{ij}=\frac{a\bar{i}}{A}\frac{b\bar{i}}{B}=\frac{ai^{*}}{A}\frac{b_{i}^{*}}{B}=\frac{e^{-\beta E_{jA}}}{Q_{A}}\frac{e^{-\beta E_{jB}}}{Q_{B}}=P_{iA}P_{iB} \end{align}
Problem 2-9 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 03:54

2-3 Show that Eq. (2-9) follows from Eq. (2-8). Note that in
deriving this result,we have writting

(1)
\begin{align} \mathit{\ln}W(a) as \mathit{A}\ln A - \mathit{A}\ln A - \sum j a_{j} \ln a_{j} +\mathit{A} \end{align}

and have considered \mathit{A}to

be a constant. Show that Eq.(2-10) is independent of the
assumption,that is derive Eq.(2-10) treating A
as \sum j
a_{j}

Answer:

(2)
\begin{align} \frac{\sigma}{\sigma a}(\ln W(a)-\varphi\sum a_{k}\mathit{E_{k}})=0\rightarrow(1) \end{align}
(3)
\begin{align} \ln W(a)=\ln A!-\ln\Pi_{k}a_{k}i=\ln A! \end{align}

from Eq (2-3)

using striling approxmation

(4)
\begin{align} \ln W(a)=A\ln A\mathit{-A-\sum a_{k}\ln a_{k}}+\sum a_{k} \end{align}

from Eq (2-1)

(5)
\begin{align} \sum a_{k=}A \end{align}

We can rewrite it as

(6)
\begin{align} \ln W(a)=A\ln A\mathit{-A-\sum a_{k}\ln a_{k}}+A \end{align}

Therefor

(7)
\begin{align} \ln W(a)=A\ln A\mathit{-\sum a_{k}\ln a_{k}} \end{align}

We can now apply it to question (1)

(8)
\begin{align} \frac{\sigma}{\sigma a}([A\ln A\mathit{-\sum a_{k}\ln a_{k}}]-\varphi\sum a_{k}\mathit{E_{k}})=0 \end{align}
(9)
\begin{align} \frac{\sigma}{\sigma a}[A\ln A\mathit{-\sum a_{k}\ln a_{k}}]-\varphi\frac{\sigma}{\sigma a}\sum a_{k}- \end{align}
(10)
\begin{align} \beta\frac{\sigma}{\sigma a}\mathit{\sum a_{k}E_{k}})=0\rightarrow(2) \end{align}
(11)
\begin{align} \frac{\sigma}{\sigma a}A\ln A-\frac{\sigma}{\sigma a}\sum a_{k}\ln a_{k}-\varphi\frac{\sigma}{\sigma a}\sum a_{k}-\beta\frac{\sigma}{\sigma a}\mathit{\sum a_{k}E_{k}})=0\rightarrow(3) \end{align}

From chin roule

(12)
\begin{align} \frac{d}{da_{j}}\sum a_{k}\ln a_{k}=...+\frac{d}{da_{j}}a\ln a_{j\neq i}+\frac{d}{da_{i}}a_{i}\ln a_{i}+... \end{align}
(13)
\begin{align} \frac{d}{da_{j}}\sum a_{k}\ln a_{k}=\ln(a)+1\rightarrow(2) \end{align}
(14)
\begin{align} -\frac{d}{da_{j}}\psi\sum a_{k}=\psi\frac{d}{daj}(a_{1}+a_{2}+...)=\varphi\rightarrow(2) \end{align}
(15)
\begin{align} -\frac{d}{da_{j}}\psi\sum a_{k}=\beta\frac{d}{da_{j}}(a_{1}E_{1}+a_{2}E2+..)=-\beta a_{j}E_{j}\rightarrow(4) \end{align}

Apply (1),(3) and (4) to (2)

-

(16)
\begin{align} ln(a)-1-\psi-\beta a_{j}E_{j}=0 \end{align}
(17)
\begin{align} -\ln(a)-1-\psi-\beta\mathbf{a}_{j}E_{j}=0 \end{align}

In the second part of
the equation we need to show that using

(18)
\begin{align} A=\sum j a_{j} \end{align}

does
not change our answer

(19)
\begin{align} \sum_{k}\frac{\partial\mathscr{\mathscr{M}A}\partial\text{hello world}\, hello\, world}{} \end{align}
(20)
\begin{align} \frac{d}{daj}(\ln W(a)-a\sum a_{k}-\beta\sum a_{k}E_{k})=\frac{d}{daj}{(\sum a_{k})\ln(\sum a_{k})-\sum \end{align}
(21)
\begin{align} a_{k}-(\sum a_{k}\ln(a_{k}))+\sum a_{k}-\varphi\sum a_{k}-\beta\sum a_{k}E_{k}} \end{align}
(22)
\begin{align} =\frac{d}{daj}{(\sum a_{k})\ln(\sum a_{k})-(\sum a_{k}\ln(a_{k})-\varphi\sum a_{k}-\beta\sum a_{k}E_{k}} \end{align}
(23)
\begin{align} =\ln(\sum a_{k})+\sum a_{k}.\frac{1}{\sum ka_{k}}--\ln(aj)-a_{j}\frac{1}{a_{j}}-\varphi-\beta\sum a_{k}E) \end{align}
(24)
\begin{align} =\ln(A)+1-\ln(a_{j})-1-\varphi-\beta E_{j} \end{align}
(25)
\begin{align} =\ln(A)-\ln(a_{j})-\alpha-\beta E_{j} \end{align}

This then gives

(26)
\begin{align} \frac{a^{*}}{A}=e^{-\alpha}e^{-\beta E_{j}} \end{align}

Which has the same form as Eq(2-10).Summing thus over j and use
Eq. (2-1) gives

(27)
\begin{align} \sum\frac{a^{*}}{A}=\sum e^{-\alpha}e^{-\beta E_{j}} \end{align}
(28)
\begin{align} 1=e^{-\alpha}\sum e^{-\beta E_{j}} \end{align}

and

(29)
\begin{align} \mathit{p_{j=}}\frac{a_{j}^{*}}{A}=\frac{e^{-\beta E_{j}}}{\sum e-\beta E_{j}} \end{align}

Where we have used Eq. (2-7) and have recovered Eq (2-12)

Problem 2-3 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 03:14

Problem 2-17

In chapter 14 we shall derive an approximate partition function
for dense gas, which is the form

(1)
\begin{align} Q(N,V,T)=\frac{1}{N!}(\frac{2\Pi mkT}{h^{2}})^{3N/2} (V-Nb)^{N}e^{aN^{2}/VkT} \end{align}

Calculate the equation of state from this partition function ?

To obtain the equation os state , we apply Eq. 2-23 to obtain the
pressure

(2)
\begin{align} \bar{\bar{P}=kT(\frac{\partial\ln Q}{\partial V}})_{N} \end{align}
(3)
\begin{align} =kT(\frac{\partial}{\partial V}\ln[\frac{1}{N!}(\frac{2\pi mkT}{h^{2}})^{3N/2}(V-Nb)^{N}e^{aN^{2}/VkT}])_{N.T} \end{align}
(4)
\begin{align} =kT(\frac{\partial}{\partial V}\{\ln[\frac{1}{N!}\frac{2\pi mkT}{h^{2}})^{3N/2}]+\ln(V-Nb)^{N}+\ln(e^{aN^{2}/VkT})\} \end{align}
(5)
\begin{align} =kT(0+\frac{N}{(V-Nb)}-\frac{aN^{2}}{V^{2}kT}) \end{align}
(6)
\begin{align} =\frac{kTN}{V-Nb}-\frac{aN^{2}}{V^{2}} \end{align}

Which is the Van der Waals equation. The energy of this gas is

(7)
\begin{align} \bar{E}=kT^{2}(\frac{\partial\ln Q}{\partial T})_{N.T} \end{align}
(8)
\begin{align} =kT^{2}(\frac{\partial}{\partial T}\ln[\frac{1}{N!}(\frac{2\pi mkT}{h^{2}})^{3N/2}(V-Nb)^{N}e^{aN^{2}/VkT}])_{N,V} \end{align}
(9)
\begin{align} =kT^{2}(\frac{\partial}{\partial T}\{\ln(\frac{1}{N!})+\ln(\frac{2\pi mkT}{h^{2}})^{3N/2}+\ln(V-Nb)^{N}+\ln e^{aN^{2}/VkT})])_{N,V} \end{align}
(10)
\begin{align} =kT^{2}(0+\frac{3}{2}N\frac{h^{2}}{2\pi mkT}\frac{2\pi mk}{h^{2}}+0-\frac{aN^{2}}{VkT^{2}}) \end{align}
(11)
\begin{align} =kT^{2}(\frac{3}{2T}-\frac{aN^{2}}{VkT^{2}}) \end{align}
(12)
\begin{align} =\frac{3}{2}NkT-\frac{aN^{2}}{V} \end{align}

The heat capacity is then

(13)
\begin{align} C_{v}=(\frac{\partial E}{\partial T})_{N,V} \end{align}
(14)
\begin{align} =(\frac{\partial}{\partial T}\frac{3}{2}NkT-aN^{2})_{N,V} \end{align}
(15)
\begin{align} =\frac{3}{2}NK \end{align}

Comparing to problem 1-30

(16)
\begin{align} (\frac{\partial E}{\partial V})_{T}=(\frac{\partial}{\partial V}\frac{3}{2}NkT-\frac{aN^{2}}{V})_{T} \end{align}
(17)
\begin{align} =\frac{aN^{2}}{V^{2}} \end{align}

which is the same

(18)
\begin{align} =T(\frac{\partial E}{\partial V})-p \end{align}
(19)
\begin{align} =T(\frac{\partial}{\partial T}\frac{kTN}{V-Nb}-\frac{aN^{2}}{V^{2}})-\frac{kTN}{V-Nb}+\frac{aN^{2}}{V^{2}} \end{align}
(20)
\begin{align} =\frac{aN^{2}}{V^{2}} \end{align}
Problem 2-17 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 02:38

Problem 2-10

Drive equation (2-31) and (2-32).

(1)
\begin{align} \bar{P}=kT(\frac{\partial\ln}{\partial V})_{N,T} \end{align}

Starting from equation 2-15

(2)
\begin{align} \bar{P}=-\frac{\sum_{j}(\frac{\partial E_{j}}{\partial V})e^{-\beta Ej}}{\sum_{j}e^{-\beta E_{j}}} \end{align}
(3)
\begin{align} (\frac{\partial Q}{\partial V})_{N,T}=\frac{\partial}{\partial V}\sum_{j}e^{-\beta E_{j}} \end{align}
(4)
\begin{align} =\sum\frac{\partial}{\partial V}e^{-\beta E_{j}} \end{align}
(5)
\begin{align} =\sum-\beta\frac{\partial E_{j}}{\partial V}e^{-\beta E_{j}} \end{align}

Eq. (2.0.1) then can be written in term of Q:

(6)
\begin{align} \bar{P}=\frac{1}{\beta}\frac{1}{Q}(\frac{\partial Q}{\partial V})_{N,T} \end{align}
(7)
\begin{align} =kT(\frac{\partial\ln(Q)}{\partial V})_{N,T} \end{align}
(8)
\begin{align} \frac{d\ln Q}{dV}=\frac{1}{Q}\frac{dQ}{dV} \end{align}
Problem 2-10 by Ebtehal Alrewaily Ebtehal Alrewaily , 17 Dec 2013 02:22
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