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Starting with Ec 7-12

(1)
\begin{align} Q=\frac{1}{N\!}\left[\frac{2m\pi kT}{h^2}\right]^{3N/2}Z_{n} \end{align}

All we have to do is to compute the classical configurational integral.

(2)
\begin{align} Z_{N}= \int e^{-\beta(\sum mgz_{i})}dx_{1}dx_{2}...dz_{1}dz_{2}...dz_{N}= A^{N}[\int_{0}^{z}e^{-\beta mgz}dz^{'} ]^N \end{align}
(3)
\begin{align} Z_{N}=A^{N}\frac{1}{(\beta mg)^{N}}(e^{-\beta mgz}-1) \end{align}

where A is the area of the cylinder
Now, we know that

(4)
\begin{align} p=kT\frac{\partial Ln Q}{\partial V} \end{align}

then

(5)
\begin{align} p=\frac{Nmg}{A}\frac{e^{-\beta mgz}}{1-e^{-\beta mgz}} \end{align}

Because

(6)
\begin{align} e^{-\beta mgz}<1 \end{align}
(7)
\begin{align} p\sim \frac{Nmg}{A}e^{-\beta mgz}(1+e^{-\beta mgz}+...) \end{align}

Finally

(8)
\begin{align} p(z)\sim p(0)e^{-\beta mgz} \end{align}
solution to exercise 7.4 by Dr Frank-FBDr Frank-FB, 21 Apr 2022 21:48
Problem 2-15
EzkerMSEzkerMS 12 Dec 2021 01:36
in discussion Chapter 2 / Solutions » Problem 2-15

Teniendo la funcion de partición de un cristal, la cual tiene al forma:

(1)
\begin{align} Q(N,V,T) = \left(\dfrac{e^{-hv/2kT}}{1-e^{-hv/kT}}\right)^{3N} \cdot e^{u_o/kT} \end{align}

Donde $hv/k=\Theta_E$ es una constante característica de un cristal, y $U_o$es la energía de sublimación de un cristal. Calcular el calor específico y muestre que a altas temperaturas uno obtiene la ley Dulong y Petit, esto es $C_v \rightarrow 3Nk$.

Primero, recordemos el calor específico tiene la forma:

(2)
\begin{align} C_v = \dfrac{\partial E}{\partial T} \end{align}

Y sabemos que la expresión de la energia promedio tiene la forma:

(3)
\begin{align} E=kT^2\left(\dfrac{\partial ln (Q)}{\partial T}\right)_{N, V} \end{align}

¨Por lo que, del logaritmo de Q, el cual es:

(4)
\begin{equation} Ln(Q)= 3N Ln[e^{-hv/2kT}]-3N Ln[1-e^{-hv/kT}]+Ln[e^{u_o/kT}] \end{equation}
(5)
\begin{equation} Ln(Q)= 3N Ln[e^{-hv/2kT}]-3N Ln[1-e^{-hv/kT}]+u_o/kT \end{equation}

Y la derivada de esto toma la forma:

(6)
\begin{align} \left(\dfrac{\partial ln (Q)}{\partial T}\right)_{N, V} =\dfrac{1}{KT^2} \left[(\dfrac{3Nhv}{2}-v_o)+3Nhv\left[ \dfrac{1}{1-e^{-hv/kT}} \right] \right] \end{align}

Multiplicando por $T^2k$ tenemos:

(7)
\begin{align} =\left[(\dfrac{3Nhv}{2}-v_o)+3Nhv\left[ \dfrac{1}{1-e^{-hv/kT}} \right] \right] \end{align}

Por lo que, derivando la expresión anterior con respecto a T

(8)
\begin{align} =\dfrac{\partial}{\partial T} \left[(\dfrac{3Nhv}{2}-v_o)+3Nhv\left[ \dfrac{1}{1-e^{-hv/kT}} \right] \right] \end{align}
(9)
\begin{align} =\dfrac{\partial}{\partial T} \left[3Nhv\left[ \dfrac{1}{1-e^{-hv/kT}} \right] \right] \end{align}
(10)
\begin{align} = \dfrac{3Nh^2 v^2e^{hv/kT} }{[(e^{-hv/kT}-1)T]^2 k} \end{align}

Y al hacer el límite de esta expresión a $T \rightarrow \infty$ se obtiene

(11)
\begin{align} C_v \rightarrow 3Nk \end{align}
Problem 2-15 by EzkerMSEzkerMS, 12 Dec 2021 01:36
Problem 2-16
EzkerMSEzkerMS 12 Dec 2021 01:30
in discussion Chapter 2 / Solutions » Problem 2-16

Teniendo la función de partición de un cristal, la cual tiene al forma:

(1)
\begin{align} Q(N,V,T, \xi) = \dfrac{[q(V,T,\xi)]^N}{N!} \end{align}

donde

(2)
\begin{align} q(V,T,\xi)= V(2 \pi m k T / h^2)^{3/2} (8 \pi ^2 I k T / h^2 ) (\dfrac{e^{-hv/2kT}}{1-e^{-hv/kT}}) (\dfrac{kT}{\mu \xi}) sinh(\dfrac{\mu \xi}{kT}) \end{align}

Donde I es el momento de inercia de la molécula; v es su frecuencia vibracional fundamental; y $\mu$ es su momento dipolar. Usando la función de partición y utilizando la relación termodinámica,

(3)
\begin{align} dA= -SdT-pdV-Md \xi \end{align}

donde $M=N \bar{\mu}$, donde $\bar{\mu}$ es el momento dipolar promedio de una molécula en la dirección del campo externo $\xi$, muestre que

(4)
\begin{align} \bar{\mu}=\mu \left[ coth \left(\dfrac{\mu \xi}{kT} \right) -\dfrac{kT}{\mu \xi}\right] \end{align}

Grafique los resultados desde $\xi =0$ hasta $\xi= \infty

Primero obtengamos dA, el cual tomará la forma:

(5)
\begin{align} dA= \dfrac{\partial A}{\partial V}dV +\dfrac{\partial A}{\partial T}dT+\dfrac{\partial A}{\partial \xi}d \xi \end{align}

Pero solo nos interesa igualar el término $\dfrac{\partial A}{\partial \xi}d \xi$, pues al igualarlo con la expresión dada del diferencial de dA, obtendremos que

(6)
\begin{align} \dfrac{\partial A}{\partial \xi} =-M \end{align}

Por lo que

(7)
\begin{align} \bar{\mu}=-\dfrac{\partial A}{\partial \xi} \cdot \dfrac{1}{N} \end{align}

Ahora A, tiene la forma:

(8)
\begin{equation} A=-kT Ln[Q] \end{equation}
(9)
\begin{align} A=-kT Ln[\dfrac{q^N}{N!}] \end{align}
(10)
\begin{equation} A=-kT NLn[q]+kTLn[N!] \end{equation}

Por lo que su derivada con respecto a $\xi$ tendrá la forma

(11)
\begin{align} \dfrac{\partial A}{\partial \xi}=-\dfrac{\partial kT NLn[q] }{\partial \xi} \end{align}

Ahora, como es $Ln [q]$ , y solo sobrevivirán aquellos términos dependientes de $\xi$, los cuales son

(12)
\begin{align} =Ln[\dfrac{kT}{\mu \xi}]+ Ln[ sinh(\dfrac{\mu \xi}{kT})] \end{align}

Entonces tendremos la expresión

(13)
\begin{align} \dfrac{\partial A}{\partial \xi}=-kTN\dfrac{\partial }{\partial \xi} [Ln[\dfrac{kT}{\mu \xi}]+ Ln[ sinh(\dfrac{\mu \xi}{kT})]] \end{align}
(14)
\begin{align} \dfrac{\partial A}{\partial \xi}=-kTN \left[\dfrac{-1}{\xi}+coth[\dfrac{\mu \xi}{k T}][\dfrac{mu}{kT}]\right] \end{align}

Multiplicando por -1/N para obtener

(15)
\begin{align} \bar{\mu}=-\dfrac{\partial A}{\partial \xi} \cdot \dfrac{1}{N}= \mu \left[ coth[\dfrac{\mu \xi}{k T}] - \dfrac{kT}{\mu \xi}\right] \end{align}
(16)
\begin{align} \bar{\mu}= \mu \left[ coth[\dfrac{\mu \xi}{k T}] - \dfrac{kT}{\mu \xi}\right] \end{align}

And looking at the graph, we can tell $\xi \rightarrow \infty$, $\bar{\mu} \rightarrow \mu$

Problem 2-16 by EzkerMSEzkerMS, 12 Dec 2021 01:30
11 -4
Richard496Richard496 26 Oct 2020 00:04
in discussion Chapter 11 / Solutions » Problem 2

determine the various thermodynamics properties of an Einstein crystal

11 -4 by Richard496Richard496, 26 Oct 2020 00:04

i wonder if this site is operating or not

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
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