thermaldynamics

第四章作业

--------------------------秦大粤【2014301020039】物基一班

problem4.1

(a)
$\frac{1}{5}$ 因为总的spin不为0
(b)
1/5
(c)
1/25

problem4.2

(a)
$\uparrow\uparrow\uparrow\downarrow,\uparrow\uparrow\downarrow\uparrow,\uparrow\downarrow\uparrow\uparrow,\downarrow\uparrow\uparrow\uparrow$,the probability that a particular spin is up is 3/4，and the probability that a particular spin is down is 1/4
(b)
the number of accessible microstate is :$C_{9}^{4}=126$,the probability that s particular spin is up is 5/9,and the prabability that a particular spin is down is 4/9

peoblem4.3

$(1/2)^5\times2$

problem4.4

(a)
$\Omega=10$
(b)
$\Omega=\frac{9!}{6!3!}=84$
(c)
3/10
(d)
1/10

problem4.5

$$S_A=\sqrt{\overline{\sum(E_{Ai}-\overline{E_A})^2}}=1.68$$

problem4.6

(a)

(b)
the probability that energy transferred from A to B:

$$P=\frac{\Omega_{tot(E_A<10)}}{\Omega_{tot}}=$$
the probability that rnergy transferred from B to A: $$P=\frac{\Omega_{tot(E_A>10)}}{\Omega_{tot}}=$$
(c)
$$\Omega_{tot}^\prime=\frac{(E+N-1)!}{E!(N-1)!}=\frac{(12+8-1)!}{12!(8-1)!}=50388$$
$$\Omega_{tot(A)}=286,\Omega_{tot(B)}=10\\\Delta S=ln(\Omega_{tot}^\prime)-ln(\Omega_{tot(A)})-ln(\Omega_{tot(B)}\approx2.87$$
(d)$E_A=6,E_A(MaxP)=6,\delta_A=2.58,\delta_b=2.58,\frac{\delta_A}{\overline{E_A}}=0.43,$相对波动较小
(e)$
(f)

problem4.7

(a)

$$S_A=kln(\Omega_A),\Omega=\frac{(E_A+N_A-1)!}{E_A!(N_A-1)!}\\so\space is\space an\space increasing\space function\space of\space E_A$$
$$S_A=kln(\Omega_b),\Omega=\frac{(E-E_B+N_B-1)!}{E_B!(N_B-1)!}\\is\space an\space increasing \space function\space of\space E_B$$
(b)
$$\frac{1}{T}正比于\frac{dS_A}{dE_A}$$
while $E_A$=3~4 the slopes are aqual
$S_{tot}$ reach the max value

problem4.8

(a)
As the number of particle increasing,the width decrease
(b)
Yes
(c)
$P_n正比于\alpha e^{-\beta\epsilon_n}$

problem4.9

(c)

$$P_s(E_s)=\frac{\Omega_B(E_tot-E_s)}{\Omega_{tot}}$$
(d)
$$P_A(E_A)=\frac{\Omega_A(E_A)\Omega_B(E_{tot}-E_A)}{\Omega_{tot}}\neq P_s(E_s)$$
from the equation of (c),we can see that $P_S(E_S)$is an decreasing funtion of $E_S$

problem4.10

$\Omega(E)=\frac{(E+N-1)!}{E!(N-1)!}$is an rapidly increasing function of E for fixed fixed N and also a rapidly increasing function of N for fixed E.

problem4.11

n=12

problem4.12

$$R_{min}=10$$

problem4.13

larger the R is,the less the difference is

problem4.14

$$g(E)\Delta E=\frac{d\Gamma(E)}{dE}\Delta E\approx 3.4*10^{22}$$

problem4.15

$$\Gamma(E,V,N)=\frac{1}{N!}(\frac{4\pi}{3})^N(\frac{V}{h^3})^N(2mE^{3/2})^N$$

problem4.16

$$\frac{\partial ln\Gamma}{\partial E}=\frac{1}{\Gamma}\frac{\partial\Gamma}{\partial E}=\frac{1}{\Gamma}g$$
thus:
$$g=\Gamma\frac{\partial ln\Gamma}{\partial E}$$
they are rapid increasing function of E,V,N

problem4.17

when it comes to be ideal gas:$E=\frac{3}{2}kT$:

$$\Delta S=S(E_2,V_2,N_2)-S(E_1,V_1,N_1)=\frac{3}{2}NKln\frac{T_2}{T_1}+NKln\frac{V_2}{V_1}$$

problem4.18

$$S_1=kln\Gamma(E)=NK[ln\frac{V}{N}+\frac{3}{2}ln\frac{mE}{3N\pi\hbar^2+\frac{5}{2}}]\\S_2=kln\Gamma(E)=NK[ln\frac{V}{N}+\frac{3}{2}ln\frac{mE}{3N\pi\hbar^2+\frac{5}{2}}]+kln\frac{\Delta E}{E}$$

problem4.19

$$S=Nk[ln\frac{V}{N}+\frac{3}{2}ln\frac{mKT}{2\pi \hbar^2}+\frac{5}{2}$$

problem4.20

$$\frac{\mu}{T}=-(\frac{\partial S}{\partial N})_{E,V}=-kln\frac{V}{N}-\frac{3}{2}kln\frac{mE}{3N\pi\hbar^2}$$
since$E=\frac{3}{2}kT$:
$$\mu=-\frac{2E}{3N}(ln\frac{V}{N}+\frac{3}{2}ln\frac{mE}{3\pi\hbar^2N}\\\mu=-kT(ln\frac{V}{N}+\frac{3}{2}ln\frac{mkT}{2\pi\hbar^2})$$

problem4.21

$$\frac{n}{N}=\frac{1}{1+e^{-\mu B/kT}}\\E=-(2n-N)\mu B=-N\mu B\frac{1-e^{-2\mu B/kT}}{1+e^{-2\mu B/kT}}=-N\mu B\space tanh\space( \mu B\beta)$$

problem4.22

$$E=(Q+\frac{1}{2}N)\hbar\omega\\ S=kln(\Omega)=kln\frac{(Q+N-1)!}{Q!(n-1)!}\\$$
thus:
$$\frac{1}{T}=(\frac{\partial S}{\partial E})_{V,N}=\frac{\partial S}{\partial Q}\frac{\partial Q}{\partial E}=kln\frac{Q+N-1}{Q}\frac{1}{\hbar\omega}\approx\frac{k}{\hbar\omega}ln(\frac{Q+N}{Q})\\ E=\frac{N\hbar\omega}{2}(\frac{e^{\hbar\omega/kT}+1}{e^{\hbar\omega/kT}-1})$$

problem4.23

the result that we obtained in table4.6 is an simple example of boltzmann distribution,where subsystem b is something like the heat bath.

problem4.24

T=300K时，trans占0.999999；T=1000K时，trans占0.985。