热力学统计物理试题(D卷)_热力学统计物理试卷

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热力学·统计物理试题（D卷）

适用于2002级本科物理学专业

(2004-2005学年度第一学期)

1.(10 points)Consider(U)=0.Show(U)=0

VT

2.(10 points)Consider C 0and(vpVpT)T0.Show Cp0

3.(20 points)Consider a chemical reaction follows that

2N232H2NH30 Show isopiestic equilibrium constant

Kp2742

21p

If the reaction follows that

N23H22NH30

calculate isopiestic equilibrium constant again.4.(20 points)Use Maxwell velocity distribution law to show the fluctuation of velocity and mean translational energy respectively follows that(v)

()

2kTm(38)232(kT)2

e

0x2xdx2432, e0x2xdx43852

5.(20 points)The electronic density of a metal is 5.91028/m.Calculate the Fermi energy, 3

Fermi velocity and degenerate preure of this free electronic gas at temperature T=0K.6.(20 points)Use canonical ensemble distribution to calculate the internal energy E, free energy F, chemical potential μ, and preure p of the ideal gas.附简答：

1.(10 points)Solution

(UV（）T=T()T =

pT)V-p;（UV)T=0;pT（pT)V(4 points)

UV

(U,T)(V,T))T（pV

=

(U,T)(p,T)(p,T)(V,T)

=0=（Up)T(4 points)

∵V

（p)T≠0;（Up)T=0(2 points)(10 points)Solution

CpCV

pVTTVT

p

(4 points)

pVTVp

T

=-1(3 points)

VpT

pV

Cp CVT

VTTppV

 C 0)T0, thusCpV  0andCv, Cp0(4 points)

Because（3.(20 points)SolutionAume NH3 with n0 mol, decomposed n0ε mol，the spare part(1-ε)n0 mol,making N2 with

1n0

n0 mol and H2 with

n0 mol.Total number is(1+ε)n0 mol.xN

n0

(1)n0

22;xH2;x NH3;(1)n0(1)n0(1)n0

Isopiestic equilibrium constant

(5 points)

K

p

1

(xN2)2(xH2)2(xNH3)

274



p2



1





1

p

(5 points)

Ifthe reaction follows that

N23H22NH

0

aume NH3 with 2n0 mol, decomposed 2n0ε mol，the spare part 2(1-ε)n0 mol, making N2 withn0 mol and H2 with3n0 mol.Total number is 2(1+ε)n0 mol.xN

n02(1)n0

;xH2

3n02(1)n0

;x NH3

2(1)n02(1)n0

;(5 points)

Isopiestic equilibrium constant

K

p

(xN2)(xH2)(xNH3)

132

p

132





2(1)



3

2(1)



(1)(1)

22

p



27

16(1)

p

(5 points)

4.(20 points)Solution

(v)2v22(5 points)

In the scope of V and dpx dpy dpz , the molecule number follows that

Vh

--

12mkT

(pxpypz)

e

dpxdpydpz

f(vx, vy,vz)dvxdvydvzmn

2kT

e

m2kT

(vxvyvz)

222

dvxdvydvz

m

4n

2kT

3e

m2kT

v

vdv

(5 points)

(v)v2



kTm

(3

)

D()d

2Vh

(2m)

3

d

(5 points)





154

(kT),22

32

(kT)

()



2

(kT)

(5 points)5.(20 points)Solution

The mean number of electron at one level ε is

when temperature T=0K: f=1ε

f=0ε>μ(0)(5 points)

4Vh

f

e



kT

1

(2m)



(0)



212

d N



(0)3

2m



NV

5.6eV

(5 points)

(0)p(0)2m

vF1.410m.s



p(0)3



NV

1

(5 points)

2.110

Pa

(5 points)

6.(20 points)Solution

(4 points)

3N

E



i1

pi

2m

1E

Z

N!h

3N

e

dq1dq3Ndp1dp3N

3N

ZV

N

2m2

N!h2

The free energy

lnZ(T, V, N)=-NkT(1lnV2mkT32F=--kT2

)Nh

pFV



NkTT,N

V

S

FV2mkT32T

Nk(ln5

V,N



Nh2

)2F

Nk(lnV2mkT325

 N 2

)V ,N Nh2

(4 points)

(4 points)(4 points)

(4 points)