[Physics, Thermal] - Solutions to Statistical Mechanics Problems

Solutions Solutions to Statistical Statistical Mechanics Mechanics Problems Problems

April 30, 2001 1. 1 ca cal. l. = 4.18 4.1866 ×107 ergs = 4.186 J. Hence, 639 kcal = 639 × 4.186 kJ = 2675 kJ. 2. For an ideal gas where where pV  = N kB T  and E  = 32 N kB T  :

        C V  V  =

C  p =

∂Q ∂T 

∂E  ∂T 

∂E  ∂T 

=

 p

V 

3 = N kB 2 ∂V  ∂T 

+p

 p

 p

3 = N kB + N kB 2 C  p − C V  V  = N kB For an adiabatic QSP process involving an ideal gas, dQ = 0 and f ro rom E  =  = C V   + pdV  = 0 V  T and dE  + C V  (C  p − C V  V  dT  + (C  V  )TdV/V  = 0 dT/T  + (γ  (γ  − 1)dV/V  1)dV/V  = 0 V γ  1 T  = constant −

or pV γ  = constant 3. For adiabatic adiabatic expansion into vacuum, vacuum, E  is a constant. Hence T 2 = T 1 +

2an {1/V 2 − 1/V 1 } 3R

4. From the definition definition of latent heat: ∆E  = L − p∆  p∆V  4

= 4.06 × 10 − 1.013 × 3.02 × 105

2

−

= 3.75 × 104 J 5. (a) Work Work done done is 1 atm cm3 = 1.013 × 105 × 10

6J

−

7

−

(b) Work done is 73 × 10

= 0.1013 J.

J.

Thus in considering macroscopic volume changes of gases, the change in surface energy can be neglected. 6. The correct correct statement statementss are: (a) The entropy change in a reversible cyclic process is zero. (b) The entropy change in a reversible QSP process is

 

dQ/T . dQ/T .

7. Rate Rate of productio production n of heat heat is I 2 R. Thus in time dt, dt, dQ = I 2 Rdt and dS  = I 2 Rdt/T . Rdt/T . 8. (a) (a) In p −V  diagram, the adiabats obeying pV γ  = K  = constant are steeper than the isothermals where pV  = N kB T  and are thus curves 2 → 3 and 4 → 1. Use the equations of state connecting  pi , V i with each other.  p2 V 2γ  = p3 V 3γ   p1 V 1γ  = p4 V 4γ   p1 V 1 = p2 V 2 1

γ 

 p3 V 3 = p4 V 4

 

 

γ 

V 4 p1 p3 V 1 Hence = V 3  p2 p4 V 2 p1 V 2 p3 V 4 but = and =  p2 V 1  p4 V 3 thus

γ −1

    V 4 V 3

V 1 V 2

=

γ −1

or V 4 /V 3 = V 1 /V 2 (b) As the gas is ideal, T i =

pi V i NkB

(c) The work done on the gas is i+1

−

  i

 pdV  = −NkB T ln V i+1 /V i isothermal stage =

−K  1 γ  (V i+1 − V i1 1 − γ  −

γ 

−

) adiabatic stage

 

pi V i V i =− { γ  − 1 V i+1

γ −1

− 1}

(d) The energy change along the isothermal stages is zero and along the adiabatic ones is 3NkB (T i+1 − T i )/2. (e) The heat absorbed is found from the differences between the change in the internal energy and work done. This must be zero for adiabatic stages. (i) For isothermal stage 1 → 2, ∆E  = ∆Q − NkB T 1 ln V 2 /V 1 = 0. Hence, ∆Q = N kB T 1 ln V 2 /V 1 . (ii) For adiabatic stage 2 → 3, ∆Q = 0. (iii) For isothermal stage 3 → 4, ∆E  = ∆Q − N kB T 3 ln V 4 /V 3 = 0. Hence, ∆Q = NkB T 3 ln V 4 /V 3 . (iv) For adiabatic stage, 4 → 1, ∆Q = 0. (f) The total work done during cycle is, as ∆E  = 0, equal to ∆Q = N kB T 1 ln V 2 /V 1 + NkB T 3 ln V 4 /V 3 . The heat absorbed at the hot stage is the first of these terms namely NkB T 1 ln V 2 /V 1 . The efficiency is then ∆W  T 3 ln V 4 /V 3 = 1+ ∆Q(1 → 2) T 1 ln V 2 /V 1 = 1 − T 3 /T 1 as V 4 /V 3 = V 1 /V 2

η=

9. To show that H  is stationary for a system in equilibrium with constant N , T , p, differentiate H  and use the second law: H  = E + pV  dH  = dE + pdV  + V dp = V dp + T dS + µdN  from the second law. Thus at constant N,p,S , then dH  = 0 or H  stationary. The other relations are proved in the same way. 2

10. To determine E and S  from equation of state for one component gas, we use a Maxwell’s relation derived from S . T dS  = dE + pdV 

                  ∂S  ∂T 

V 

∂ 2 S  ∂V ∂T 

1 ∂E  = T  ∂T 

=

∂S  ∂V 

,

V 

1 ∂ 2 E  T  ∂V ∂T 

=

∂E  ∂V 

+

T 

p T 

1 ∂ 2 E  1 ∂p + T  ∂V ∂T  T  ∂T 

−

or p +

     

1 ∂E  = T  ∂V  T 

= T 

T 

1 ∂E  T 2 ∂V 

∂p ∂T 

−

T 

= T 

V 

V 

p T 2

∂S  ∂V 

T 

This gives a differential equation relating the change in E  changes with V  at fixed T . Apply this to Van der Waals gas: an2 ( p + 2 )(V  − nb) = nRT  V  ∂E  nRT   p + = ∂V  T  V  − nb

    ∂E  ∂V 

an2 = 2 V  T 

an2 3 E  = − + c(T ) → nRT as V   → ∞ V  2 3 an2 Hence E  = nRT  − . 2 V  Here we have used the fact that integration constant is independent of  V  and can only depend on T . The dependence on this is known for large V  where ideal behaviour is valid. Similarly, S  can be found:

    ∂S  ∂V 

=

T 

∂p ∂T 

= V 

nR V  − nb

Integrating S  = nR ln(V  − nb) − nR ln n + nc(T ) The last two terms are constants of integration and written like this to ensure that S is extensive. Now use

  ∂S  ∂T 

V 

 

1 ∂E  3 = = nR T  ∂T  V  2T  3 S  = nR ln T  + nd(V ) 2

Where d(V ) is a constant of integration wrt T . Comparing these expressions gives 3 S  = nR ln(V /n − b) + nR ln T  + nc 2 

Where c is independent of both V  and T . 3



11. dE  = T dS  − f d l F   = E  − TS....(1) dF  = −f dl − SdT 

                             ∂ 2 F  ∂l∂T 

Hence −

∂E  ∂l

Now

From (1)

∂ 2 E  = ∂l∂S 

∂T  ∂l

∂S  ∂l

= −f  + T 

T 

∂f  ∂S 

=

S 

∂f  ∂T 

=

l

∂f  ∂T 

=

l

= −f  + T 

T 

∂T  ∂S 

l

l

T 

∂f  ∂T 

l

∂f  ∂T 

T  c l l

1 ∂f  cl ∂T 

dl

=

dT  ∆T  = ≈ T  T 

Hence for isentropic small change

∂S  ∂l

=

l

12. G = E  − T S + pV  dG = dE  − T dS  − SdT  + pdV  + V dp = −SdT  + V dp + µdN  ∂G S  = − ∂T   p,N 

 

V  =

  ∂S  ∂p

  ∂G ∂p

Hence T,N 

=− T,N 

  ∂V  ∂T 

 p,N 

13. To show that the Joule-Thomson coefficient vanishes for an ideal gas. Recall, αJT  = −

V  (1 − αT ) C  p

whereas expansion coefficient for ideal gas is

 

1 ∂V  α= V  ∂T 

=

 p,N 

N kB = 1/T   pV 

Hence, αJT  vanishes. 14. Number of distinct arrangements of  n atoms occupying N  interstitial sites is Ω1 =

N ! n!(N  − n)!

which is also number of ways, Ω 2 , of arranging N  − n atoms on lattice sites. Hence entropy is



N ! S  = kB ln n!(N  − n)!



2

∼ 2kB (N ln N  − n ln n − (N  − n)ln(N  − n)) F  = E  − T S  = n + 2kB T (n ln n + (N  − n)ln(N  − n)) + const

4

Minimising F  wrt n gives /2kB T 

−

n = Ne when   kB T .

15. The probabilities of occupying the three energy levels are: /kB T 

−

z=e

1 1 + 2z + z 2 2z  p2 = 1 + 2z + z 2 z2  p3 = 1 + 2z + z 2 E  = N ( p1  + p2 2 + p3 3)  p1 =

= N ( + p2  + p3 2) 2z = N ( + ) 1+z ∂E  2N 2 z = ∂T  kB T 2 (1 + z)2

 

16. The gases have same numbers of particles and and final temperature is same as initial one as no energy is added to the system. Hence, as each gas occupies a volume 2V  in the final state, then from the Sackur-Tetrode formula: 3 S (initial) = 2N kB ln T  + 2 V  2NkB ln + 2Nc N  3 S (final) = 2N kB ln T  2 2V  +2N kB ln + 2Nc N  ∆S  = 2N kB ln 2 



This is greater than zero and hence the gases irreversibly expand into the larger volume. If  gases are identical, then we have 2N  particles occupying volume 2V  which has final entropy: 3 S (final) = 2N kB ln T  + 2 2V  2NkB ln + 2N c 2N  = S (initial) 

In this case, no extra entropy arises from ‘mixture’. 17. Number of arrangements of  n Ge atoms around C is arrangements are:

4! (4−n)!n! .

(1 − x)4  p0 = Z  3 /k T  4x(1 − x) e  p1 = Z  2 6x (1 − x)2 e 2/k T   p2 = Z  −

−

5

B

B

Probabilities of realising these

4x3 (1 − x)e  p3 = Z  x4 e  p4 =

3/kB T 

−

4/kB T 

−

Z 

Here, Z  = (1 − x)4 + 4x(1 − x)3 e

/kB T 

−

+ 6x2 (1 − x)2 e

2/kB T 

−

+ 4x3 (1 − x)e 3/k T  + x4 e 4/k T  = (x + (1 − x)e /k T )4 −

B

−

−

At 700 C, kB T  = 8 × 973 × 10 5 = .07784 eV. Hence e are: .3763, .4166, .1729, .0319, .0022. ◦

−

/kB T 

−

B

B

= .2767. Thus Z  = 2.6571 and pi

This assumes equilibrium is achieved. In practice this may not be the case as the diffusion energies of the atoms are so large that initial distributions are frozen in. hω/kB T  ¯

−

18. Ratio is approximately e

= 0.03.

19. N ! n!(N  − n)! ln Ω = N ln N  − n ln n − (N  − n)ln(N  − n) ln Ω(2N, 2n) = 2N ln 2N  − 2n ln 2n Ω=

−2(N  − n)ln2(N  − n) = 2 ln Ω + 2N ln 2 − 2n ln 2 − 2(N  − n) l n 2 = 2ln Ω This shows S  = kB ln Ω is extensive function. 20. Let n1 , n2 be number of right and left pointing links and l = L/a,z = l/n. L = (n1 − n2 )a 1 1 n = n1 + n2 n1 = (l + n) n2 = (n − l) 2 2 n! W  = n1 !n2 ! S  = kB ln W  = kB (n ln n − n1 ln n1 − n2 ln n2 )

= kB (n ln n −

n ((1 + z) ln((1 + z)n/2) + (1 − z) ln((1 − z)n/2)) 2 n ∼ kB (n ln 2 − {(1 + z)z − (1 − z)z + ..}) 2 = nkB (ln2 − z 2 /2),

where we have Taylor expanded about z = 0. Hence F  = nkB T (z 2 /2 − ln2)

 

τ  = −

∂F  ∂L

= kB T 

L na2

21. Energy levels are E k = kµB with k = −1, 0, 1. The statistical weight is Ω= 6

N ! , n1 !n2 !n3 !

F  =



nk E k − kB T  ln Ω

k

=



nk E k − kB T {N ln N  −

k

 

nk ln nk }.

k

M inimise f  = F  − α(

nk − N )

k

with respect to α and n k

   ∂f  ∂α

  ∂f  ∂n k

=

nk − N  = 0,

k

= E k + kB T ln nk + kB T  − α = 0, E k /kB T 

−

nk = N e where Z  =

E k = kµB,

k = 0,



/Z,

E k /kB T 

−

e

+ 1,

−1 x = µB/kB T 

Z  = 1 + 2 cosh x 2N µ sinh x M  = 1 + 2 cosh x 2 → 2N µ B/3kB T  as T  → ∞

→ Nµ as T  → 0 To align spins, need µB ∼ kB T . Hence B ∼ 14 Tesla (140 kG).

7

.