3.20 Materials at Equilibrium (SMA 5111): Solutions to Problem Set 5 (2003)

Problem Set 5 Solutions - McQuarrie Problems 3.20 MIT Dr. Anton Van Der Ven Fall 2001 Fall 2003 Problem 3-4 We have to derive the thermodynamic properties of an ideal monatomic gas from the following:
= eq

3 2mkT 2 
= e and q = V h2
is the partition function for the grand canonical ensemble, where T, V,  are fixed. The characteristic potential for the grand canonical ensemble is the ’’grand canonical potential’’  = E
T S
N =
pV (since E = T S
pV + N ) d =
SdT
pdV
N d The thermodynamic properties for the grand canonical ensemble are:
 ∂ S=
∂T N,V
 ∂ p=
∂V T ,
 ∂ N =
∂ T ,V The grand canonical potential is related to
according to  =
kT ln
but
= e q
so,
 =
kT  q
=
kT  2mkT h2  32 V e  (see table 3-1) =
(kT ) 5 2
2m h2  32 V  exp    kT Starting with N, N =
 ∂ ∂  T ,V =  2mkT  32 h2 V  exp   kT  and p p=
 ∂ ∂V  T , = kT  2mkT  32 h2 exp   kT  Putting those together we can get the ideal gas equation of state, namely N p = kT V Now S,

3
3
 3 5 5 2m 2 2m 2
 ∂  2 + (kT ) = k  (kT ) 2 S=
V e V e 2 ∂T N,V h2 h2 kT 2 1 ((3-8-1)) 3
3 2mkT 2  2mkT 2  Ve
V e h2 T h2 
 
3 5 2mkT 2   S= k
Ve 2 h2 T 5 S = k 2

S= 5  k
2 T  N N ((3-8-2)) But from (3-8-1) we can get 
32 2mkT e = NV h2 
3  2mkT 2 V  =
kT  ln h2 N
1 
Putting this into (3-8-2) we get 
 
 3   3   5 5 2mkT 2 V 2mkT 2 V S= k + k ln N = N k ln e 2 + ln 2 h2 N h2 N S = N k ln  2mkT h2  32 V N 5  e2  This is the same expression as that obtained in the canonical ensemble (see Chapter 5). This is due to the equivalence of ensembles when N is very large. Problem 3-10 We are dealing with the isothermal-isobaric ensemble this time, with the partition function for an ideal monatomic gas given to us in the problem as  N 3 5 (2m) 2 (kT ) 2 = ph3  The isothermal-isobaric is for fixed (N, T, P ).  The characteristic potential for this ensemble is the Gibbs free energy G = E
T S + pV dG =
SdT + V dP + dN  The thermodynamic properties are:
 ∂G ∂T p,N
 ∂G V = ∂p T ,N
 ∂G = ∂N p,T S=
 G is related to the isothermal-isobaric partition function  according to (see Table 3-1):   3 5 (2m) 2 (kT ) 2 G =
kT ln  =
N kT ln ph3 2   3 5 3 2 G =
N kT ln (2m) + ln (kT )
ln h
ln p 2 Starting with V, V =  ∂G ∂p  T ,N = NkT p which is the ideal gas equation of state. Now for  :   3 5
 ∂G (2m) 2 (kT ) 2 = = kT ln ∂N p,T ph3 
 3 2mkT 2  =
kT ln kT +kT ln p h2 o And now for S,  3 5 (2m) 2 (kT ) 2 5 S=
+ Nk = N k ln 3 ph 2 p,N 
 3  5 2mkT 2 V S = N k ln + N k ln e 2 2 h N
∂G ∂T   S = N k ln  2mkT h2  32 V N 5  e2  These are the same expressions as obtained in the grand canonical ensemble (Problem 3-8) and in the canonical ensemble (see chapter 5). This is due to the equivalence of ensembles in the thermodynamic limit, i.e N is very large such that fluctuations are negligible. Problem 3-12 When looking at fluctuations, we derived   2  E
E 1 P (E) ≈ √ exp
2kT 2 Cv 2kT 2 Cv For an ideal monatomic gas (derived explicitly in chapter 5) 3 E = N kT 2 ∂E 3 Cv = = Nk ∂T 2 We are now asked what is the probability that the N -particle system will sample an energy that differs by 10
4 % from the average energy, E = 32 N kT ? (We can let N = NA = 6.022  1023M )

4   3 10 E = 10
6 E = 10
6 N kT E = 100 2 
6 2 9 2 2 2 2 10 N k T (E) 3 
6 2 4 = = 10 N 3 2 2 2kT Cv 4 2kT  2 N k So now we can go back to P (E ),   1 3 
6 2 √ N P (E) ≈ exp
10 4 kT 3N 3   1 √ exp
4.5  1011 kT 3N This is an extremely small probability which validates our earlier assumptions. P (E) ≈ Problem 3-18 Derive an expression for the fluctuation of the pressure in the canonical ensemble.. We know the pressure in state i is
 ∂Ei pi =
∂V and by definition     ∂Ei  i
∂V exp
E kT   p= i  i exp
E kT i 2 Fluctuation is defined as  p2 = (p
p) = p2
p2 . Using the methods developed in class: Step 1: Multiply both sides by the partition function   
Ei pQ = pi exp kT i Step 2: Get derivative with respect to mechanical variable’s conjugate.   
Ei ∂ ∂ pi exp (pQ) = ∂V ∂V kT i    ∂Q ∂p ∂
Ei p pi exp +Q = ∂V ∂V ∂V kT i        ∂ ∂p ∂
Ei
Ei p pi exp exp +Q = ∂V kT ∂V ∂V kT i i   
       
    ∂pi  ∂Ei ∂p ∂Ei
Ei 1
Ei
Ei 1 p pi exp 
 
 exp +Q = exp + kT ∂V kT ∂V ∂V kT kT ∂V kT i i i Step 3: Divide through by the partition function    ∂E  1      
E       ∂pi i i i i i pi exp
E p exp
E 
exp 
∂E   ∂p ∂V kT kT ∂V kT ∂V kT Q ∂V i i i + = + Q Q Q Q
 2 p ∂p ∂pi p2 + = + kT ∂V ∂V kT Rearranging a little... 

 ∂p ∂p i  p2 = p2
p2 = kT
∂V ∂V  ∂pi ∂V 1 kT   has no immediate macroscopic interpretation, it must be calculate in any specific case and depends  2  i on the particular spectrum of ∂∂VE2i . This conclusion holds for all generalized forces in the form Ai = ∂E ∂a , where a is an extensive displacement conjugate to A. Hence, we cannot make an unqualified assertion that fluctuations in all kinds of external forces will be small. Note: 4 Compare this with fluctuations in extensive quantities such as E, H, or N which can be expressed in terms of thermodynamic response variables suchs as heat capacities or compressibilities. Specific calculations of the fluctuations in p of a perfect gas by Fowler is estimated as (p
p2 ) ≈ 5  10
12 p2 for a cubic centimeter of gas under standard conditions. This is approximately molecules in the gas. 1 2 n3 , where n is the number of (Source: The Principles of Statistical Mechanics, Richard C Tolman, Oxford University Press, first edition 1938) Problem 3-24 2 Show that H 2
H = kT 2 Cp in an N, p, T ensemble. N, P, T fixed means we are working in the isothermal-isobaric ensemble..  The partition function in this ensemble is  = e
Ej
pV V j Where the Ej ’s are the energies of the system when it has volume V. We also remember that H = E + pV. Using the methods developed in class: Step 1: Multiply both sides by the partition function  (Ej + pV ) e
Ej
pV H = V,j Step 2: Get the temperature derivative at constant (N, P ) (The conjugate variable to H in this case)  
  1  1 ∂H 2 +H 2 (Ej + pV ) e
Ej
pV  = (Ej + pV ) e
Ej
pV ∂T N,P kT kT 2 V,j V,j Step 3: Divide through by the partition function   

Ej
pV 2
Ej
pV pV e (E + ) j V,j ∂H 1 1 V,j (Ej + pV ) e + H = kT 2 ∂T N,P kT 2   H2 H or 2 H 2
H = kT 2  but we know ∂H ∂T  N,P
∂H ∂T  = Cp . So, 2 H 2
H = kT 2 Cp 5 N,P Problem 3-26  Show that ∂ ∂N  2 V,T V =
N 2  ∂p ∂V  .. N,T From Gibbs-Duhem we have SdT
V dp + N d = 0 At constant T , we then get
We can use the chain rule and get:
d dp  d dp
=  = d dN V N 
V,T dN dp  V,T Using partial derivative manipulation


 dN dN dV =
dp V,T dV P,T dp N,T  dV  V But for a single component system dN p,T = the molar volume = N and we get

 dN N dV =
dp V,T V dp N,T Putting this all together we get
d dN  V,T V2 =
2 N
dp dV  N,T Problem 4-2 Show that 6N V  h2 12mkT  32 given in table 4-1 is very large for electrons in metals at T = 300K.  Take Na-metal having the following properties - stable in the bcc crystal structure with lattice constant a = 4.23  10
10m - two Na atoms per bcc unit cell - number of valance electrons per Na atom = 1 - valance electrons in Na can be considered nearly free So we can get the following values to substitute into the original equations: N 2 = 3 V (4.23  10
10 ) h = 6.6262  10
34 J  s J k = 1.3807  10
23 K me = 9.1095  10
31 kg Putting those all together we get 6N V
h2 12mkT  32 = 1524 >> 1 Therefore Boltzmann statistics cannot be applied to electrons in metals. Must use Fermi-Dirac statistics. 6 Problem 4-6 S= ∞    xn1 1 xn2 2 N=0 {nj } with n1 and n2 = 0, 1, and 2. The Let’s consider   {nj }    means {nj } n1 n2 with the restriction that n1 + n2 = N. x1n1 xn2 2 for several values of N. N = 0 → possible combinations of n1 and n2 are 0 and 0.   =⇒ =1 {nj } N = 1 → possible combinations of n1 and n2 are n1 1 0   =⇒ n2 0 1 N 1 1 x1n1 x2n2 = x1 + x2 {nj } N = 2 → possible combinations of n1 and n2 are n1 0 1 2 =⇒   n2 2 1 0 N 2 2 2 xn1 1 xn2 2 = x22 + x1 x2 + x21 {nj } N = 3 → possible combinations of n1 and n2 are n1 1 2 =⇒   n2 2 1 N 3 3 xn1 1 xn2 2 = x1 x22 + x21 x2 {nj } N = 4 → possible combinations of n1 and n2 are n1 2 =⇒   n2 2 N 4 x1n1 x2n2 = x21 x22 {nj } For N > 4 →   {nj } x1n1 x2n2 = 0 because n1 + n2 � 4 Putting everything together we get   x1n1 x2n2 = 1 + x1 + x2 + x22 + x1 x2 + x21 + x1 x22 + x12 x2 + x21 x22 N {nj } 7 Now lets consider S= 2    1 + xk + xk2 k=1    S = 1 + x1 + x21 1 + x2 + x22 = 1 + x1 + x2 + x22 + x1 x2 + x21 + x1 x22 + x21 x2 + x21 x22 This expression contains the exact same terms as that obtained with    N {nj } xn1 1 x2n2 . Problem 4-8 We need to show this ( remember the upper (lower) signs is for Fermi-Dirac (Bose-Einstein) : S =
k  [nj ln nj  (1 nj ) ln (1 nj )] j Start with the partition function and go from there. Becuase of the equivalence of ensembles in the thermodynamic limit, we can calculate the entropy using the ensemble that offers the most mathematical convenience. For Fermions   ∂ ln
  or Bosons, this is the grand canonical ensemble. S = k ln
+ kT ∂T V,
=  j ln
=  1 +
e   1 � εj kT  � εj  ln 1 +
e kT j ∂ ln
= ∂T  � εj 
e kT ∂ ∂T 1 
e j � εj kT =   εj
 kT 2  exp 
(εj
) kT  � εj j    � εj  S=k  ln 1 
e kT +  j 1 
e kT   ε

e
εj jkT  1 
e
εj  To make this a little easier to manipulate we can write this shorthand by making the following substitutions. Let: u = 1 
e
εj and v =
e
εj u = 1  v and u v nj = u So we now have S=k  { ln u
nj ln v} = k  j S=k  { ln u
nj ln nj
nj ln u} j {(1
nj ) ln u
nj ln nj } = k j Remembering that u v=1 
(1
nj ) ln j v=1 S =k  j S =
k {
(1
nj ) ln (1  j {nj ln nj  (1 8
  1
nj ln nj u nj )
nj ln nj } nj ) ln (1 nj )} Problem 4-12 N-distinguishable independent particles, each of which can be in state +εo or -εo . N+ = number of particles with energy +εo N
= number of particles with energy
εo with N+ + N
= N The total energy is given as: E = N+ εo
N
εo = 2N+ εo
N εo   We now have to evaluate the partition function Q by summing exp
E over levels and compare it to the result kT Q = q N ..  We know
 
Ej Q= exp kT j where j labels a state in which the system can reside.  Instead of summing over the states that the system can be in, we can also sum over the possible energy levels, making sure we take account of the degeneracy of each energy level, i.e.
 
E Q= (E) exp kT E where (E) is the number of states with energy E.  For this system we have already stated the allowed energy levels are E = 2N+ εo
N εo where N+ can vary from 0 → N.  For each allowed energy level E, there are (E) possible states compatible with this energy. Since the N particles are distinguishable, N! N! (E) = = N+ !N
! N+ (N
N+ )! This represents the number of ways that N+ particles out of N can be in the + state.  So now we can write   N! Q= (E) exp (
E ) =⇒ exp [
 (2N+ εo
N εo )] Since E is uniquely a function of N+ N+ (N
N+ )! E N+ This can be rewritten in a nicer form if we remember the binomial expansion n  n!  x n 1 (1 + x)n = n !(n
n1 )! n =0 1 1 So we have  Q = eNεo   N+
 2εo if we let x = e or  
2ε N+ N! o  e N+ (N
N+ )! and rewrite using the binomial expansion we have N  Q = eNεo 1 + e
2εo  N Q = eεo + e
εo Now we need to compare this to Q = q N , where q is the single particle partition function defined as 9 q=  e
εl l where l labels the single particle states. In this example there are two single particle states, with energy -εo and +εo . =⇒ q = e
εo + eεo and  N Q = q N = e
εo + eεo The last part of this problem asks us to calculate and plot the heat capacity for this system. We know Cv = and E is given as   ∂ ln e
εo + eεo ln Q 2 = kT N E = kT ∂T ∂T

εo  εo
εo εo εo
2e e
eεo 2e N ε E = kT 2 N kT
εo kT = o e + eεo e
εo + eεo 2∂ E =
N εo tanh (εo ) and Cv is then Cv = Plotting Cv Nk  vs. kT εo  ε 2 ε  ∂E o o 2 = Nk sech ∂T kT 2 kT  10 ∂E ∂T Problem 5-4 Calculate the entropy of Ne at 300K and 1 atm.  The entropy of an ideal gas (eqn 5-20): 
S = N k ln 2mkT h2  32 5 V e2 N  Note this is neglecting electronic excitations (see Chapter 5). Some data: m = 3.351  10
26 kg k = 1.3807  10
27 Jk h = 6.6262  10
34 J  s p = 1 atm = 1.013  105 P a Putting that together we can get V kT = = 4.0889  10
26 N p
3 2mkT 2 = 8.852  1031 h2 and   S = N k ln 4.41  107  N  S = k ln 4.41  107 where N = the number of Ne atoms. We are now asked to estimate the translational degeneracy  -From our study of fluctuation theory, we found that the fluctuation in energy of a thermodynamic system (with N very large) is exceedingly small. -Therefore, the energy of the gas is essentially always very close to E (see discussion on page 63 of McQuarrie) and we can use the expression for the entropy in the microcanonical ensemble. S = k ln   Compare this with where  is the degeneracy at fixed energy E.  N  S = k ln 4.41  107 and we get  N  = 4.41  107 which makes sense since it should be large because N is on the order of 1023. Problem 5-9 What is the DeBroglie wavelength of Arÿ at 298K?
= Use: h2 2mkT m = 6.634  10
26 kg k = 1.3807  10
27 Jk h = 6.6262  10
34 J  s 11  12 and we get  = 1.6  10
11 m  Now compare this with the inter-atomic distance The volume per Ar atom is V kT = N p V = 4.06  10
26 m3 . The interatomic distance ∼ with p = 1.013  105 Pa. So N (See page 83 on the relevance of this result.) 12  V  13 N = 3.0  10
9 >>