Thermodynamics and an Introduction to Thermostatistics 5, Exercises of Chemistry

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      Problem Set 5 Solutions - McQuarrie Problems 3.20 MIT Dr. Anton Van Der Ven Problem 3-4 Fall 2003 We have to derive the thermodynamic properties of an ideal monatomic gas from the following:
= eq
3 22mkT
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
Nd 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
(see table 3-1) but
= eq
so, 3 3 2mkT 2m2 25 2
kT 
kT  V e = (kT ) V exp= q
= 2 2h h kT Starting with N, 3 2 2 2mkT N =
∂ = V  exp kT∂ hT,V ((3-8-1)) and p 3 2 2 ∂      p =
∂V = kT 2mkT exp kThT, Putting those together we can get the ideal gas equation of state, namely N p = kT V Now S, ∂ 5 2m 2m S =
= k kT ) V e + (kT ) V e 2 2 kT 2∂T 2 h hN,V 1 3 2 3 2 5 2 3 2


   

 [

 ] ︸ ︷︷ ︸
 [
 ] [
 ]) { [  
 ]} [ ]   [


 [ ]  =  = 3 2 3 2 5 2mkT  2mkT S k V e V e= 2 22 h T h 3 25  2mkT S = k
V e h22 T N 5  S = k
N ((3-8-2)) 2 T But from (3-8-1) we can get 3 22mkT e = NV
1 h2 2mkT 3 2 V  =
kT  ln h2 N Putting this into (3-8-2) we get 3 3 5 2mkT V 2mkT V2 2 e 5 2 + ln S k + ln N = Nk ln k= 2 22 h N h N 5 2 3 2 2 VS = Nk ln 2mkT  e h N 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 2 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 (2m) (kT ) ]N53 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 S =
∂T p,N ∂G V = ∂p T,N ∂G ∂N p,T  G is related to the isothermal-isobaric partition function  according to (see Table 3-1): (2m) (kT ) 53 G =
kT ln  =
NkT ln ph3 2 2 2 ∑∑ ∑
 ∑ ∑
   ∑ ∑ 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 2)(p
p ≈ 5 10
12 2p 1for a cubic centimeter of gas under standard conditions. This is approximately , where n is the number of 2 3n molecules in the gas. (Source: The Principles of Statistical Mechanics, Richard C Tolman, Oxford University Press, first edition 1938) Problem 3-24 2 2Show that H2
H = kT C p 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 + pV ) e
Ej
pV H = (Ej V,j Step 2: Get the temperature derivative at constant (N, P ) (The conjugate variable to H in this case)   ∂H 1 + pV ) e
Ej
pV +H  (Ej  = 1 (Ej + pV )2 e
Ej
pV ∂T kT 2 kT 2 N,P V,j V,j Step 3: Divide through by the partition function ∂H 1 V,j (Ej + pV ) e
Ej
pV 1 V,j (Ej + pV ) 2 e
Ej
pV + H = ∂T kT 2  kT 2 N,P ︸ ︷︷ ︸ ︸ ︷︷ ︸ H H or
2 2 2
 = N,P H H kT ∂H ∂T   but we know ∂H ∂T N,P = pC . So,
2 2 2= pH H kT C 2 5 =
=
=
=
   






  



  
 Problem 3-26 2 2 ∂p ∂V ∂ VShow that ..∂N N V,T N,T From Gibbs-Duhem we have SdT
V dp + N d = 0 At constant T , we then get d V = dp N We can use the chain rule and get: d d dN = dp dN V,T dp V,T Using partial derivative manipulation dN dN dV dp V,T dV P,T dp N,T But for a single component system dV p,T = the molar volume = V and we get dN N dN N dV V dp N,T dp V,T Putting this all together we get d V 2 dp N 2dN dVV,T N,T Problem 4-2 3 2h V 12mkT 26NShow that given in table 4-1 is very large for electrons in metals at T = 300 K.  Take Na-metal having the following properties
10 .- stable in the bcc crystal structure with lattice constant a = 4 23  10 m - 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 = V (4 23  10
10)3 . .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 h2 3 2 = 1524 >> 1 V 12mkT Therefore Boltzmann statistics cannot be applied to electrons in metals. Must use Fermi-Dirac statistics. 6 ∑ ∑ ∑ ∑∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑∑          Problem 4-6 ∞  n n 21S = x1 x2 N=0 {nj } with n1 and n2 = 0 1 2, , and . The means with the restriction that n1 + n2 = N. 1{nj } n n2 n n 21Let’s consider x x for several values of N. 1 2 {nj } N = 0→ possible combinations of n1 and n2 are 0 and 0. =⇒ = 1 {nj } N = 1→ possible combinations of n1 and n2 are n2 Nn1 1 0 1 0 1 1 n n 21=⇒ x x2 = x1 + x21 {nj } N = 2→ possible combinations of n1 and n2 are N 2 2 2 2 2 1 1 2 + x1 =⇒ x x2 = x2 + x x {nj } N = 3→ possible combinations of n1 and n2 are n2n1 0 2 1 1 2 0 n n 21 n2 Nn1 1 2 3 2 1 3 n n 2 2 21=⇒ x x2 = x x 2 + x x 1 1 1 2 {nj } N = 4→ possible combinations of n1 and n2 are n2 Nn1 2 2 4 n n 2 2 21=⇒ x x2 = x x 21 1 {nj } n n 21For N > 4→ x1 x = 0 because n1 + n2 � 42 {nj } Putting everything together we get n n 2 2 2 2 2 21 2 1 2 + x1 + x x 2 + x x x1 x = 1 + x1 + x2 + x2 + x x 2 1 1 2 + x x 21 N {nj } 7 ∑        
εlq = e l where l labels the single particle states. In this example there are two single particle states, with energy -εo and + ε .o =⇒ q = e
εo + e εo and
ε oQ = qN = e + eε o N ∂EThe last part of this problem asks us to calculate and plot the heat capacity for this system. We know Cv = ∂T and E is given as εo∂ ln Q 2 ∂ ln e
εo + e E = kT 2 = kT N ∂T ∂T εo e
εo
εo εo

εoe
eεo 2 kT 2 E = kT N kT 2
εo = Nεo ee
εo + eεoe + eε o E =
Nε o tanh ( ε o) and Cv is then ∂E  εo  Cv = = Nk εo 2 sech 2 ∂T kT 2 kT kTPlotting Cv vs.Nk εo 10 [
 ]
   [ ]   [ ]    
    = 2 2 Problem 5-4 Calculate the entropy of Ne at 300K and 1 atm.  The entropy of an ideal gas (eqn 5-20): 53 2mkT V e S = Nk ln h2 N Note this is neglecting electronic excitations (see Chapter 5). Some data: m = 3 351  10
26 kg . k = 1 3807  10
27 . k J .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 2mkT 3 2 = 8 852  1031 . h2 and S = Nk 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 = ln  k where  is the degeneracy at fixed energy E. Compare this with S = k ln 4 41  107 . and we get N  = 4 41  10 7 . N 23which makes sense since it should be large because N is on the order of 10 . Problem 5-9 What is the DeBroglie wavelength of Arÿ at 298K? h2 2mkT Use: m = 6 634  10
26 kg . 1 2 k = 1 3807  10
27 . k J .h = 6 6262  10
34 J  s 11     >>  and we get  = 1 6  10
11m.  Now compare this with the inter-atomic distance The volume per Ar atom is V kT = N p with p .= 1 013 10 5 Pa. So V N .= 4 06 10
26 3 m . The interatomic distance ∼ .= 3 0 10
9 N (See page 83 on the relevance of this result.) 1 3 V 12


  
   ∑
    = ∂ ∂N V,T,H ∂ p =
∂V T,N,H and from stat mech we have from above =
kTN ln [exp [  H ] + exp [
 H ]] o o Lets get the entropy, S ∂ S =
∂T V,N,H  H
o (exp [  H ]
exp [
 H ]) kT 2 o o S = kN ln (exp [  H ] + exp [
 H ]) + kTN o o exp [  H ] + exp [
 H ]o o S = kN {ln (exp [  H ] + exp [
 H ])
 tanh (  H )}o o o o Now for the magnetization ∂ exp [  H ]
exp [
 H ]oM =
= kTN o = N tanh (  H )o o∂H o exp [  H ] + exp [
 H ]V,N,T o o M = N tanh (  H )o o The energy E = E
TS
HM E = + TS + HM = 0 However if you define a quantity called the internal magnetic energy (which is a quantity analogous to the enthalpy in the T, p, N ensemble) EH = E
HM you can get EH =
N H tanh (  H )o o The last part of this problem asks you to determine the behavior of the energy and entropy as T → 0. EH (T → 0) =
N H o lim S = lim kN {ln (exp [  H ] + exp [
 H ])
 tanh (  H )} = 0 o o o o →∞ →∞ (S T → 0) = 0 is in accordance with the third law of thermodynamics. Problem 3 (a) N M =  n io i=1 (see solution to Problem 2 - Method 1) ∂ exp [  H ]
exp [
 H ]oM =
= kTN o = N tanh (  H )o o∂H o exp [  H ] + exp [
 H ]V,N,T o o M = N tanh (  H )o o 3 ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ 〈
 〈 〉 (b) The partition function for the N, V, H, T fixed ensemble is ( s = exp [
 Es
M H )] s where s = states We want to determine the fluctuations in the extensive quantity M. Use the 3-step procedure developed in class. Step 1: Multiply both sides by the partition function M = Ms exp [
 (Es
M H )] states s Because Es is always constant (see comment in Problem 2) we can therefore arbitrarily set it to zero and write ) [ )]) N N N M =  n i exp  H nio o n1 ,n 2,...nN i=1 i=1 Step 2: Get derivative with respect to mechanical variable’s conjugate. ) [ )])2N N N ∂M ∂  + M =  o ni exp  H ni∂H ∂H o n1 ,n 2,...nN i=1 i=1 2{ ) [ )])} ) [ )]) N N N N N N ∂M +M   ni exp  H ni =  o ni exp  H ni∂H o o o n1 ,n 2 ,...nN i=1 i=1 n1 ,n 2 ,...nN i=1 i=1 Step 3: Divide through by the partition function ∂M 2 + M = M 2 ∂H or we can write it like this ( M) 2 〉 = M2
M 2 = 1 ∂M  ∂H with: ∂M ( ) = N 2 1
tanh2 (H ) ∂H ,N (c) As  → ∞, tanh ( H ) → 1. Therefore MT →0 = N and ( M )2 = 2 {1
1} = 0. In other T →0 words, the ground state with all the spins aligned has no fluctuations. 4 ∑
 ∑ ∑ ∑

  = Problem 4 From the first and second law we have dE = T dS
pdV + HdM + dN E = T S
pV
HM + N (Euler Form) We are working with fixed N, T, V, M in this problem. We need to make a Legendre transform because T is a controlling variable instead of E. S
E =
F = ln Q k Q = exp [
E state] states F =
kT ln Q From the differential form of F, namely dF =
SdT
pdV + HdM + dN we get that ∂F H = ∂M T,V,N We are working under constant magnetization so N M =  ni = o (n+
n
)o i=1 where n+ = number of up spins and n
= number of down spins. That means that the sum in the expression of Q must be performed over only those states with fixed M (i.e. fixed n+ and n
) Also the atoms in the system do not interact, meaning that the energy is independent of the number and arrangement of up/down spins and is therefore constant = Eo. Q = exp [
E states] = exp [
E o] =  exp [
E o] states with states with magnetization M magnetization M where  is the number of states that are consistent with a magnetization M =  (n+
n
) .o N ! n+! (N
n+)! N ! Q = exp [
Eo] n+! (N
n+)! N ! N ! F =
kT ln exp [
Eo] = Eo
kT ln n+! (N
n+)! n+! (N
n+)! using Stirling’s approximation, F = Eo
kT {(N ln N
N)
n+ ln n+ + n+
(N
n+) ln ( N
n+) + ( N
n+)} but we know from above that M +  N M =  (n+
n
) =  (2 n+
N) =⇒ n+ = o o o 2o we can substitute this back into our expression for F to get F as a function of M 


 M +  N M +  N  N
M  N
Mo
oF = Eo
kT N ln N
o ln ln o 2 2 2 2o o o o 5