University:
Massachusetts Institute of Technology
Course:
3.20 | Materials at Equilibrium (SMA 5111)
Academic year:

2024
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159

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20
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krisjnatx

Problem Set 4 Solutions 3.20 MIT Dr. Anton Van Der Ven Fall 2002 Problem 1-22 For this problem we need the formula given in class (McQuarie 1-33) for the energy states of a particle in an three-dimensional infinite well, namely  nxnynz  h 2 n 2  n 2  n 2  , n x , n y , n z  1, 2. . . . x y z 8ma 2 Now we can make a table n x n y n z n 2x  n 2y  n 2z Degeneracy Problem 1-29 1 1 1 3 1 1 2 6 1 2 1 6 2 1 1 6 1 2 2 9 2 1 2 9 2 2 1 9 1 1 3 11 1 3 1 11 3 1 1 11 1 3 3 3 Start with the differential form for E dE  TdS  pdV E V We can use a Maxwell relation on  T S V T S V T S V T p from FT, V T  p T V So, E V T T p T V  p Problem 1-41 Show that x  x  2  x 2  x 2 x  x  2  x 2  2x x  x 2  x 2  2x x  x 2  x 2  2 x  x   x 2  x 2  x 2 Problem 1-43 Here we have to plot the Gaussian px  to see what happens as   0 1  2 exp  x x  2 2 2 for several values of  As   0, the function becomes sharper and sharper (remember the area under the curve is contrained to be 1, as we will see in problem 1-44a). Thus, the Gaussian approaches a delta function. Problem 1-44 Gaussian distribution is px  (a) show    pxdx  1    Let 2 1 exp  x  x  2 2  2 2 1 exp  x  x  2 2  2 dx u  x  x , then du  2 dx 2 So we now have    1 expu 2 du  And using standard integral tables or the math software, it can be show that this integral is equal to 1. (b) n th central moment for n  0, 1, 2, and 3 For n  0 x  x  0  1 see part a For n  1 x  x       x  x   pxdx    x  x  x  x  2 exp  2 2  2 dx Let u  x  x , then du  2 dx 2 Then x  x   1     u  expu 2 du  1  0   u  expu 2 du  1    0 u  expu 2 du I II For I let v  u and dv  du. Then  0 I 1  1  u  expu 2 du    0 v  expv 2 dv  II Thus, 1     u  expu 2 du  0 For n  2 x  x  2  Let    x  x   pxdx     x  x  2 x  x  2 exp  2 2  2 dx u  x  x , then du  2 3 x  x   1     2 2 u 2 expu 2 du  dx 2 2 2     u 2 expu 2 du Now lets take a crack at this integral....    u 2 expu 2 du We have to do this by parts. Remembering how to do that.... yx   y  x  yx      y  x  yx     yx  For our case, let y   2u expu 2 du and x  u 2 y  expu 2  and x    1 2 So,    u 2 expu 2 du  u expu 2  2       1 2 0  expu 2   2 Which leaves us with, x  x  2   2 For n  3 3 x  x      x  x  3  pxdx     x  x  3 x  x  2 exp  2 2  2 dx As before, let u  x  x , then du  2 dx 2 Then, 3 2 x  x   2     u 3 expu 2  2 Since e u is a symmetric function around the origin and u 3 is an antisymmetric function around the origin, the integral of the product of the two functions is zero. 3 x  x   0 1 0  2 (c) lim px  lim 0 exp  x x  2 2 2 The delta function is defined as  x  x , the delta function     x  a  xdx  a and   xdx  1 So lets see if this works for our case lim  0 Let u  xa  2 and du    1 exp  x  a 2 2  2 2  xdx dx  2 lim 1 0     expu 2     2 a Since   0 1     expu 2   a  a     expu 2  a  Problem 1-49 Maximize WN 1 , N 2 , . . . . N m   N! m Nj!  j1 under the contraints that  N j  N and  E j N j  . Since the natural log is a monotonic function, we can maximize lnW. M  ln WN 1 , N 2 , . . . . N m   ln Using Sterling approximation, this can be written as N! m Nj!  j1 N ln N  N   N j ln N j   N j j j But the maximization must be constrained therefore we intoduce Lagrange multipliers M  N ln N  N   N j ln N j  N j  j j  Nj  N   EjNj   N Remembering  N j  N and taking the derivative of M with respect to N j M N j   ln N j  1    E j  0  ln N j  1    E j ln N j     E j , with    1   N j  exp   expE j  Problem 1-50 We want to show that the maximum of WN 1 , N 2 , . . . N m   N! m  Nj! j1 occurs for N 1  N 2  N s     N s Maximize M  ln W  N ln N  N   N j ln N j   N j subject to the constraint  N j  N. So we must use Lagrange multipliers: M  N ln N  N   N j ln N j   N j    Nj  N M   ln N  1    0 j N j ln N j  1    N j  exp1   But now we must determine  s s j1 j1  N j   exp1    S exp1    N  exp1    So Nj  N s N s Problem 1-51 Here we use Lagrange multipliers again with the constraint  j P j  1. N M    P j ln P j   j1 N  Pj  1 j1 Maximize M P j   ln P j  1    0 P j  exp1   Determine  N  P j  N exp1    1 j1 exp1    1 N Thus Pj  1 N Problem 2-1 From stat mech we get E V And from thermo we have N,  p   p N,V E V N,T T p T N,V  p To show why   cont  T lets see what happens if we let   T where  is a constant E V N,  T p T  p N,V or E V N,T but from a Maxwell relation we know E V T p T N,T p T N,V  S V  T S V N,T N,V  p N,T and we can write  p dE  TdS  PdV This statement violates the second law of thermodynamics because it implies that Q Q  TdS  T  dS Problem 2-2 Given n! n!n  n 1 ! n  is n 1  n 1 ?  n 1 is the value of n 1 that maximizes n. From problem 1-50, we know that n 1  n2 .  n 1 is given by n  n1 n1  n 1 0 n  n 1 0 n! n 1 !nn 1 ! n! n 1 !nn 1 ! We also know that (given in the problem): n y  1  x n   xn 1 n 1 0 n! n 1 !n  n 1 ! If we take the derivative of y with respect to x we get n  y  n1  x n1   n1xn n 1 0 11 n! n 1 !n  n 1 ! If we let x  1: n n2 n1   n1 n 1 0 Furthermore, n! n 1 !n  n 1 ! 2-2-1 n  2n  n! n 1 !n  n 1 ! n 1 0 So if we put this all together back in (2-2-1) we get n  n1 n1  n! n 1 !nn 1 ! n 1 0 n  n1  n2 n  n  n 1 2 2 n! n 1 !nn 1 ! n 1 0 n 1  n 1 Problem 2-5 Show that S  k  P j ln P j We have the following three relations already S  kT  ln Q T N,V exp E j kT Pj   k ln Q Q and Q  exp j E j kT Starting with S we can write Q  k ln Q  kT S  kT Q Q T S k Q S k Q   ln j j Ej kT exp  j exp E j kT Ej kT 2 exp E j kT E j kT  k ln Q exp E j kT  k ln Q  k ln Q E j kT exp E j S  k  ln exp kT j Q k ln Q P j Since  P j  1 S  k  P j  ln exp j E j kT  k ln Q j E j kT S  k  P j  ln exp j  k  P j  ln j 1 Q E j kT exp S  k  P j ln  Pj Q j S  k  j P j ln P j Problem 2-8 Q    E j expE j  j  E j expE j  E  j  Q E   Q  Q   ln Q   ln Q  Problem 2-10 First derive E  kT 2  ln Q T N,V starting from the result of problem 2-8, namely E   ln Q  From the chain rule  ln Q T E  We know   1 kT so T   kT 2 , thus  ln Q T E  kT 2  ln Q T We can check this by taking kT 2 kT 2  ln Q T kT N,V Q T  kT 2 N,V  2 T N,V T  N,V Q E  exp  kTj .  kT  Q N,V  2 T E  exp  kTj N,V Q E  exp  kTj Q  Ej kT 2 The kT 2 cancels and we are left with kT 2  ln Q T N,V  E  E j exp  kTj Q definition of E So we have shown again that indeed E  kT 2  ln Q T . Now for the relation involving p p   p j exp  Q Ej kT    E j V exp  Ej kT Q p  kT Q V Q  kT  ln Q V  kT V  E  exp  kTj j N,T Q N,T Problem 2-11 Starting with F  kT ln Q (Remember F  A S   F T S  kT  ln Q T V,N p   F V p  kT  ln Q V V,N  k ln Q Eqn. 2-33 T,N Eqn. 2-32 T,N F  E  TS  E  F  TS E  kT ln Q  T  kT E  kT 2  ln Q T  ln Q T V,N  k ln Q Eqn. 2-31 V,N Problem 2-13 E For a particle confined to a cube of length a we are asked to show p j  23 Vj . We can start with the equation for the energy states of a particle in an three-dimensional infinite well, namely Ej  h 2 n 2  n 2  n 2  , n x , n y , n z  1, 2. . . . x y z 8ma 2 1 Remembering that a 3  V or a  V 3 we can write E j in terms of V 2 3 2 E j  h V n 2x  n 2y  n 2z  8m E j pj   V 5 2 3 3 2 2  2 h V n 2x  n 2y  n 2z   2  1  h V n 2x  n 2y  n 2z  8m 3 8m 3 V Ej So we have 2 Ej 3 V pj  and taking the ensemble average gives us p  2 E 3 V Problem 2-14 3N 2 2mkT h2 QN, V, T  1 N! VN For p ,  ln Q V p  kT ln Q  ln T.N  3N ln 2mkT 2 h2 p  kTN V 1 N!  N ln V Now for E E  kT 2  ln Q T N,V  kT 2 3N 2 2mk h2 2mkT h2 E  3N kT 2 Ideal gas equation of state is obtained when Q  fTV N  ln Q  ln fT  N ln V p  kT  ln Q V T.N  kT ln fT V p  NkT V  kT T,N N ln V V Problem 2-15 We are given Q as Q substituting in   3N h 2kT exp h kT exp 1 exp U o kT h k Q 3N exp   2T 1  exp  T  exp U o kT To find c v we need to use the following two relations E  kT 2  ln Q T cv  N,V E T Using your favorite math package (Maple), we can get E as E 1 2 Uo e  kT 3Nk e Uo kT  12Nk e U o  k kT  9Nk e 2  kU o kT  1  e T  2U o e 2 and c v as cv  E T  3 Now we can take the lim c v and we find that T kN 2 e T  T 2 1  e T 2 Uo kT  4U o e U o  k kT  2U o e 2  kU o kT  k 2 e T N lim 3 T T 2 1  e  T 2  3Nk

3.20 Materials at Equilibrium (SMA 5111): Solutions to Problem Set 4

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