Descripción: Statistical Mechanics - Pathria Solutions...
Statistical Mechanics - Homework Assignment 3 Alejandro G´omez Espinosa∗ March 4, 2013
Pathria 3.41 A system of N spins at a negative temperature (E > 0) is brought into contact with an ideal-gas thermometer consisting of N’ molecules. What will the nature of their state of mutual equilibrium be? Will their commom temperature be negative or positive, and in what manner will it be affected by the ratio N 0 /N ? If the system of N spins has negative temperature its energy must be bounded. Then, if it is brought into contact with a sistem of N’ molecules, the equilibrium will be reach when both temperatures are the same. But since the thermometer does not have bounded energy, the temperature in equilibrium should be greater than zero. Then, by conservation of energy we can write: ε 3 − N ε tanh + N 0 kT 0 = ET kT 2
(1)
where we can see that the ratio N 0 /N will not be affected by this result. Pathria 4.8 Determine the grand partition function of a gaseous system of magnetic atoms (with J = 21 and g = 2) that can have, in addition to the kinetic energy, a magnetic potential energy equal to µB H or −µB H, depending on their orientation with respect to an applied magnetic field H. Derive an expression for the magnetization of the system, and calculate how much heat will be given off by the system when the magnetic field is reduced from H to zero ar constant volume and constant temperature. In this case, the Hamiltonian of the system is given by: p2i ± µB H 2mi
Hi =
(2)
Then, using the partition function for one particle and taking only the positive magnetic potential energy: Q+ 1
Z = = = =
∗
e
3 −βE d p
h3
e−βµB H
Z
h3 V e−βµB H h3 V
d3 q
e−βµB H h3
Z =
e
−β
p2 +µB H 2m
d3 p d3 q h3
p2
e−β 2m d3 p d3 q Z ∞ p2 e−β 2m d3 p −∞
r
2m √ π β
[email protected]
1
3
=
2mπ βh2
3/2
V eβµB H
The total partition function for one particle must be: − Q1 = Q+ 1 + Q1 2mπ 3/2 −βµB H 2mπ 3/2 −βµB H = Ve + Ve βh2 βh2 2mπ 3/2 −βµB H −βµB H = V e + e βh2 2mπ 3/2 = 2V cosh(−βµB H) βh2
Therefore, the partition function for the system of N particles: QN
=
QN 1 1 = N! N!
2mπ βh2
3N/2
(2V )N coshN (−βµB H)
Thus, the grand partition function, given by (4.3.15) is: Q =
∞ X
z N QN
N =0
∞ X zN = N!
N =0
2mπ βh2
!N
3/2
2V cosh(−βµB H)
!N 2mπ 3/2 2V z cosh(−βµB H) = βh2 N =0 ! 2mπ 3/2 = exp 2V z cosh(−βµB H) βh2 ∞ X 1 N!
For the magnetization: ∂ ∂F = kT ln QN = − ∂H T ∂H !N 3/2 ∂ 1 2mπ = kT ln 2V cosh(−βµB H) ∂H N! βh2 ! ! ∂ 2mπ 3/2 1 = N kT ln 2V cosh(−βµB H) − ln N ! ∂H βh2 N ! ∂ 2mπ 3/2 = N kT ln 2V + ln cosh(−βµB H) ∂H βh2
M
∂ (ln cosh(−βµB H)) ∂H sinh(−βµB H) = −N kT βµB = −N µB tanh(−βµB H) cosh(−βµB H)
= N kT
2
Finally, the heat in the system at constant volume and temperature is given by Q = T ∆S. Therefore, let us begin by calculating the entropy of the system: ∂F S = − ∂T V,N !N ∂ 2mπ 3/2 1 = k T ln 2V cosh(−βµB H) ∂T N! βh2 3/2 !N 2mπkT 1 µB H ∂ T ln 2V cosh − = k 2 ∂T N! h kT ! ! 2mπkT 3/2 µB H ∂ T N ln 2V cosh − − T ln N ! = k ∂T h2 kT ∂ 3 2mπkT µB H 2V = k T N ln + T ln cosh − − T ln ∂T 2 h2 kT N! µ H sinh − µB H B kT 3 2mπkT 3N µB H 2V − ln = k N ln + + ln cosh − + 2 µ H 2 h 2 kT N! B kT cosh − kT 3 2mπkT 3 N! µB H µB H µB H = N k ln + N k + k ln cosh − + tanh − (3) 2 h2 2 2V kT T kT Then, for the final state, H = 0: 3 Sf = N k ln 2
2mπkT h2
+
3N k 2
(4)
The difference in entropy (∆S) is given by the subtraction of (3) and (4), and finally the heat relation is find as: Q = T (Sf − Si ) N! µB H µB H µB H = −T k ln cosh − + tanh − 2V kT T kT N! = −β ln cosh (−βµB H) − µB H tanh (−βµB H) 2V Pathria 4.10 A surface with N0 adsorption center has N (≤ N0 ) gas molecules adsorbed on it. Show that the chemical potential of the adsorbed molecules is given by µ = kT ln
N (N0 − N )a(T )
(5)
where a(T ) is the partition function of a single adsorbed molecule. Solve the problem by constructing the grand partition function as well as the partition function of the system. (Neglect the intermolecular interaction among the adsorbed molecules.) Since a(T ) is the partition function of a single molecule, then the total partition function must be: QN =
QN aN (T ) 1 = N! N! 3
(6)
and the grand partition function: Q=
N X N =0
zN
aN (T ) N!
(7)
To calculate the chemical potential, µ=
∂F ∂N
(8)
let us compute first the free energy: F
aN (T ) N! = −kT ln aN (T ) − ln N ! = −kT ln
= −kT (N ln a(T ) − ln N !)
Plugging this result into (8): ∂ (−kT (N ln a(T ) − ln N !)) ∂N N! = −kT ln a(T ) − ln N0 − N N! = kT − ln a(T ) + ln N0 − N N = kT ln (N0 − N )a(T )
µ =
Pathria 5.6 Determine the values of the degeneracy discriminant (nλ3 ) for hydrogen, helium, and oxygen at NTP. Make an estimate of the respective temperature ranges where the magnitude of this quantity becomes comparable to unity and hence quantum effects become important. The thermal wavelenght is: nλ3 =
nh3 N P h3 P h3 h3 = = = V (2πmkT )3/2 kT (2πmkT )3/2 (2πmkT )3/2 (2πm)3/2 (kT )5/2
(9)
Then, For H:
nλ3 =
For He:
nλ3 =
For O:
nλ3 =
(6.63 × 10−34 Js)3 (101325P a) = 2.64 × 10−5 (2π(1.67 × 10−27 kg))3/2 )(1.38 × 10−23 m2 kgs−2 K −1 )293K)5/2 (6.63 × 10−34 Js)3 (101325P a) = 3.33 × 10−6 (2π(6.64 × 10−27 kg))3/2 )(1.38 × 10−23 m2 kgs−2 K −1 )293K)5/2 (6.63 × 10−34 Js)3 (101325P a) = 4.40 × 10−7 (2π(2.56 × 10−26 kg))3/2 )(1.38 × 10−23 m2 kgs−2 K −1 )293K)5/2
In this cases, the temperature must be in the range of approximate 10−3 K to this quantity becomes comparable to unity. Since this quantity is very small compare to the physical world, this effects will become important when some particles reaches a temperature close to this value.
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