Lecture 9
Overview
(Ch. 1-3)
Format of the first midterm: three problems with multiple
questions. Total: 100 points.
• The Ideal Gas Law, calculation of W, Q and dS for
various ideal gas processes.
• Einstein solid and two-state paramagnet, multiplicity and
entropy, the stat. phys. definition of T, how to get from the
multiplicity to the equation of state.
Only textbook and cheat-sheets (handwritten!) are allowed.
No homeworks and lecture notes.
DO NOT forget to bring your calculator!
Problem 1
V
V2
V1
3
2
1
T1
T2 T
One mole of a monatomic ideal gas goes through a
quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in
the Figure. T1 and T2 are given.
(a) (10) Calculate the work done by the gas. Is it
positive or negative?
(b) (20) Using two methods (Sackur-Tetrode eq. and
dQ/T), calculate the entropy change for each stage and
for the whole cycle, Stotal. Did you get the expected
result for Stotal? Explain.
(c) (5) What is the heat capacity (in units R) for each
stage?
Problem 1 (cont.)
(a) 1 – 2
2–3
V  T  P = const (isobaric process)
V = const (isochoric process)
W12  P V2  V1   R T2  T1   0
W23  0
V1

3 – 1 T = const (isothermal process) W31  PdV  R T1
V2
Wtotal  W12  W31  RT2  T1   R T1 ln
V1
dV
V1
T1

R
T
ln

R
T
ln
0
1
1
V V
V2
T2
2
T
T1
T
 R T1  2  1  ln 2   0
T2
T1 
 T1
Problem 1 (cont.)
V
V2
V1
(b)
3
2
1
Sackur-Tetrode equation:
S U , V , N   R ln
V 3
U
 R ln  k B ln f N , m 
N 2
N
Vf 3
Tf
 V f 3 Tf 
S  R ln
 R ln
 R  ln
 ln 
Vi 2
Ti
 Vi 2 Ti 
1–2
T1
2–3
3–1
S cycle 
V  T  P = const (isobaric process)
T2 T
V = const (isochoric process)
T = const (isothermal process)
5
T
T 3
T
R ln 2  R ln 2  R ln 2  0
2
T1
T1 2
T1
S12 
S 23 
5
T
R ln 2
2
T1
3
T
3
T
R ln 1   R ln 2
2
T2
2
T1
S31  R ln
V1
T
  R ln 2
V2
T1
as it should be for a quasistatic cyclic process
(quasistatic – reversible),
because S is a state function.
Problem 1 (cont.)
V
V2
V1
(b)
3
dS
2
Q
- for quasi-static processes
T
V  T  P = const (isobaric process)
1–2
T2
CP dT 5
T
 R ln 2
T
2
T1
T1
 Q  CP dT S12  
1
T1
2–3
T2 T
V = const (isochoric process)
T1
CV dT
3
T
  R ln 2
T
2
T1
T2
 Q  CV dT S 23  
dU  Q
3–1
T = const (isothermal process)
V
dU  0
Q  WON
S cycle 
V
1 1
RT1 1 dV
V1
T2
S31   PdV 

R
ln


R
ln
T1 V2
T1 V2 V
V2
T1
5
T
T 3
T
R ln 2  R ln 2  R ln 2  0
2
T1
T1 2
T1
Problem 1 (cont)
(c)
V
3
V2
Q  C d T
2
1–2
V1
Let’s express both Q and dT in terms of dV :
3
5
C  C P  CV  R  R  R  R
2
2
1
T1
V  T  P = const (isobaric process)
T2 T
2–3
V = const (isochoric process)
3
C  CV  R
2
3–1
T = const (isothermal process), dT = 0 while Q  0
C 
At home: recall how these results would be modified for diatomic and polyatomic gases.
One mole of a monatomic ideal gas goes through a
quasistatic three-stage cycle (1-2, 2-3, 3-1) shown in the
Figure. Process 3-1 is adiabatic; P1 , V1 , and V2 are given.
Problem 2
P
P1
1
(a) (10) For each stage and for the whole cycle, express the
work W done on the gas in terms of P1, V1, and V2.
Comment on the sign of W.
2
(b) (5) What is the heat capacity (in units R) for each stage?
3
V1
V2
V
(c) (15) Calculate Q transferred to the gas in the cycle; the
same for the reverse cycle; what would be the result if Q
were an exact differential?
(d) (15) Using the Sackur-Tetrode equation, calculate the
entropy change for each stage and for the whole cycle,
Stotal. Did you get the expected result for Stotal? Explain.
Problem 2 (cont.)
P
P1
2
1
3
V1
(a)
V2
V
1–2
P = const (isobaric process)
2–3
V = const (isochoric process)
3–1
 W12  P1 V2  V1   0
 W23  0
adiabatic process

P1 V1
1  1
1 
 W31    P(V )dV     dV   P1 V1

  1
 1 
V
1


V
V
2
 1

V2
V2
V1

 

V1
5
3   V1 
   P1 V1 1   
3
2  V2 

2/3

0


P
Problem 2 (cont.)
(c)
1–2
P1
P = const (isobaric process)
2
3
5  P V PV  5
 Q12  CP T2  T1   R  2 2  1 1   P1 V2  V1   0
2  R
R  2
2–3
1
V1
V2
V = const (isochoric process)
5


3


3
3
V
 Q23  CV T3  T2   V2 P3  P1   P1V1  P3V2  P1V2  1   1  0
 V2 

2
2


 Q31  0
3 – 1 adiabatic process


5


  V1  3 
5
3
 Q   Q12   Q23  P1 V2  V1   P1V2     1
2
2
  V2 



For the reverse cycle:
 Qreverse   Q
If Q were an exact differential, for a cycle Q should be zero.
V
Problem 2 (cont.)
P
P1
(d)
2
1
3
V1
V2
V
Sackur-Tetrode equation:
S U , V , N   R ln V 
3
R ln U  k B ln f  N 
2
Vf 3
Tf
 V f 3 Tf 
S  R ln
 R ln
 R  ln
 ln 
Vi 2
Ti
V
2 Ti 
i

1–2
V  T  P = const (isobaric process)
S12 
2–3
V = const (isochoric process)
3–1
S cycle 
T3  T3 P3
3
S 23  R ln   
2
T2  T2 P1

V 
P3  P1  1 
 V2 

5
V
R ln 2
2
V1
 3
V 5
V
  R  ln 1  R ln 1
V2 2
V2
 2
Q = 0 (quasistatic adiabatic = isentropic process)
5
V 5
V
R ln 2  R ln 2  0
2
V1 2
V1
S31  0
as it should be for a quasistatic cyclic process
(quasistatic – reversible),
because S is a state function.
Problem 3
10
20
Start with the definition:
40
C
Q
dT
3
Q  dU  W  RdT  PV dV
2
C (V ) 
30
Calculate the heat capacity of one mole of an ideal
monatomic gas C(V) in the quasi-static process shown
in the Figure. P0 and V0 are given.
Q 3
dV
 R  P(V )
dT 2
dT
 V
P0
V  P0 1  
V0
 V0 
PV  RT
P  P0 
P0 dV  V  P0V
1   
R  V0  R
P
P0
we need to find
the equation of
this process
V=V(T)
 V
P0V 1    RT
 V0 
 dV  P0 dV 
V
 
 
1  2   dT
R 
V0 
 V0 
 V 
3
dV 3
V



C (V )  R  P(V )
 R  R1   /1  2 
2
dT 2
V0 
 V0  
0
V0
dV R
1

dT P0 
V 
1  2 
V0 

5
V 
R   4 
2
V0 

C (V ) 
V
1 2
V0
V
P
Problem 3 (cont.)
50
P0
S=const
adiabat
Does it make sense?
5
V 
R   4 
2
V0 
C (V )  
V
1 2
V0
C/R
T=const
isotherm
C (V )   at V  V0 / 2
the line touches an isotherm
2.5
0
1.5
0
1/2
5/8
1
V/ V0
V0/2
5V0/8
V0
C (V )  0 at V  5V0 / 8
the line touches an adiabat
V
Problem 4
(10) The ESR (electron spin resonance) set-up can detect the minimum difference in
the number of “spin-up” and “spin-down” electrons in a two-state paramagnet
N-N =1010. The paramagnetic sample is placed at 300K in an external
magnetic field B = 1T. The component of the electron’s magnetic moment
along B is  B =  9.3x10-24 J/T. Find the minimum total number of electrons
in the sample that is required to make this detection possible.
N
 E2  E1 

 exp  
N
k BT 

E2  E1  2 B B

 2  B B 
  1010
N   N   N  1  exp  
 k BT  



 2  B B 
 2  B B 
1010
 

N   N   N  1  exp  
1  exp  
k
T
k
T
 2  B B  
B
B

 



 

1

exp

 k BT  

 2 B B 
 2  9.3 1024 1 
  exp  
  0.9955
exp  
 23
 1.38 10  300 
 k BT 
2 1010
N  N 
 4,4 1012
0.0045
- the high-T limit
Problem 5


 
Consider a system whose multiplicity
N
f N /2

U
,
V
,
N

f
N
V
U
is described by the equation:
where U is the internal energy, V is the volume, N is the number of particles in the
system, Nf is the total number of degrees of freedom, f(N) is some function of N.
(a)
(10) Find the system’s entropy and temperature as functions of U. Are these
results in agreement with the equipartition theorem? Does the expression for the
entropy makes sense when T  0?
(b)
(c)
(5) Find the heat capacity of the system at fixed volume.
(a)
(15) Assume that the system is divided into two sub-systems, A and B; sub-system
A holds energy UA and volume VA, while the sub-system B holds UB=U-UA and
VB=V-VA. Show that for an equilibrium macropartition, the energy per molecule is
the same for both sub-systems.
S  k B ln   k B ln f N   Nk B ln V 
T
2U
N f kB
Nf
U
k BT
2
Nf
k B ln U
2
1  S 
Nf

  kB
T  U  N ,V
2U
- in agreement with the equipartition theorem
When T  0, U  0, and S  -  - doesn’t make sense. This means that the expression
for  holds in the “classical” limit of high temperatures, it should be modified at low T.
Problem 5 (cont.)
(b) CV   Q    U   dU  TdS  PdV   T  S   T NkB f 1  f NkB
2T 2
 T  N ,V  T  N ,V
 T  N ,V
(c)
 U ,V , N   f N  V NU 3 N / 2
 U , V , N   VAN A V  VA  B U A3 N A / 2 U  U A 
N
3NB / 2

3N A 3 N A / 21
U  U A 3 N B / 2  3N A U A3 N A / 2 U  U A 3 N B / 21  0

UA
U A
2
2
3N A
U  U A   3N B U A  0
2
2
UA UB

N A NB