Lecture 6. Entropy of an Ideal Gas (Ch. 3)
Today we will achieve an important goal: we’ll derive the equation(s) of state
for an ideal gas from the principles of statistical mechanics. We will follow the
path outlined in the previous lecture:
Find  (U,V,N,...) – the most challenging step
S (U,V,N,...) = kB ln  (U,V,N,...)
  S (U , V , N ,...) 

T  

U


1
Solve for U = f (T,V,N,...)
So far we have treated quantum systems whose states in the configuration
(phase) space may be enumerated. When dealing with classical systems with
translational degrees of freedom, we need to learn how to calculate the
multiplicity.
Multiplicity for a Single particle
- is more complicated than that for an Einstein solid, because it depends on three
rather than two macro parameters (e.g., N, U, V).
Example: particle in a onedimensional “box”
-L
L
The total number of ways of filling up the cells in phase space is the
product of the number of ways the “space” cells can be filled times
the number of ways the “momentum” cells can be filled
px
-L
L x
px
-px
The number of microstates:
x

Q.M.
L px
2 L  p x

x  p x
h
   space p
Quantum mechanics (the
uncertainty principle) helps
us to numerate all different
states in the configuration
(phase) space:
x  px  h
Multiplicity of a Monatomic Ideal Gas (simplified)
For a molecule in a three-dimensional box: the state of the
molecule is a point in the 6D space - its position (x,y,z) and its
momentum (px,py,pz). The number of “space” microstates is:
For N molecules:
 V 
 space   3 
 x 
 space 
N
n
There is some momentum distribution of molecules in an ideal
gas (Maxwell), with a long “tail” that goes all the way up to p =
(2mU)1/2 (U is the total energy of the gas). However, the
momentum vector of an “average” molecule is confined within
a sphere of radius p ~ (2mU/N)1/2 (U/N is the average energy
per molecule). Thus, for a single “average” molecule:
4
 p3
3
p 
p x  p y  p z
 V  p3 
 V Vp 
  
  space p   3

3
3
 h 
 x  px 
However, we have over-counted the multiplicity,
because we have assumed that the atoms are
distinguishable. For indistinguishable quantum
particles, the result should be divided by N! (the
number of ways of arranging N identical atoms in a
given set of “boxes”):
p
p
N
The total number of microstates
for N molecules:
V
V
 3
x  y  z x
indistinguishable
N
1  V Vp 

N !  h 3 
N
pz
More Accurate Calculation of N (I)
Momentum constraints:
py
px
1 particle -
px2  p y2  pz2  2mU
2 particles -
p12x  p12y  p12z  p22x  p22y  p22z  2mU
The accessible momentum volume for N particles = the “area” of a 3N-dimensional hypersphere p
2 3 N / 2 3 N 1 N =1
" area" 
 3N 
 1 !

 2

1 V N  2 mU 
N 


2
N ! 3N / 2!  h

Monatomic
ideal gas:
(3N degrees
of freedom)
r
3N / 2
2p


2 3 / 2 2
r  1 / 2!  / 2  4 r 2
1 / 2!
The reason why m matters: for a
given U, a molecule with a larger
mass has a larger momentum, thus a
larger “volume” accessible in the
momentum space.
 e5 / 2 V  4  m U  3 / 2 
indistinguishable U ,V , N    3

 
 h N  3N  
U   U f N / 2
N
2p
f N- the total # of “quadratic” degrees of freedom
pz
More Accurate Calculation of N (II)
For a particle in a box (L)3: (Appendix A)
py
h
px2  p 2y  pz2
h2
2
2
2

p

x
E  n x , n y , nz  

n  n y  nz 
2  x
2L
px
2m
8mL
nx , n y , nz  0
If p>>p, the total degeneracy (multiplicity)
of 1 particle with energy U is:
1 S p 1  2 L 
V
p
p
1 U   3

S

p

S
p 



3
3
3
3
3
3
2  p 
2 h
h
3
p
3
If p>>p, the total degeneracy (multiplicity) of N
indistinguishable particle with energy U is:
1 VN
p
N U  
S
p 
3N  3N
N! h
Plug in the “area” of the hyper-sphere:
 e V  4 mU 
indistinguishable U ,V , N    3


h
N
3
N



5/ 2
3/ 2
N

   2p 

Entropy of an Ideal Gas
S U ,V , N   kB ln  U ,V , N 
 V  4 m U 3/ 2  5
S ( N ,V ,U )  Nk B ln   2
   Nk B  ln  2p 
 N  3h N   2
The Sackur-Tetrode equation: (Monatomic ideal gas)
3/ 2


 V  4 m U   5 

S ( N ,V ,U )  N k B ln  




2
N
3
h
N
2
 


  

V
 Nk B ln 
 N
U 
 
N
3/ 2
 3
V 3
U
5
 4 m  
  Nk B   ln  2    Nk B ln  Nk B ln   ( N , m)
N 2
N
 3h  
3
 2
an average volume an average energy
per molecule
per molecule
In general, for a gas of
polyatomic molecules:
S ( N ,V , T )  N k B ln
V f
 N k B ln T   ( N , m)
N 2
f  3 (monatomic), 5 (diatomic), 6 (polyatomic)
Problem
Vf
Tf
f
S ( N ,V , T )  N k B ln  N k B ln
Vi 2
Ti
For each gas:
The temperature
after mixing:
U1 T1   U 2 T2   U total T f 
Tf 
H2 :
Two cylinders (V = 1 liter each) are
connected by a valve. In one of the
cylinders – Hydrogen (H2) at P = 105
Pa, T = 200C , in another one –
Helium (He) at P = 3·105 Pa,
T=1000C. Find the entropy change
after mixing and equilibrating.
5
3
N1k BT1  N 2 k BT2
2
2
5
3
N1k B  N 2 k B
2
2
S H 2  N k B ln 2 
S total  S H 2  S He
Tf
5
N k B ln
2
T1
k
 N1  N 2  k B ln 2  B
2
5
3
3
5

N1k BT1  N 2 k BT2   N1k B  N 2 k B T f
2
2
2
2


5P1  3P2
P
P
5 1 3 2
T1
T2
He :
S He  N 2 k B ln 2 
Tf
Tf 

 3N 2 ln 
5 N1 ln
T
T2 
1

Tf
3
N 2 k B ln
2
T2
S total  0.67 J/K
Entropy of Mixing
Consider two different ideal gases (N1,
N2) kept in two separate volumes (V1,V2)
at the same temperature. To calculate the
increase of entropy in the mixing process,
we can treat each gas as a separate
system. In the mixing process, U/N
remains the same (T will be the same
after mixing). The parameter that
changes is V/N:
3/ 2

  V  4 m U   5 

S ( N ,V ,U )  N k B ln  




2
N
3
h
N
2





 

V 
V 
S total S A  S B

 N1 ln    N 2 ln  
kB
kB
 V1 
 V2 
if N1=N2=1/2N , V1=V2=1/2V
Stotal N  V  N  V 
 ln 
  ln 
  N ln 2
kB
2 V / 2  2 V / 2 
The total entropy of the system is greater after mixing – thus,
mixing is irreversible.
Gibbs “Paradox”
V 
V 
S total S A  S B

 N1 ln    N 2 ln  
kB
kB
 V1 
 V2 
- applies only if two gases are different !
If two mixing gases are of the same kind (indistinguishable molecules):
 V V 
V 
V 
S total / k B   Stotal  S A  S B  k B   N1  N 2  ln  1 2   N1 ln  1   N 2 ln  2 
 N1  N 2 
 N1 
 N2 
 V N1 
 V N2 
 N1 ln 

N
ln
2


0
 N V1 
 N V2 
if
N1 N 2 N


V1 V2 V
Stotal = 0 because U/N and V/N available for each molecule remain the same after mixing.
 V  4  m U 3 / 2  5
S ( N ,V ,U )  N k B ln  
   N kB
2
N
3
h
N
  2
 
1 V N 2 3 N / 2
N 
N! h3 N  3N 

!
2



2mU

3N
Quantum-mechanical indistinguishability is important! (even though this equation
applies only in the low density limit, which is “classical” in the sense that the distinction
between fermions and bosons disappear.
Problem
Two identical perfect gases with the same pressure P and the same number of
particles N, but with different temperatures T1 and T2, are confined in two vessels,
of volume V1 and V2 , which are then connected. find the change in entropy after
the system has reached equilibrium.
3/ 2
 V  4  m U 3 / 2  5
 V  4 m 3
  5
S ( N ,V ,U )  N k B ln  
k BT    N k B
   N k B  N k B ln  
2
2
N
3
h
N
2
N
3
h
2
 
  2
 
 
5
5
V
V
3/ 2 
3/ 2 
Si  S1  S 2  N k B ln  1 T1    N k B  N k B ln  2 T2    N k B
N
 2
N
 2
5
 V  V 
3/ 2 
S f  2 N k B ln  1 2 T f    2 N k B
 2N
 2
Tf 
T1  T2
2
- prove it!
2


 V1  V2   2 N 2 

 V1  V2 2  3  T f  
T f 3
S
 ln 
 ln 

  ln 

  ln 
3/ 2
3/ 2 
N kB
2
N
V
V
4
V
V
2
T
T
 1 2 

1 2
 T1  T2  
 1 2 


2
 V1  V2 2  3  T1  T2 2  
T1  T2  5  T1  T2  
ln 
 ln 
  ln 
   PV1  V2   2 Nk B


4
V
V
2
4
T
T
2
2
4
T
T


1 2
1 2
1 2






at T1=T2, S=0, as it should be (Gibbs paradox)
An Ideal Gas: from S(N,V,U) - to U(N,V,T)
 U ,V , N   f N  V NU f N / 2
Ideal gas:
(fN degrees of freedom)
S U , V , N  
1 S f Nk B


T U 2 U

f
U
V
Nk B ln  Nk B ln   ( N , m)
2
N
N
U ( N ,V , T ) 
f
N k BT
2
- the “energy”
equation of state
- in agreement with the equipartition theorem, the total energy should be
½kBT times the number of degrees of freedom.
The heat capacity for
a monatomic ideal gas:
f
 U 
CV  
  N kB
 T V , N 2
Partial Derivatives of the Entropy
We have been considering the entropy changes in the processes where two
interacting systems exchanged the thermal energy but the volume and the number of
particles in these systems were fixed. In general, however, we need more than just
one parameter to specify a macrostate, e.g. for an ideal gas
S  S U,V , N   kB ln  U,V , N 
When all macroscopic quantities S,V,N,U are allowed to vary:
 S
dS  
 U

S
 d U  
 N ,V
 V

 S 
 d V  
 d N

N
 N ,U

U ,V
We are familiar with the physical meaning only
one partial derivative of entropy:
Today we will explore what happens if we let the V
vary, and analyze the physical meaning of the
other two partial derivatives of the entropy:
1
 S 




U

V , N T
 S 


 V U , N
Mechanical Equilibrium
and Pressure
UA, VA, NA
e.g. ideal gas
UB, VB, NB
kB N P
 S 




V
T
 V U , N
S AB
Let’s fix UA,NA and UB,NB , but allow V to vary
(the membrane is insulating, impermeable for
gas molecules, but its position is not fixed).
Following the same logic, spontaneous
“exchange of volume” between sub-systems
will drive the system towards mechanical
equilibrium (the membrane at rest). The
equilibrium macropartition should have the
largest (by far) multiplicity  (U, V) and
entropy S (U, V).
In mechanical equilibrium:
S AB S A S B S A S B




0
VA VA VA VA VB
 VA   VB 
SA
SB
VAeq
VA
The stat. phys.
definition of pressure:
S A S B

V A VB
- the volume-per-molecule should be the same
for both sub-systems, or, if T is the same, P must
be the same on both sides of the membrane.
S 
S 



P T 
 

  V U , N   V U , N
 S 

/ 
  U V , N
The “Pressure” Equation of State for an Ideal Gas
The “energy” equation of state (U  T):
1  S 
f
1

  N kB
T  U V , N 2
U
Ideal gas:
(fN degrees of freedom)
S ( N ,V , T )  N k B ln
f
U  N kB T
2
V f
 N k B ln T   ( N , m)
N 2
The “pressure” equation of state (P  T):
N kB
 S 
P T 
 T
V
 V U , N
PV  N kBT
- we have finally derived the equation of state of an ideal gas from first principles!
Thermodynamic identity I
Let’s assume N is fixed,
 S 
 S 
dS  
 dU  
 dV
 U  N ,V
 V  N ,U
1
 S 

 
 U V , N T
thermal equilibrium:
P
 S 



 V U , N T
mechanical equilibrium:
1
P
 dS  dU  dV
T
T
i.e.
dU  TdS  PdV
Quasi-Static Processes
d U  Q  W
d U  T d S  P dV
(all processes)
(quasi-static processes with fixed N)
Thus, for quasi-static processes :
Q  T d S
Q
T
Q
T
P
Comment on State Functions :
dS
dS
- is an exact differential (S is a state function).
Thus, the factor 1/T converts Q into an exact
differential for quasi-static processes.
Quasistatic adiabatic (Q = 0) processes: d S  0
S  0
Q  0
V
 isentropic processes
The quasi-static adiabatic process with an ideal gas :
PV   const
VT  1  const
- we’ve derived these equations from the 1st
Law and PV=RT
On the other hand, from the Sackur-Tetrode equation for an isentropic process :
S  0  VT
f /2
 const
VT
1
 1
 const
Problem:
(all the processes are quasi-static)
(a) Calculate the entropy increase of an ideal gas in an isothermal process.
(b) Calculate the entropy increase of an ideal gas in an isochoric process.
You should be able to do this using (a) Sackur-Tetrode eq. and (b) d S 
T
f

dT  dV 
V
2

Q  dU  PdV  Nk B 
 U ,V , N   f N  V NU f N / 2
ST const
Vf
 Nk B ln 
 Vi



Q
T
Q
 f dT dV 
dS 
 Nk B 


T
V 
2 T

S  NkB ln g N VT f / 2
SV const
 Tf
f
 NkB ln 
2
 Ti




Let’s verify that we get the same result with approaches a) and b) (e.g., for T=const):
 Vf
N k BT
Q  W  
dV  N k BT ln 
V
 Vi
V
Vf
Since U = 0,
Q

  S 
T

(Pr. 2.34)
Problem:
A bacterias of mass M with heat capacity (per unit mass) C, initially at temperature
T0+T, is brought into thermal contact with a heat bath at temperature T0..
(a) Show that if T<<T0, the increase S in the entropy of the entire system (body+heat
bath) when equilibrium is reached is proportional to (T)2.
(b) Find S if the body is a bacteria of mass 10-15kg with C=4 kJ/(kg·K), T0=300K,
T=0.03K.
(c) What is the probability of finding the bacteria at its initial T0+T for t =10-12s over
the lifetime of the Universe (~1018s).
(a)
ΔS body 
 T0 
Q
CdT 

  0


C
ln




T T  T T
 T0  T 
T  T
T0
0
T0
ΔS heat bath 
0
Q

T0

T0
 CdT 
T0  T
T0

CT
0
T0
2
 T0  T  
 C  T 
T
2 3
  ln 1      
  0
ΔS total  ΔS body  ΔS heat bath  C
 C ln 

 ...  
T0
2
3
 2  T0 
 T0  
(b)
2
C  T  4 103 110 15 J / K  0.03 
 20
 
ΔS total  

  2 10 J / K
2  T0 
2
 300 
2
Problem (cont.)
(b)
 for the (non-equilibrium) state with Tbacteria = 300.03K is greater than 
in the equilibrium state with Tbacteria = 300K by a factor of
T0
T0  T
 Stotal 
 2 10 20 J / K  1450
630
  exp 

 exp 

e

10
 23

 1.38 10 J / K 
 kB 
The number of “1ps” trials over the lifetime of the Universe:
1018
 1030
12
10
Thus, the probability of the event happening in 1030 trials:
 # events probability of occurrence of an event   1030 10630  0