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
The total number
of microstates:
-L
L x
px
-px
The number of microstates:
x

Q.M.
L p x
L p x

x  p x

   space p

L p x
L

x  p x x
Quantum mechanics (the
uncertainty principle) helps
us to numerate all different
states in the configuration
(phase) space:
x  px  
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  p3 
 

   space p   3
3 
3

  
 x  p x 
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
N
1  V  p3 

 indistinguishable  
3
N!   
N
pz
More Accurate Calculation of p
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!  

Monatomic
ideal gas:
(3N degrees
of freedom)
r
3N / 2
2p

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.
 e 3m V  U 
indistinguishable U ,V , N   
 
3
h
N
N

5/ 2
U   U f N / 2

2 3 / 2 2
r  1 / 2!  / 2  4 r 2
1 / 2!
3/ 2



N
2p
f N- the total # of “quadratic” degrees of freedom
Entropy of an Ideal Gas
1 V N  2 mU 
N 


2
N! 3N / 2!  

3N / 2
2p
 V  4  m U 3 / 2  5
S ( N ,V ,U )  N k B ln  
   N k B  ln 2p 
2
 N  3h N   2
Monatomic
ideal gas:
3/ 2


 V  4 m U   5 

S ( N ,V ,U )  N k B ln  




2
N
3
h
N
2
 


  

the SackurTetrode
equation
 V  U 3 / 2  3
5
V 3
U
 4  m 
 N k B ln      N k B   ln 

N
k
ln

N
k
ln
  ( N , m)

B
B

2
N
N
2
3
3
h
N
2
N



   
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  S A  S B   N1  N 2  ln  1 2   N1 ln  1   N 2 ln  2 
 N1  N 2 
 N1 
 N2 
 V N1 
 V N2 


  0
 N1 ln 
 N 2 ln 

 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
other two parameters (V and N) 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
 S 


 N U ,V
Thermodynamic Identity for dU(S,V,N)

if monotonic as a function of U
(“quadratic” degrees of freedom!), may
be inverted to give
S  S U,V , N 
U  U S,V , N 
 U 
 U 
 U 
dU  
 dS  
 dV  
 dN

S

V

N

V , N

S ,N

 S ,V
 U

 S
compare with

  T
 N ,V
 S

 U
 U

 V

1


 N ,V T
d U  T d S  P dV   d N

   P
 N ,S
pressure
 U 

  
  N V , S
chemical potential

shows how much the system’s energy
changes when one particle is added to the
system at fixed S and V. The chemical
potential units – J.
- the so-called thermodynamic
identity for U
This holds for quasi-static processes (T, P,  are well-define throughout the system).
The Exact Differential of S(U,V,N)
d U  T d S  P dV   d N
 S
dS U , V , N   
 U
dS 
1
P

d U  dV  d N
T
T
T

S 
 S 
 d U  
 d V  
 d N
 N ,V
  V  N ,U
  N U ,V
The coefficients may
be identified as:
 S

 U

1
 
 N ,V T
S 
P



  V  N ,U T
 S 


  
T
  N V ,U
Again, this holds for quasi-static processes (T and P are well defined).
Type of
interaction
Exchanged
quantity
Governing
variable
Formula
thermal
energy
temperature
1  S 


T  U V , N
mechanical
volume
pressure
P  S 


T  V U , N
diffusive
particles
chemical
potential

 S 
 

T
 N U ,V
connection
between
thermodynamics
and statistical
mechanics
Mechanical Equilibrium
and Pressure
UA, VA, NA
UB, VB, NB
P k N
 S 
  B


V
 V U , N T
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).
- the sub-system with a smaller volume-per-molecule
(larger P at the same T) will have a larger S/V, it will
expand at the expense of the other sub-system.
In mechanical
equilibrium:
S AB
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
Ideal gas:
(fN degrees of freedom)
S ( N ,V , T )  N k B ln
V f
 N k B ln T   ( N , m)
N 2
The “energy” equation of state (U  T):
1  S 
f
1

  N kB
T  U V , N 2
U
f
U  N kB T
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!
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
 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
V
VT  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 body 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
An example of a non-quasistatic adiabatic process
Caution: for non-quasistatic adiabatic processes, S might be non-zero!!!
P
2
S = const
along the
isentropic
line
2*
1
Vf
Vi
Pr. 3.32. A non-quasistatic compression. A cylinder with
air (V = 10-3 m3, T = 300K, P =105 Pa) is compressed (very
fast, non-quasistatic) by a piston (S = 0.01 m2, F = 2000N,
x = 10-3m). Calculate W, Q, U, and S.
U   Q   W
U  TS  PV
V
quasistatic adiabatic  isentropic
W 
Vf


 P(V )dV  PV  const  PiVi

Vi
PiVi  1
1  PiVi  Vi




  1 V f  1 Vi  1    1  V f

 1

PiVi 
x 

1


1



  1 
x 


 PiVi




holds for all processes,
energy conservation
quasistatic, T and P are welldefined for any intermediate state
Q = 0 for both

non-quasistatic adiabatic
Vf
1
V V  dV
i
 1

 1


x
 1 105 Pa  10 3 m3  10  2  1 J
x
Vf
 W   P  dV  P Vi  V f 
Vi
 2 105 Pa 10 2 m 2 10 3 m 2
 2J
The non-quasistatic process results
in a higher T and a greater entropy
of the final state.
S U , V , N   N k B ln V 
Direct approach:
Q0
adiabatic quasistatic  isentropic
f
N k BT
N k B T  
V
2
V
VT
Vf
f /2
f
N k B ln U  k B ln f  N 
2
U   W
 const
f
2 Vf
S  N k B ln
 N k B ln
0
Vi 2
f Vi
f
N k B T   PV
2
2/ f
T f  Vi 
 
Ti  V f 
adiabatic non-quasistatic
Vf
Uf
f
S  N k B ln
 N k B ln
Vi 2
Ui
V f  Vi f U f  U i  Pi Vi 
5 2 
2
 N kB 



10





V
2
U
T
2
250


i
i
i


105 Pa  10 -3 m3  10 -2
1


J/K
300K
300
V
 10  2
V
U  W  2 J
Ui 
f
5
5
N k BTi  Pi Vi  105 Pa 10-3 m 3  250 J
2
2
2
P
2
 U =  Q = 1J
To calculate S, we can consider any quasistatic
process that would bring the gas into the final state (S is
a state function). For example, along the red line that
coincides with the adiabat and then shoots straight up.
Let’s neglect small variations of T along this path ( U <<
U, so it won’t be a big mistake to assume T  const):
1
Vf
Vi
S  0 (adiabata) 
V
Q
T

2J  1J
1

J/K
300K
300
The entropy is created because it is an irreversible,
non-quasistatic compression.
P
2
U = Q = 2J
For any quasi-static path from 1 to 2, we must have the
same S. Let’s take another path – along the isotherm
and then straight up:
1
isotherm: S 
T
1
Vf
Vi
V
Total gain of entropy:
Vf
Pi Vi
P
(
V
)
dV


T
Vi
Pi Vi  V f 
dV

V V T ln  Vi 
i
Vf
Pi Vi x
105 Pa 10-3 m 3 10-2
1



J/K
T x
300K
300
U Q
2J
2
“straight up”:
S 
 

J/K
T
T 300K 300
1
2
1
S  
J/K
J/K 
J/K
300
300
300
P
The inverse process, sudden expansion of an ideal gas
(2 – 3) also generates entropy (adiabatic but not
quasistatic). Neither heat nor work is transferred: W =
Q = 0 (we assume the whole process occurs rapidly
enough so that no heat flows in through the walls).
2
3
1
Vf
Vi
V
Because U is unchanged, T of the ideal gas is unchanged.
The final state is identical with the state that results from a
reversible isothermal expansion with the gas in thermal
equilibrium with a reservoir. The work done on the gas in
the reversible expansion from volume Vf to Vi:
Wrev  N k BT ln
Vf
Vi
The work done on the gas is negative, the gas does positive work on the piston in an
amount equal to the heat transfer into the system
Qrev  Wrev  0
S 
Qrev  Wrev
V
P V x
1

 N k B ln i  i i

J/K
T
T
Vf
Ti x 300
Thus, by going 1  2  3 , we will increase the gas entropy by
S123 
2
J/K
300