Lecture 8. Thermodynamic Identities (Ch. 3)
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:
S  kB ln  U ,V , N 
We are familiar with only one partial derivative of entropy:
1
 S 



 U V , N T
Today we will explore what happens if we let the other two parameters (V
and N) vary, and analyze the physical meaning of two partial derivatives of
the entropy:
 S 


 V U , N
 S 


 N U ,V
General Approach:
S  S U,V , N 

if monotonic as a function of U
(“quadratic” degrees of freedom!), may
be inverted to give
U  U S,V , N 
When all macroscopic quantities S,V,N,U are allowed to vary:
 S 
S 
 S 
 d U  
 d V  
 d N
dS  
  U  N ,V
  V  N ,U
  N U ,V
 U 
 U 
 U 
dU  
 dS  
 dV  
 dN
 S V , N
 V  S , N
 N  S ,V
So far, we have abbreviated one of these partial derivatives:
The other partial derivatives are:
 U

 S

  T
 N ,V
 U

 V

   P
 N ,S
pressure
 S

 U

1


 N ,V T
 U 

  
  N V , S
chemical potential
Thermodynamic Identities
With these abbreviations:
d U  T d S  P dV   d N
- the so-called
thermodynamic identity
 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.
 is an intensive variable, independent of the size of the system (like P, T, density).
Extensive variables (U, N, S,V...) have a magnitude proportional to the size of the
system. If two identical systems are combined into one, each extensive variable is
doubled in value.
The thermodynamic identity holds for the quasi-static processes (T, P,  are welldefine throughout the system)
The 1st Law for quasi-static processes (N = const):
 S 
S 
 S 
 d U  
 d V  
 d N
dS  
  U  N ,V
  V  N ,U
  N U ,V
d U  T d S  P dV
dS 
1
P

d U  dV  d N
T
T
T
This identity holds for small changes S provided T and P are well defined.
The coefficients may
be identified as:
 S

 U

1
 
 N ,V T
S 
P



  V  N ,U T
 S 


  
T
  N V ,U
Quasi-static Adiabatic Processes
Let’s compare two forms of the 1st Law :
d U  T d S  P dV
d U  Q  W
(quasi-static processes, N is fixed, and
P is uniform throughout the system)
(holds for all processes)
Q  T d S
Thus, for quasi-static processes :
Quasistatic adiabatic (Q = 0) processes:
In this case,
d U  T d S  P dV
dS
Q
T
d S  0  isentropic processes
reduces to
d U   P dV
- the change in internal energy is due to a “purely mechanical” compression (or
expansion) that involves no change in entropy. (No entropy change = no heat flow).
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
V
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
holds for all processes,
energy conservation
U  TS  PV
quasistatic, T and P are welldefined for any intermediate state
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




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.
Direct approach:
S U , V , N   N k B ln V 
Q0
adiabatic quasistatic  isentropic
f
N k BT
N k B T  
V
2
V
VT
f /2
f
N k B ln U  k B ln f  N 
2
U   W
 const
Vf
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 T f  Ti  Pi Vi
 N kB 


2 Ti 
Ti
 Vi
105 Pa 10-3 m 3 10-2
1


J/K
300K
300
V
 10  2
V
U  W  2 J
Ui 
5 2 

2

10



2 250 

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 adiabata 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 2J  1J
1


J/K
T
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
Comment on State Functions
dS
Q
T
is an exact differential (S is a state function). Thus,
the factor 1/T converts Q into an exact differential.
U, S and V are the state functions, Q and W are not. Specifying an initial and final states
of a system does not fix the values of Q and W, we need to know the whole process
(the intermediate states). In math terms, Q and W are not exact differentials and we
need to use  instead of d for their increments. Analogy: in classical mechanics, if a force
is not conservative (e.g., friction), the initial and final positions do not determine the work,
the entire path must be specified.
y
 a  Ax x, y dx  Ay x, y dy
z(x1,y1)
z(x2,y2)
x
- it is an exact differential if it is
the difference between the values of some (state) function
z(x,y) at these points:
 a  zx  dx, y  dy   zx, y 
Ax x, y  Ay x, y 

y
x
 z 
 z 
d z    dx    dy
 x  y
 y  x
A necessary and sufficient condition for this:
If this condition
holds:
e.g., for an ideal gas:
z x, y 
y
T
f

Q  dU  PdV  Nk B  dT  dV 
V
2

Ax x, y  
z x, y 
x
Ay x, y  
- cross derivatives
are not equal
Entropy Change for Different Processes
The partial derivatives of S play very important roles because they determine
how much the entropy is affected when U, V and N change:
Type of
interaction
Exchanged
quantity
Governing
variable
thermal
energy
temperature
1  S 


T  U V , N
volume
pressure
P  S 


T  V U , N
particles
chemical
potential
mechanical
diffusive
Formula

 S 
 

T

N

U ,V
The last column provides the connection between statistical physics and
thermodynamics.
Temperature
UA, VA, NA
UB, VB, NB
To discuss the physical meaning of these
partial derivatives of S, let’s consider an
ideal gas isolated from the environment, and
consider two sub-systems, A and B,
separated by a membrane, which, initially, is
insulating, impermeable for gas molecules,
and fixed at a certain position. Thus, the
sub-systems are isolated from each other.
(We have already considered such a system
when we discussed T, but we want to
proceed with P and )
Assume that the membrane becomes thermally conducting (allow exchange U while
V and N remain fixed). The system will evolve to the state of equilibrium that is
characterized by a maximum entropy. One of the sub-systems can increase its
energy only at the expense of the other sub-system, and it’s crucial that the subsystem receiving energy increases its entropy more than the donor loses. Thus, the
sub-system with a larger S/U should
S AB S A S B
S A S B
receive energy, and this process will continue




0

U

U

U

U

U
until SA/UA and SB/UB become the same:
A
A
A
A
B
This was our logic when
we defined the stat.
phys. temperature as:
1
 S 

 
 U V , N T
- the energy should flow from
higher T to lower T; in thermal
equilibrium, TA and TB should be
the same.
Mechanical Equilibrium
and Pressure
UA, VA, NA
UB, VB, NB
P k N
 S 
  B


V
 V U , N T
Let’s now 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 Equation(s) of State for an Ideal Gas
 U ,V , N   g N  V NU f N / 2
Ideal gas:
(fN degrees of freedom)

S U ,V , N   N k B ln g N V U f / 2

The “energy” equation of state (U  T):
1  S 
f
1

  N kB
T  U V , N 2
U
U
f
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!
In other words, we can calculate the thermodynamic information for an isolated
system by counting all the accessible microstates as a function of N, V, and U.
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.
(c) What is the process in which the entropy remains constant?
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 
T
Q
 f dT dV 
dS 
 dU  PdV  Nk B 


T
V 
2 T

 U ,V , N   f N  V NU f N / 2
(a)
Vf
S  NkB ln 
 Vi
(c)
S  0  VT f / 2  const
Q  0



an adiabatic process
Q
S  NkB ln g N VT f / 2
(b)
 Tf
f
S  NkB ln 
2
 Ti




VT  1  const
a quasistatic adiabatic process
= an isentropic process
Diffusive Equilibrium
and Chemical Potential
UA, VA, SA
UB, VB, SB
Let’s fix VA and VB (the membrane’s position is
fixed), but assume that the membrane becomes
permeable for gas molecules (exchange of both
U and N between the sub-systems, the
molecules in A and B are the same ).
For sub-systems in
diffusive equilibrium:
 S AB 


0

U
A V A , N A

 S A  SB

 N A  NB
 S AB 


0
 N A U A , V A

 S 

 
T
 N U ,V
S
In equilibrium,
TA  TB
 A  B
NA
 S 

 N U ,V
  T 
UA
- the chemical
potential
Sign “-”: out of equilibrium, the system with the larger S/N will get more
particles. In other words, particles will flow from from a high /T to a low /T.
Chemical Potential: examples
Einstein solid:
 U 

  N V , S
  
consider a small one, with N = 3 and q = 3.
 ( N  3, q  3) 
q  N  1 !  10
q !( N  1) !
let’s add one more oscillator:
S ( N  3, q  3)  kB ln 10
S ( N  4, q  3)  kB ln 20
1

d U  d N To keep dS = 0, we need to decrease the
energy, by subtracting one energy quantum.
T
T
 U 
  
  
Thus, for this system
  N S
 U 
 
  N S
  
dS 
Monatomic ideal gas:
3/ 2

5
   4 m  

5/ 2
S ( N ,V ,U )  N k B ln V 
U

ln
N




2
3
h
2





 

3/ 2
 V  2 m

VT 3 / 2 
 S 
 
  T 
   k BT ln   2 k BT     k BT ln  g m 

N 
 N U ,V
 
 N  h

At normal T and P, ln(...) > 0, and  < 0 (e.g., for He,  ~ - 5·10-20 J ~ - 0.3 eV.
Sign “-”: usually, by adding particles to the system, we increase its entropy.
To keep dS = 0, we need to subtract some energy, thus U is negative.
The Quantum Concentration
3/ 2
3/ 2
2


 V  2 m

 h3


h
P 


   k BT ln 
   k BT ln   2 k BT    k BT ln n

3/ 2
5/ 2 

k BT  
 
  2  mkBT  
 N  h
 2m

n=N/V – the concentration of molecules
0
The chemical potential increases with the density of the gas or with
its pressure. Thus, the molecules will flow from regions of high
density to regions of lower density or from regions of high pressure
to those of low pressure .
3/ 2
2
 

 n 

h


  k BT ln  
  k BT ln n

n 
  2  mkBT  
 Q
 2 m

nQ   2 k BT 
 h

dB
3/ 2

when n
increases
when n  nQ,   0
- the so-called quantum concentration (one particle per
cube of side equal to the thermal de Broglie wavelength).
When nQ >> n, the gas is in the classical regime.
h
h
 
p
mkBT
nQ 
1
dB 3
 mkBT 


2
 h 
3/ 2
At T=300K, P=105 Pa , n << nQ. When n  nQ, the quantum statistics comes into play.
Ideal Gas in a Gravitational Field
Pr. 3.37. Consider a monatomic ideal gas at a height z above sea level, so each
molecule has potential energy mgz in addition to its kinetic energy. Assuming that
the atmosphere is isothermal (not quite right), find  and re-derive the barometric
equation.
U  U kin  Nmgz
note that the U that appears in the Sackur-Tetrode
equation represents only the kinetic energy
d U  T d S  P dV   d N
3/ 2
 V  2 m
 dU 
 
   ( z  0)  mgz  k BT ln   2 k BT    mgz
  
 
 N  h
 d N  S ,V
In equilibrium, the two chemical potentials must be equal:
3/ 2
3/ 2
 V  2 m
 V  2 m
 
 
 k BT ln 
 2 k BT    mgz  k BT ln 
 2 k BT  
 
 
 N ( z )  h
 N (0)  h
kBT ln N ( z)  mgz  kBT ln N (0)
N ( z )  N ( 0) e

mgz
k BT
S and CP
Q  T d S - this is a useful result that removes the restriction of constant volume.
Tf

For P = const processes, Q = CPdT
S P   CP dT
Ti
T
Pr. 3.29. Sketch a graph of the entropy of H20 as a function of T at P = const.
S P  C P dT
T

C
 S 

  P
 T  P T
S
At T 0, the graph goes to 0 with zero slope. After
initial fast increase of the slope, the rate of the S
increase slows down (CP – almost const). When
solid melts, there is a large S at T = const,
another jump – at liquid–gas phase transformation.
“Curving down” in both liquid and gas phases –
because CP  const.
ice
water
vapor
T
Future Directions
Although the microcanonical ensemble is conceptually simple, it is not
the most practical ensemble. The major problem is that we must specify
U - isolated systems are very difficult to realize experimentally, and T
rather than U is a more natural independent variable.
Ensemble
Macrostate
microcanonical
U, V, N
(T fluctuates)
canonical
T, V, N
(U fluctuates)
grand
canonical
U, V, 
(N fluctuates)
Probability
Pn 
S U ,V , N   kB ln 
1

En
1  kB T
Pn  e
Z
1 
Pn  e
Z
Thermodynamics
 En   N n 
kB T
F T ,V , N   kB T ln Z
 T ,V ,    kB T ln Z