Mechanical Equilibrium and Pressure
UA, VA, NA
UB, VB, NB
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
S A S B

VA 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.
for ideal gas
æ ¶S ö
kN P
=
ç ÷ =
è ¶V øU ,N V T
Relation between entropy and pressure:
æ ¶S ö
æ ¶S ö
æ ¶S ö
P ºT ç ÷ =ç ÷ /ç ÷
è ¶V øU,N è ¶V øU,N è ¶U øV,N
This definition implies that you are holding both the internal energy and the
number of particles constant in taking the derivative. This can be applied to
the expression for the entropy of a monoatomic ideal gas:
This relationship is just the ideal gas law! But the ideal gas law can be
obtained from just Newton's laws, which give an expression for average
pressure of a gas. That along with the kinetic temperature gives the form of
the gas law. So, looking at it from the other direction, the ideal gas law offers
confirmation of the relationship between pressure and entropy.
We have finally derived the equation of state of an ideal
gas from first principles!
Thermodynamic identity
Let’s assume N is fixed:
In thermal equilibrium:
In mechanical equilibrium:
 S 
 S 
dS  
dU



 dV
 U  N ,V
 V  N ,U
1
 S 




U

V , N T
P
 S 

 

V

U , N T
1
P
 dS  dU  dV
T
T
dU  TdS  PdV
This useful summary relationship called the thermodynamic identity makes
use of the power of calculus and particularly partial derivatives. It may be
applied to examine processes in which one or more state variables is held
constant, e.g., constant volume, constant pressure, etc. The thermodynamic
identity holds true for any infinitesmal change in a system so long at the
pressure and temperature are well defined. It is presumed that the number of
particles is constant (i.e., you are dealing with the same system before and
after the change).
Quasistatic Processes
d U  Q  W
d U  T d S  P dV
(quasistatic processes with fixed N)
(all processes)
Thus, for quasistatic processes :
We see that the above relation is valid even if the work is being done, provided
that the pressure is well-defined and uniform throughout the system. In particular,
for the adiabatic process
adiabatic + quasistatic = isentropic
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 for quasi-static processes.
S  0
P
Q  0
V
From the Sackur-Tetrode equation for an isentropic process :
S  0  VT
f /2
 const
VT
1
 1
 const
Example
Bacterias of mass M with heat capacity (per unit mass) C, initially at
temperature T0+T, arebrought 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
 0

T  T
T  T
T0
T0
0
0
Q

T0
æ T0 + DT ö
DT
DStotal = DSbody + DSheat bath = C
- C ln ç
÷
T0
è T0 ø
ΔS heat bath 

T0
 CdT 
T0  T
T0

CT
0
T0
é
a 2 a 3 ù C æ DT ö
» êln (1+ a ) = a - + -...ú = ç ÷ > 0
2 3
ë
û 2 è T0 ø
2
2
(b)
(c)
C  T  4 103 110 15 J / K  0.03 
 20
 
ΔS total  

  2 10 J / K
2  T0 
2
 300 
2
 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 
Thus, the probability of the event happening in 1030 trials:
 # events probability of occurrence of an event   1030 10630  0