1Fundamentals of statistical mechanics

II Statistical Physics 1.2 Pressure, volume and the first law of thermodynamics
So far, our system only had one single parameter the energy. Usually, our
systems have other external parameters which can be varied. Recall that our
“standard” model of a statistical system is a box of gas. If we allow ourselves
to move the walls of the box, then the volume of the system may vary. As we
change the volume, the allowed energies eigenstates will change. So now Ω, and
hence S are functions of energy and volume:
S(E, V ) = k log Ω(E, V ).
We now need to modify our definition of temperature to account for this depen-
dence:
Definition
(Temperature)
.
The temperature of a system with variable volume
is
1
T
=
S
E
V
,
with V fixed.
But now we can define a different thermodynamic quantity by taking the
derivative with respect to V .
Definition
(Pressure)
.
We define the pressure of a system with variable volume
to be
p = T
S
V
E
.
Is this thing we call the “pressure” any thing like what we used to think of
as pressure, namely force per unit area? We will soon see that this is indeed the
case.
We begin by deducing some familiar properties of pressure.
Proposition.
Consider as before two interacting systems where the total volume
V
=
V
1
+
V
2
is fixed by the individual volumes can vary. Then the entropy of
the combined system is maximized when T
1
= T
2
and p
1
= p
2
.
Proof. We have previously seen that we need T
1
= T
2
. We also want
dS
dV
E
= 0.
So we need
dS
1
dV
E
=
dS
2
dV
E
.
Since the temperatures are equal, we know that we also need p
1
= p
2
.
For a single system, we can use the chain rule to write
dS =
S
E
V
dE +
S
V
E
dV.
Then we can use the definitions of temperature and pressure to write
Proposition (First law of thermodynamics).
dE = T dS p dV.
This law relates two infinitesimally close equilibrium states. This is sometimes
called the fundamental thermodynamics relation.
Example.
Consider a box with one side a movable piston of area
A
. We apply
a force F to keep the piston in place.
= F
dx
What happens if we move the piston for a little bit? If we move through a
distance d
x
, then the volume of the gas has increased by
A
d
x
. We assume
S
is
constant. Then the first law tells us
dE = pA dx.
This formula should be very familiar to us. This is just the work done by the
force, and this must be
F
=
pA
. So our definition of pressure in terms of partial
derivatives reproduces the mechanics definition of force per unit area.
One has to be cautious here. It is not always true that
p
d
V
can be equated
with the word done on a system. For this to be true, we require the change to
be reversible, which is a notion we will study more in depth later. For example,
this is not true when there is friction.
In the case of a reversible change, if we equate
p
d
V
with the work done,
then there is only one possible thing
T
d
S
can be it is the heat supplied to
the system.
It is important to remember that the first law holds for any change. It’s just
that this interpretation does not.
Example.
Consider the irreversible change, where we have a “free expansion”
of gas into vacuum. We have a box
gas
vacuum
We have a valve in the partition, and as soon as we open up the valve, the gas
flows to the other side of the box.
In this system, no energy has been supplied. So d
E
= 0. However, d
V >
0,
as volume clearly increased. But there is no work done on or by the gas. So in
this case,
p
d
V
is certainly not the work done. Using the first law, we know
that
T dS = p dV.
So as the volume increases, the entropy increases as well.
We now revisit the concept of heat capacity. We previously defined it as
d
E/
d
T
, but now we need to decide what we want to keep fixed. We can keep
V
fixed, and get
C
V
=
E
T
V
= T
S
T
V
.
While this is the obvious generalization of what we previously had, it is not a
very useful quantity. We do not usually do experiments with fixed volume. For
example, if we do a chemistry experiment in a test tube, say, then the volume is
not fixed, as the gas in the test tube is free to go around. Instead, what is fixed
is the pressure. We can analogously define
C
p
= T
S
T
p
.
Note that we cannot write this as some sort of
E
T
.