4Classical thermodynamics

II Statistical Physics

4.5 Thermodynamic potentials

So far, the state of gas is specified by

p

and

V

. These determine our other

quantities

T, E, S

. But we could instead use other variables. For example, when

drawing the T S plots, we are using T and S.

We can choose any 2 of

p, V, E, T, S

to specify the state. Recall that we had

dE = T dS − p dV.

So it is natural to regard

E

=

E

(

S, V

), or

S

=

S

(

E, V

). Taking derivates of

these expressions, we find

∂E

∂S

V

= T,

∂E

∂V

S

= −p.

In statical mechanics, we defined

T

and

p

this way, but here we derived it as a

consequence. It is good that they agree.

We now further differentiate these objects. Note that since derivatives are

symmetric, we know

∂

2

E

∂V ∂S

=

∂

2

E

∂S∂V

.

This tells us that

∂T

∂V

S

= −

∂p

∂S

V

.

Equations derived this way are known as Maxwell equations.

There are 3 more sets. We can, as before, define the Helmholtz free energy by

F = E − T S,

and then we have

dF = −S dT − p dV.

So we view F = F (T, V ).

What does this

F

tell us? For a reversible change at constant temperature,

we have

F (B) − F (A) = −

Z

B

A

p dV.

We now use the fact that the change is reversible to write this as

R

B

A

¯dW

, which

is the work done on the system. So the free energy measures the amount of

energy available do work at constant temperature.

Using this, we can express

∂F

∂T

V

= −S,

∂F

∂V

T

= −p.

Again taking mixed partials, we find that

∂S

∂V

T

=

∂p

∂T

V

.

We obtained

F

from

E

by subtracting away the conjugate pair

T, S

. We can

look at what happens when we mess with another conjugate pair, p and V .

We first motivate this a bit. Consider a system in equilibrium with a reservoir

R

of fixed temperature

T

and pressure

p

. The volume of the system and the

reservoir R can vary, but the total volume is fixed.

We can consider the total entropy

S

total

(E

total

, V

total

) = S

R

(E

total

− E, V

total

− V ) + S(E, V ).

Again, since the energy of the system is small compared to the total energy of

the reservoir, we can Taylor expand S

R

, and have

S

total

(E

total

, V

total

) = S

R

(E

total

, V

total

) −

∂S

R

∂E

V

E −

∂S

R

∂V

E

V + S(E, V ).

Using the definition of T and p, we obtain

S

total

(E

total

, V

total

) = S

R

(E

total

, V

total

) −

E + pV − T S

T

.

The second law of thermodynamics says we want to maximize this expression.

But the first term is constant, and we are assuming we are working at fixed

temperature and pressure. So we should minimize

G = F + pV = E + pV − T S.

This is the Gibbs free energy. This is important in chemistry, since we usually

do experiments in a test tube, which is open to the atmosphere. It is easy to

find that

dG = −S dT + V dp.

So the natural variables for G are G = G(T, p). We again find

S = −

∂G

∂T

p

, V =

∂G

∂p

T

.

Using the symmetry of mixed partials again, we find

−

∂S

∂p

T

=

∂V

∂T

p

.

We now look at the final possibility, where we just add

pV

and not subtract

T S

.

This is called the enthalpy

H = E + pV.

Then we have

dH = T dS + V dp.

So we naturally view H = H(S, p). We again have

T =

∂H

∂S

p

, V =

∂H

∂p

S

.

So we find

∂T

∂p

S

=

∂V

∂S

p

.

For the benefit of mankind, we collect all the Maxwell relations in the following

proposition

Proposition.

∂T

∂V

S

= −

∂p

∂S

V

∂S

∂V

T

=

∂p

∂T

V

−

∂S

∂p

T

=

∂V

∂T

p

∂T

∂p

S

=

∂V

∂S

p

.

Ideal gases

We now return to gases. We previously defined ideal gases as those whose atoms

don’t interact. But this is a microscopic definition. Before we knew atoms

existed, we couldn’t have made this definition. How can we define it using

classical thermodynamics?

Definition

(Ideal gas)

.

An ideal gas is a gas that satisfies Boyle’s law , which

says

pV

is just a function of

T

, say

pV

=

f

(

T

), and Joule’s law, which says

E

is a function of T .

These are true for real gases at low enough pressure.

Let’s try to take this definition and attempt to recover the equation of state.

We have

dE = T dS − p dV.

Since E is just a function of T , we know that

∂E

∂V

T

= 0.

In other words, plugging in the formula for dE, we find

T

∂S

∂V

T

− p = 0.

Using the Maxwell equations, we find

T

∂p

∂T

p

− p = 0.

Rearranging this tells us

∂p

∂T

V

=

p

T

.

Using Boyle’s law pV = f(T ), this tells us

f

0

(T )

V

=

f(T )

T V

.

So we must have

f(T ) = CT

for some C. So we have

pV = CT.

Since the left hand side is extensive, we must have

C ∝ N

. If we don’t know

about atoms, then we can talk about the number of moles of gas rather than

the number of atoms. So we have

C = kN

for some k. Then we obtain

pV = NkT,

as desired. In statistical physics, we found the same equation, where

k

is the

Boltzmann constant.

Carnot cycle for the ideal gas

Let’s now look at the Carnot cycle for ideal gases.

– On AB, we have dT = 0. So dE = 0. So ¯dQ = −¯dW . Then we have

Q

H

=

Z

B

A

¯dQ = −

Z

B

A

¯dW =

Z

B

A

p dV,

where we used the fact that the change is reversible in the last equality.

Using the equation of state, we have

Q

H

=

Z

B

A

NkT

V

dV = NkT

H

log

V

B

V

A

.

Similarly, we have

Q

C

= −NkT

C

log

V

C

V

D

.

–

Along

BC

, since the system is isolated, we have

¯dQ

= 0. So d

E

=

¯dW

=

−p

d

V

. From Joule’s law, we know

E

is just a function of temperature.

So we have

E

0

(T ) dT = −

NkT

V

dV.

We now have a differential equation involving

T

and

V

. Dividing by

NkT

,

we obtain

E

0

(T )

NkT

dT = −

dV

V

.

Integrating this from B to C, we find

Z

T

C

T

H

E

0

(T )

NkT

dT = −

Z

C

B

dV

V

.

So we find

log

V

C

V

B

=

Z

T

H

T

C

E

0

(T )

NkT

dT.

Note that the right hand side depends only on the temperatures of the hot

and cold reservoirs. So we have

log

V

D

V

A

= log

V

C

V

B

.

In other words, we have

V

D

V

A

=

V

C

V

B

.

Alternatively, we have

V

C

V

D

=

V

B

V

A

.

Finally, we calculate the efficiency. We have

η = 1 −

Q

C

Q

H

= 1 −

T

C

T

H

,

as expected. Of course, this must have been the case.

We can again talk about heat capacities. We again define

C

V

=

∂E

∂T

V

= T

∂S

∂T

V

.

We can also define

C

p

= T

∂S

∂T

p

.

We can use the Maxwell relations to find that

∂C

V

∂V

T

= T

∂

2

p

∂T

2

V

,

∂C

p

∂p

T

= −T

∂

2

V

∂T

2

p

.

More interestingly, we find that

C

p

− C

V

= T

∂V

∂T

p

∂p

∂T

V

.

For an ideal gas, we have pV = NkT . Then we have

∂V

∂T

p

=

Nk

p

,

∂p

∂T

V

=

Nk

V

.

Plugging this into the expression, we find

C

p

− C

V

=

T (Nk)

2

pV

= Nk.

So for any ideal gas, the difference between

C

p

and

C

V

is always the same. Note

that C

V

and C

p

themselves need not be constant.