4Classical thermodynamics

II Statistical Physics

4.4 Entropy

We’ve been talking about the second law in the previous chapter all along, but

we didn’t mention entropy! Where does it come from?

Recall that in the Carnot cycle, we had

η = 1 −

Q

C

Q

H

= 1 −

T

C

T

H

.

So we get

Q

H

T

H

=

Q

C

T

C

.

But

Q

H

is the heat absorbed, and

Q

C

is the heat emitted. This is rather

asymmetric. So we set

Q

1

= Q

H

T

1

= T

H

Q

2

= −Q

C

T

2

= T

C

.

Then both Q

i

are the heat absorbed, and then we get

2

X

i=1

Q

i

T

i

= 0.

We now consider a more complicated engine.

V

p

C

B

D

A

E

F

G

We want to consider AEFGCD as a new engine. We saw that Carnot ABCD had

Q

AB

T

H

+

Q

CD

T

C

= 0. (1)

Similarly, Carnot EBGF gives

Q

EB

T

H

+

Q

GF

T

M

= 0. (2)

But we have

Q

AB

= Q

AE

+ Q

EB

, Q

GF

= −Q

F G

.

So we can subtract (1) − (2), and get

Q

AE

T

H

+

Q

F C

T

M

+

Q

CD

T

C

= 0.

But this is exactly the sum of the

Q

T

for the Carnot cycle AEFGCD. So we again

find

3

X

i=1

Q

i

T

i

= 0.

If we keep doing these corner cuttings, then we can use it to approximate it any

simple closed curve in the plane. So we learn that

I

¯dQ

T

= 0

for any reversible cycle. This allows us to define a new function of state, called

the entropy.

Definition (Entropy). The entropy of a system at A = (p, V ) is given by

S(A) =

Z

A

0

¯dQ

T

,

where 0 is some fixed reference state, and the integral is evaluated along any

reversible path.

We next want to see that this entropy we’ve defined actually has the properties

of entropy we’ve found previously.

It is immediate from this definition that we have

T dS = ¯dQ.

We also see that for reversible changes, we have

−p dV = ¯dW.

So we know that

dE = T dS − p dV,

which is what we previously called the first law of thermodynamics.

Since this is a relation between functions of state, it must hold for any

infinitesimal change, whether reversible or not!

Since the same first law holds for the entropy we’ve previously defined, this

must be the same entropy as the one we’ve discussed previously, at least up to a

constant. This is quite remarkable.

Historically, this was how we first defined entropy. And then Boltzmann

came along and explained the entropy is the measure of how many states we’ve

got. No one believed him, and he committed suicide, but we now know he was

correct.

Let’s compare an irreversible Ivor and the Carnot cycle between

T

H

and

T

C

.

We assume the Carnot cycle does the same work as Ivor, say W . Now we have

W = Q

0

H

− Q

0

C

= Q

H

− Q

C

.

By Carnot’s theorem, since Carnot is more efficient, we must have

Q

0

H

> Q

H

.

Note that we have

Q

0

H

T

H

−

Q

0

C

T

C

=

Q

0

H

T

H

+

Q

H

− Q

C

− Q

0

H

T

C

=

Q

H

T

H

−

Q

C

T

C

+ (Q

0

H

− Q

H

)

1

T

H

−

1

T

C

= (Q

0

H

− Q

H

)

1

T

H

−

1

T

C

.

The first term is positive, while the second term is negative. So we see that the

sum of

Q

T

is negative.

The same method of “cutting off corners” as above shows that

I

¯dQ

T

≤ 0

for any cycle. This is called Clausius inequality.

Now consider two paths from

A

to

B

, say

I

and

II

. Suppose

I

is irreversible

and II is reversible.

V

p

A

B

II

I

Then we can run II backwards, and we get

Z

I

¯dQ

T

−

Z

II

¯dQ

T

=

I

¯dQ

T

≤ 0.

So we know that

Z

I

¯dQ

T

≤ S(B) − S(A).

Let’s now assume that

I

is adiabatic. In other words, the system is isolated and

is not absorbing any heat. So

¯dQ

= 0. So the left hand side of the inequality is

0. So we found that

S(B) ≥ S(A)

for isolated systems! This is precisely the way we previously stated the second

law previously.

If the change is reversible, then we obtain equality. As an example of this,

we look at the Carnot cycle. We can plot this in the

T S

plane. Here the Carnot

cycle is just a rectangle.

S

T

A B

CD

We see that the periods of constant entropy are exactly those when the system

is adiabatic.

We can look at some examples of non-reversible processes.

Example. Suppose we have a box of gas:

gas

vacuum

On the left, there is gas, and on the right, it is vacuum. If we remove the

partition, then the gas moves on to fill the whole box. If we insulate this good

enough, then it is adiabatic, but it is clearly irreversible. This is not a quasi-static

process.

In this scenario, we have

¯dQ

= 0 =

¯dW

. So we know that d

E

= 0, i.e.

E

is

constant. Except at low temperatures, we find that the temperature does not

change. So we find that

E

is a function of temperature. This is Joule’s law. We

can plot the points in a T S plane:

S

T

initial final

Note that it doesn’t make sense to draw a line from the initial state to the final

state, as the process is not quasi-static.