2Classical gases

II Statistical Physics

2.2 Monoatomic ideal gas

We now begin considering ideal gases.

Definition

(Ideal gas)

.

An ideal gas is a gas where the particles do not interact

with each other.

Of course, this is never true, but we can hope this is a good approximation

when the particles are far apart.

We begin by considering a monoatomic ideal gas. These gases have no

internal structure, and is made up of single atoms. In this case, the only energy

we have is the kinetic energy, and we get

H =

p

2

2m

.

We just have to plug this into our partition function and evaluate the integral.

We have

Z

1

(V, T ) =

1

(2π~)

3

Z

d

3

p d

3

q e

−βp

2

/2m

=

V

(2π~)

3

Z

d

3

p e

−βp

2

/2m

Here

V

is the volume of the box containing the particle, which is what we obtain

when we do the d

3

q integral.

This remaining integral is just the Gaussian integral. Recall that we have

Z

dx e

−ax

2

=

r

π

a

.

Using this three times, we find

Proposition. For a monoatomic gas, we have

Z

1

(V, T ) = V

mkT

2π~

2

3/2

=

V

λ

3

,

where we define

Definition

(Thermal de Broglie wavelength)

.

The thermal de Broglie wavelength

of a gas at temperature T is

λ =

r

2π~

2

mkT

.

If we think of the wavelength as the “size” of the particle, then we see that

the partition function counts the number of particles we can fit in the volume

V

.

We notice that this partition function involves

~

, which is a bit weird since we

are working classically, but we will see that the

~

doesn’t appear in the formulas

we derive from this.

The generalization to multiple particles is straightforward. If we have

N

particles, since the partition function is again multiplicative, we have

Z(N, V, T ) = Z

N

1

= V

N

λ

−3N

.

There is a small caveat at this point. We will later see that this is not quite right.

When we think about the quantum version, if the particles are indistinguishable,

then we would have counted each state

N

! times, which would give the wrong

answer. Again, this doesn’t affect any observable quantity, so we will put this

issue aside for the moment, until we get to studying the entropy itself, in which

case this N! does matter.

We can similarly define the pressure to be

p = −

∂F

∂V

T

=

∂

∂V

(kT log Z)

T

.

Then plugging our partition function into this definition, we find

p =

NkT

V

.

Rearranging, we obtain

Proposition (Ideal gas law).

pV = NkT.

Notice that in this formula, the

λ

has dropped out, and there is no dependence

on ~.

Definition

(Equation of state)

.

An equation of state is an equation that relates

state variables, i.e. variables that depend only on the current state of the system,

as opposed to how we obtained this system.

The ideal gas law is an example of an equation of state.

Let’s now look at the energy of the ideal gas. We can similarly compute

hEi = −

∂ log Z

∂β

V

=

3

2

NkT = 3N

1

2

kT

.

This is a general phenomenon. Our system has

N

particles, and each particle

has three independent directions it can move in. So there are 3

N

degrees of

freedom.

Law

(Equipartition of energy)

.

Each degree of freedom of an ideal gas contributes

1

2

kT to the average energy.

In the next section, we will study gases with internal structure, hence internal

degrees of freedom, and each such degree of freedom will again contribute

1

2

kT

to the average energy.

Of course, this law requires some hidden assumptions we do not make precise.

If we add a degree of freedom

s

with a term

s

5.3

log

(2

s

+ 1) in the Hamiltonian,

then there is no reason to believe the contribution to the average energy would

still be

1

2

kT

. We will also see in the next section that if the degree of freedom

has some potential energy, then there will be even more contribution to the

energy.

There are other quantities of the gas we can compute. We know the average

energy of a single particle is

hp

2

i

2m

=

3

2

kT.

So we have

hp

2

i ∼ mkT.

Thus, for a single particle, we have

|p| ∼

√

mkT ,

and so

λ ∼

h

|p|

.

This is the usual formula for the de Broglie wavelength. So our thermal de

Broglie wavelength is indeed related to the de Broglie wavelength.

Finally, we can compute the heat capacity

C

V

=

∂E

∂T

V

=

3

2

Nk.

Boltzmann’s constant

Recall that Boltzmann’s constant is

k = 1.381 ×10

−23

J K

−1

This number shows that we were terrible at choosing units. If we were to invent

physics again, we would pick energy to have the same unit as temperature, so

that

k

= 1. This number

k

is just a conversion factor between temperature and

energy because we chose the wrong units.

But of course, the units were not chosen randomly in order to mess up our

thermodynamics. The units were chosen to relate to scales we meet in everyday

life. So it is still reasonable to ask why

k

has such a small value. We look at the

ideal gas law.

pV

T

= Nk.

We would expect when we plug in some everyday values for the left hand side,

the result would be somewhat sensible, because our ancestors were sane when

picking units (hopefully).

Indeed, we can put in numbers

p = 10

5

N m

−2

V = 1 m

3

T = 300 K,

and we find that the LHS is ∼ 300.

So what makes

k

such a tiny number is that

N

is huge. The number of

particles is of the order 10

23

. Thus, for Nk to have a sensible value, k must be

tiny.

The fact that

k

is small tells us that everyday lumps of matter contain a lot

of particles, and in turn, this tells us that atoms are small.

Entropy

The next thing to study is the entropy of an ideal gas. We previously wrote

down

Z = Z

N

1

,

and briefly noted that this isn’t actually right. In quantum mechanics, we know

that if we swap two indistinguishable particles, then we get back the same state,

at least up to a sign. Similarly, if we permute any of our particles, which are

indistinguishable, then we get the same system. We are over-counting the states.

What we really should do is to divide by the number of ways to permute the

particles, namely N!:

Z =

1

N!

Z

N

1

.

Just as in the constant

h

in our partition function, this

N

! doesn’t affect any of

our observables. In particular,

p

and

hEi

are unchanged. However, this

N

! does

affect the entropy

S =

∂

∂T

(kT log Z).

Plugging the partition function in and using Stirling’s formula, we get

S = Nk

log

V

Nλ

3

+

5

2

.

This is known as the Sackur-Tetrode equation.

Recall that the entropy is an extensive property. If we re-scale the system by

a factor of α, then

N 7→ αN, V 7→ αV.

Since

λ

depends on

T

only, it is an intensive quantity, and this indeed scales

as

S 7→ αS

. But for this to work, we really needed the

N

inside the logarithm,

and the reason we have the

N

inside the logarithm is that we had an

N

! in the

partition function.

When people first studied statistical mechanics of ideal gases, they didn’t

know about quantum mechanics, and didn’t know they should put in the

N

!.

Then the resulting value of

S

is no longer extensive. This leads to Gibbs paradox .

The actual paradox is as follows:

Suppose we have a box of bas with entropy

S

. We introduce a partition

between the gases, so that the individual partitions have entropy

S

1

and

S

2

.

Then the fact that the gas is not extensive means

S 6= S

1

+ S

2

.

This means by introducing or removing a partition, we have increased or decreased

the entropy, which violates the second law of thermodynamics.

This

N

!, which comes from quantum effects, fixes this problem. This is a

case where quantum mechanics is needed to understand something that really

should be classical.

Grand canonical ensemble

We now consider the case where we have a grand canonical ensemble, so that

we can exchange heat and particles. In the case of gas, we can easily visualize

this as a small open box of gas where gas is allowed to freely flow around. The

grand ensemble has partition function

Z

ideal

(µ, V, T ) =

∞

X

N=0

e

βµN

Z

ideal

(N, V, T )

=

∞

X

N=0

1

N!

e

βµ

V

λ

3

N

= exp

e

βµ

V

λ

3

Armed with this, we can now calculate the average number of particles in our

system. Doing the same computations as before, we have

N =

1

β

∂ log Z

∂µ

V,T

=

e

βµ

V

λ

3

.

So we can work out the value of µ:

µ = kT log

λ

3

N

V

.

Now we can use this to get some idea of what the chemical potential actually

means. For a classical gas, we need the wavelength to be significantly less than

the average distance between particles, i.e.

λ

V

N

1/3

,

so that the particles are sufficiently separated. If this is not true, then quantum

effects are important, and we will look at them later. If we plug this into the

logarithm, we find that µ < 0.

Remember that µ is defined by

µ =

∂E

∂N

S,V

.

It might seem odd that we get energy out when we add a particle. But note that

this derivative is taken with

S

fixed. Normally, we would expect adding a particle

to increase the entropy. So to keep the entropy fixed, we must simultaneously

take out energy of the system, and so µ is negative.

Continuing our exploration of the grand canonical ensemble, we can look at

the fluctuations in N, and find

∆N

2

=

1

β

2

log Z

ideal

= N.

So we find that

∆N

N

=

1

√

N

→ 0

as N → ∞. So in the thermodynamic limit, the fluctuations are negligible.

We can now obtain our equation of state. Recall the grand canonical potential

is

Φ = −kT log Z,

and that

pV = −Φ.

Since we know log Z, we can work out what pV is, we find that

pV = kT

e

βµ

V

λ

3

= NkT.

So the ideal gas law is also true in the grand canonical ensemble. Also, from

cancelling the

V

from both sides, we see that this determines

p

as a function of

T and µ:

p =

kT e

βµ

λ

3

.