Monoatomic classical ideal gas

The **monoatomic classical ideal gas**, in statistical mechanics, is an approximation that models gas as a large number of independent particles with no internal structure, at a high enough temperature that quantum correlations don't affect the gas and at a low enough temperature that relativistic corrections don't affect the gas.

## Energy Levels

The energy levels of an ideal gas are given by the requirement for an integer number of wavelengths of the particles to fit inside the container. This means that

k
\_

=

(

2
π

L
x

I
x

,

2
π

L
y

I
y

,

2
π

L
z

I
z

)

{\displaystyle \underline{k} = \left( \frac{2\pi}{L\_x} I\_x , \frac{2\pi}{L\_y} I\_y, \frac{2\pi}{L\_z} I\_z \right)}

, and therefore that the interval between adjacent wavenumber levels is

Δ
k

\_

=

(

2
π

L
x

,

2
π

L
y

,

2
π

L
z

)

{\displaystyle \underline{\Delta k} = \left( \frac{2\pi}{L\_x} , \frac{2\pi}{L\_y}, \frac{2\pi}{L\_z} \right)}

.

The momentum levels of the gas are therefore found by the de Broglie relation to be

p
\_

=

k
\_

ℏ

{\displaystyle \underline{p} = \underline{k} \hbar}

.

The energy of a particle in classical mechanics is

ϵ

p
\_

=

p
\_

2

2
m

{\displaystyle \epsilon\_{\underline{p}} = \frac{\underline{p}^2}{2m}}

, so the energy levels in terms of wavenumber are

ϵ

k
\_

=

ℏ
2

k
\_

2

2
m

{\displaystyle \epsilon\_{ \underline{k} } = \frac{\hbar^2 \underline{k}^2}{2m}}

, with the allowable wavenumber levels are as above.

## Partition Function

The single-particle partition function is found by summing over all of these energy levels as

Z
1

=

∑
k

e

−
β

ϵ
k

{\displaystyle Z\_1 = \sum\_k e^{-\beta \epsilon\_k}}

. If the distance between adjacent wavenumber levels is known (as they are), this can be approximated as an integral, because the wavenumbers in a classical ideal gas are likely to be very large; the sum can be written as

Z
1

=

∑
k

L
x

L
y

L
z

(

2
π

)

3

e

−
β

ϵ

k
\_

Δ
k

\_

3

≈

∫
k

V

(

2
π

)

3

e

−
β

ϵ

k
\_

d
3

k
\_

{\displaystyle Z\_1 = \sum\_k \frac{L\_x L\_y L\_z}{\left(2 \pi \right)^3} e^{-\beta \epsilon\_{\underline{k}}} \underline{\Delta k}^3 \approx \int\_k \frac{V}{\left(2 \pi \right)^3} e^{-\beta \epsilon\_{\underline{k}}} \text{d}^3 \underline{k}}

.

Substituting in the known energy levels gives

Z
1

=

∫
k

V

(

2
π

)

3

e

−
β

ℏ
2

k
\_

2

2
m

d
3

k
\_

{\displaystyle Z\_1 = \int\_k \frac{V}{\left(2 \pi \right)^3} e^{-\beta \frac{\hbar^2 \underline{k}^2}{2m}} \text{d}^3 \underline{k}}

. This integral can be converted into spherical coordinates by the transformation

d
3

k
\_

=
4
π

k
2

d
k

{\displaystyle \text{d}^3 \underline{k} = 4\pi k^2 \text{d}k}

 and then solved to give the partition function

Z
1

=

V

m

3

ℏ
3

2
π
β

3

=

V

m

k
B

T

3

ℏ
3

2
π

3

{\displaystyle Z\_1 = \frac{V \sqrt{m}^3}{\hbar^3 \sqrt{2 \pi \beta}^3} = \frac{V \sqrt{m k\_B T}^3}{\hbar^3 \sqrt{2 \pi}^3}}

. This can be simplified by defining the thermal wavelength

λ

th

=
ℏ

2
π
β

m

{\displaystyle \lambda\_{\text{th}} = \hbar \sqrt{\frac{2 \pi \beta}{m}}}

.

# Z

Z
1
N

N
!

{\displaystyle Z = \frac{Z\_1^N}{N!}}

, with the factor of

N
!

{\displaystyle N!}

 because the particles in the monoatomic classical ideal gas are indistinguishable, and therefore

# Z

1

N
!

(

V

λ

th

3

)

{\displaystyle Z = \frac{1}{N!} \left( \frac{V}{\lambda\_{\text{th}}^3} \right)}

.

## Free Energy

The Helmholtz free energy of a classical ideal gas is found from the partition function by

# F
−

k
B

T
ln
⁡
# Z
−
N

k
B

T

(

1
−
ln
⁡

(

N

ℏ
3

2
π

3

V

m

k
B

T

3

)

)

{\displaystyle F = -k\_B T \ln Z = -N k\_B T \left( 1 - \ln \left( \frac{N \hbar^3 \sqrt{2 \pi}^3}{V \sqrt{m k\_B T}^3} \right) \right) }

.

## Pressure

The pressure, and consequently the equation of state, can be found by differentiating the free energy with respect to volume. This gives

# p
−

∂
F

∂
V

=

∂

∂
V

(

N

k
B

T

(

1
−
ln
⁡

V
0

V

−
1

)

)

{\displaystyle p = - \frac{\partial F}{\partial V} = \frac{\partial}{\partial V} \left( N k\_B T \left( 1 - \ln V\_0 V^{-1} \right) \right) }

 and hence

# p

N

k
B

T

V

{\displaystyle p = \frac{N k\_B T}{V}}

. This is also the equation of state for an ideal gas.

## Entropy

The entropy can be found by differentiating the free energy with respect to temperature. This gives

# S
−

∂
F

∂
T

=

∂

∂
T

(

N

k
B

T

(

1
−
ln
⁡

T
0

3
2

T

3
2

)

)

{\displaystyle S = -\frac{\partial F}{\partial T} = \frac{\partial}{\partial T} \left( N k\_B T \left( 1 - \ln T\_0^{\frac{3}{2}} T^{\frac{3}{2}} \right) \right)}

 and hence

# S
N

k
B

(

5
2

−
ln
⁡

(

N

ℏ
3

2
π

3

V

m

k
B

T

3

)

)

{\displaystyle S = N k\_B \left( \frac{5}{2} - \ln \left( \frac{N \hbar^3 \sqrt{2 \pi}^3}{V \sqrt{m k\_B T}^3} \right) \right) }

. This expression is called the Sakur-Tetrode equation.