Module 4 – Thermodynamics

One area of physics where statics plays an important role is Thermodynamics where

we consider how energy is distributed between large numbers of particles.

Thermodynamics plays an essential role in the development of our understanding

of matter and hence the development of Modern Physics. You should have already

been introduced to Thermodynamics and the Kinetic Theory of Gases in your

University Physics I course and will have additional course(s) in this area as you

progress through your degree. Unfortunately, this material is near the end of the

University Physics I course and students tend to be rushed with exams and projects

so they don’t spend the time necessary to master this important material. You

should therefore review this material in your University Physics I textbook. For

Fundamentals of Physics 10th Ed. by Halliday & Resnick, this is Chapters 18

through 20. The material in this module uses Statistical concepts to extend H&R’s

discussions in Chapter 19 on Kinetic Theory. It is essential for understanding the

discussions on Thermal Radiation in the next module as well as work on Quantum

Statistical distributions in a module later in the course.

I. Maxwell - Boltzmann Distribution

A. The energy distribution function for a classical system of identical particles that

are far enough apart to be distinguishable is called the Maxwell-Boltzmann

distribution.

B. Formula

kT)ε(eA)ε( f

where is the energy of the particle

k is the Boltzmann constant = 8.617 x 10-5 eV/K

T is the temperature in Kelvin

A is the normalization constant

1

2 3

4 5

C. Graph

A particle exponentially prefers a lower energy state when it is available.

D. Thermal Energy

kT has the units of energy! It is a measure of the thermal energy of the system.

etemperaturroomateV401

Tk

It is the ratio of the energy of the state to the thermal energy that determines the

probability for the particle filling an allowed energy state.

E. Temperature Dependence

As shown in the next graph, the probability of finding a particle in a higher

energy state increases as the temperature increases. The environment serves as an

energy source.

In order to conserve the total probability (area under the curve), the normalization

constant must decrease as the temperature increases.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6

Pro

ba

bilit

y D

en

sit

y

Energy/Etherm

II. 1-D Mono-Atomic Ideal Gas

We will first use the Maxwell-Boltzmann to analyze a system of mono-atomic

ideal gas molecules moving only in the x-direction. We will expand our work to

3-D in the next section.

Goals: 1) Show that the energy distribution function is a Gaussian.

2) Show that the average energy is (1/2)kT {Equipartition Theorem}

A. The energy of a free particle is just kinetic so

2

2

1xx vmε

B. Energy Distribution Function

Plugging our particle’s energy into the Maxwell-Boltzmann distribution gives us

T)k2/v(m

x/kT)(ε

x

2

xee AA)(ε x

x

f .

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 1 2 3 4 5 6 7 8 9 10

Pro

bab

ilit

y D

en

sit

y

Energy/Etherm

kT = 2 eV

kT = 1 eV

We now rewrite our equation in a more convenient form as

])(2/[)v(

x

22

x m

Tk

x eA)(ε

f .

We now recognize this as a Gaussian distribution from the Statistics Module you

just completed an average and standard deviation given by

0avx and mTkσ .

Using our knowledge of the Gaussian distribution, we also know the normalization

constant Ax (Test Your Knowledge Problem from Statistics Module) and can write

the normalized energy distribution as

])(2/[)v( 22

x m

Tk

x eTkπ2

m)(εΦ

.

C. Average Energy

Using the definition for an average and our energy distribution function, we have

x

dveεε

]2

)

m

Tk(2/[

2)

xv(

xTkπ2

mx .

x

dvevmε

]2

)

m

Tk(2/[

2)

xv(

2

x2

1

Tkπ2

mx

We will rewrite the equation so that we can use substitution to solve the integral.

x

dvevε

Tk2

m

2])

m

T2k/[

xv(

2

xπ2

mx

We also note that the integrand is an even function so

0

Tk2

m

2])

m

T2k/[

xv(

2

xπ

m

xdvevεx .

Using the substitution relations

mTk2

vη x

2

m

Tk22x ηv

xdvTk2

mdη

We have the equation

0

η2

π

Tk2 dηeηε2

x .

From Table D-1 on page 585 of Rohlf’s Textbook or using the Integral tables in a

Math Handbook, we find that

4

πdηeη

0

η22

.

Thus, we find the equipartition result that

24

π

π

Tk2xε

Tk

The result that we have just obtained is an extremely important relationship. It says

that the temperature of a substance containing one-dimensional mono-atomic atoms

(free particles) is proportional to the average energy one of these molecules

possesses. If we multiply this average energy by the number of molecules in the

substance, we have obtained the total microscopic energy stored in the substance.

The number of molecules can be found using a scale to find the mass of the

substance, the gram molecular weight of the substance, and Avogadro’s number.

Thus, we can find the microscopic energy contained in a substance with a

thermometer and a scale. Now our present result is not yet practical as it is one

dimensional motion and our world is three dimensional so we will extend it.

III. Extending To 3-D

We now consider extending our mono-atomic ideal gas work to three dimensions.

A. The energy of the particle is still kinetic and is given as

2

2

12

2

12

2

12

2

1zyx mvmvmvvmε

B. Energy Distribution Function

Substituting the kinetic energy into the Maxwell-Boltzman distribution

T)k2/]

zv

yv

x(m[v

T)k2/v(m222

2

ee AA)(ε

f .

By the law of exponents this reduces to

T)k2/z

v(mT)k2/y

v(mT)k2/x

v(m 222

eeeA)(ε

f .

We can now rewrite our normalization constant A as the product of three

normalization constants (Ax, Ay, Az). Thus, our equation becomes

T)k2/v(mA

T)k2/v(mA

T)k2/v(mA)(ε

2

z

z

2

y

y

2

x

xeeef

Thus, we see that our 3-D energy distribution function is just the product of three

1-D energy distribution functions!! Inserting our results from the previous section,

we have

)(ε)(ε)(ε)(εzyx

ffff

0vvv zyx

mTkσ and

Tkπ2m

AAA zyx .

T)k2/v(mT)k2/v(mT)k2/v(m)(ε

2

z

2

y2

x2

3

eeeTkπ2

mf

We now see a very import relationship between adding energy to a system and what it does

to the Maxwell-Boltzmann energy distribution:

IMPORTANT

Adding Energy Terms leads to Multiplying Energy Distribution Functions.

C. Average Energy

You can now use the distribution function to find the average energy for the 3-D

mono-atomic molecule and you will find that it is 3

2 𝑘𝑇. This is just three times

our 1-D result. This is the origin of the Equipartition Theorem!!

D. Average Velocity

If you have a container of gas sitting on a table, we know that the average velocity

of the gas molecules must be zero or else the center of mass of the container would

not be at rest. We can now use the Maxwell-Boltzmann distribution to show that

our intuition is correct.

Using the definition of an average, the average velocity is calculated by integrating

over all of possible velocity components.

⟨�⃗⃗� ⟩ = ∫ �⃗⃗� 𝒇(𝒗𝒙, 𝒗𝒚, 𝒗𝒛) 𝒅𝒗𝒙𝒅𝒗𝒚𝒅𝒗𝒛

⟨�⃗⃗� ⟩ = ∫ ∫ ∫ (𝒗𝒙�̂� + 𝒗𝒚𝒋̂ + 𝒗𝒛�̂�)∞

−∞

∞

−∞

∞

−∞

(𝒎

𝟐𝝅𝒌𝑻)𝟑/𝟐

𝒆−𝒎(𝒗𝒙𝟐+𝒗𝒚

𝟐+𝒗𝒛𝟐)/𝒌𝑻𝒅𝒗𝒙𝒅𝒗𝒚𝒅𝒗𝒛

⟨�⃗⃗� ⟩ =

�̂�∫ √𝒎

𝟐𝝅𝒌𝑻𝒗𝒙𝒆

−𝒎(𝒗𝒙

𝟐)𝒌𝑻 𝒅𝒗𝒙 ∫ √

𝒎

𝟐𝝅𝒌𝑻𝒆−

𝒎(𝒗𝒚𝟐)

𝒌𝑻 𝒅𝒗𝒚 ∫ √𝒎

𝟐𝝅𝒌𝑻𝒆−

𝒎(𝒗𝒛𝟐)

𝒌𝑻 𝒅𝒗𝒛

∞

−∞

∞

−∞

∞

−∞

+𝒋̂ ∫ √𝒎

𝟐𝝅𝒌𝑻𝒆−

𝒎(𝒗𝒙𝟐)

𝒌𝑻 𝒅𝒗𝒙 ∫ √𝒎

𝟐𝝅𝒌𝑻𝒗𝒚𝒆

−𝒎(𝒗𝒚

𝟐)

𝒌𝑻 𝒅𝒗𝒚 ∫ √𝒎

𝟐𝝅𝒌𝑻𝒆−

𝒎(𝒗𝒛𝟐)

𝒌𝑻 𝒅𝒗𝒛

∞

−∞

∞

−∞

∞

−∞

+�̂�∫ √𝒎

𝟐𝝅𝒌𝑻𝒗𝒙𝒆

−𝒎(𝒗𝒙

𝟐)𝒌𝑻 𝒅𝒗𝒙 ∫ √

𝒎

𝟐𝝅𝒌𝑻𝒆−

𝒎(𝒗𝒚𝟐)

𝒌𝑻 𝒅𝒗𝒚 ∫ √𝒎

𝟐𝝅𝒌𝑻𝒗𝒛𝒆

−𝒎(𝒗𝒛

𝟐)𝒌𝑻 𝒅𝒗𝒛

∞

−∞

∞

−∞

∞

−∞

We see that each component of the vector average is the product of three integrals.

We also know that two of the integrals in each component are equal to one as we

have a normalized distribution function.

⟨�⃗⃗� ⟩ = �̂� ∫ √𝒎

𝟐𝝅𝒌𝑻𝒗𝒙𝒆

−𝒎(𝒗𝒙

𝟐)

𝒌𝑻 𝒅𝒗𝒙 +∞

−∞𝒋̂ ∫ √

𝒎

𝟐𝝅𝒌𝑻𝒗𝒚𝒆

−𝒎(𝒗𝒚

𝟐)

𝒌𝑻 𝒅𝒗𝒚∞

−∞+ �̂� ∫ √

𝒎

𝟐𝝅𝒌𝑻𝒗𝒛𝒆

−𝒎(𝒗𝒛

𝟐)

𝒌𝑻 𝒅𝒗𝒛∞

−∞

Each of the three remaining integrals have integrands which are odd functions and

are being integrated over symmetrical limits. Thus, each of these integrals are

zero!!!

⟨�⃗⃗� ⟩ = �̂� 𝟎 + 𝒋 ̂𝟎 + �̂� 𝟎 = 𝟎

When working physics problems, you want to keep an eye out for integration tricks

like taking advantage of normalized distribution functions and integrating an odd

integrand over symmetrical limits. Physics problems quickly become very

mathematical and these tricks will save you considerable time and reduce the

number of math errors. Furthermore, textbook authors in advanced courses assume

you know the tricks and will use them without warning.

E. Speed Distribution

While the average velocity of a gas molecule is zero, the average speed of a gas

molecule is not zero since speed is a non-negative scalar quantity. To do problems

involving speed, we need to go back and write the un-normalized Maxwell-

Boltzmann distribution function in terms of spherical coordinates where the

particles speed is the radial component. It is common practice to work physics

problems in spherical coordinates, but many students may not have dealt with this

calculus previously. The un-normalized Maxwell-Boltzmann distribution function

for velocity in spherical coordinates is simply

𝑓(𝑣, 𝜃, 𝜑) = 𝐴𝑒−𝑚𝑣2/2𝑘𝑇

The problem is now finding the normalization constant, A, by integrating over the

volume of speed space in spherical coordinates rather than Cartesian coordinates as

we did for velocity components. We start by drawing a sphere of radius 𝑟 = 𝑣 as

shown below:

Changing the particle’s velocity coordinates by infinitesimal amounts, 𝑑𝑣, 𝑑𝜑, and

𝑑𝜃 creates a volume cube whose three sides are shown in green, red, and purple

and whose lengths are 𝑑𝑣, 𝑣𝑑𝜃, and 𝑣 𝑠𝑖𝑛(𝜃)𝑑𝜑. Using the formula for the volume

of a cube, we have

𝑑𝜏 = 𝑣2𝑠𝑖𝑛(𝜃)𝑑𝑣𝑑𝜃𝑑𝜑

We now integrate our distribution over all possible speed space to find out

normalization constant.

1 = ∫ 𝐴𝑣2𝑒−𝑚𝑣2/2𝑘𝑇𝑑𝑣 ∫ 𝑠𝑖𝑛(𝜃)𝜋

0

∞

0

𝑑𝜃 ∫ 𝑑𝜑2𝜋

0

𝑣 𝑠𝑖𝑛(𝜃)𝑑𝜑

𝑣 𝑑𝜃

𝑑𝑣

The last two integrals are 2 and 2π respectively so we have

1 = 4𝜋𝐴 ∫ 𝑣2𝑒−𝑚𝑣2/2𝑘𝑇𝑑𝑣∞

0

𝐴 = 1

4𝜋 ∫ 𝑣2𝑒−𝑚𝑣2/2𝑘𝑇𝑑𝑣∞

0

If we substitute𝑥 = √𝑚𝑘𝑇

2 𝑣, we get the integral into a form found in integral tables

and in Rohlf’s textbook

𝐴 = 1

4𝜋 (2𝑘𝑇𝑚 )

3/2

∫ 𝑥2𝑒−𝑥2𝑑𝑥

∞

0

= 1

4𝜋 (2𝑘𝑇𝑚 )

3/2√𝜋4

𝐴 = 1

(2𝜋𝑘𝑇𝑚 )

3/2= (

𝑚

2𝜋𝑘𝑇)3/2

Plugging our normalization constant back into our Maxwell-Boltzmann distribution

function for velocity in spherical coordinates:

𝑓(𝑣, 𝜃, 𝜑) = (𝑚

2𝜋𝑘𝑇)3/2

𝑒−𝑚𝑣2/2𝑘𝑇

To find the Maxwell-Boltzmann distribution for the speed of the particles, we must

integrate the dependence on direction (𝜃, 𝜑) from the velocity dependent

distribution as follows:

𝑓(𝑟) = ∫ ∫ (𝑚

2𝜋𝑘𝑇)3/2

𝑒−𝑚𝑣2/2𝑘𝑇𝑣 𝑠𝑖𝑛(𝜃)𝑑𝜑 𝑣𝑑𝜃2𝜋

0

𝜋

0

𝑓(𝑟) = (𝑚

2𝜋𝑘𝑇)3/2

𝑣2𝑒−𝑚𝑣2/2𝑘𝑇 ∫ 𝑠𝑖𝑛(𝜃)𝑑𝜃 ∫ 𝑑𝜑2𝜋

0

𝜋

0

𝑓(𝑟) = (𝑚

2𝜋𝑘𝑇)3/2

𝑣2𝑒−𝑚𝑣2/2𝑘𝑇(2)(2𝜋)

This gives us the Maxwell-Boltzmann distribution for the speed of the particle as

𝑓(𝑟) = 4𝜋 (𝑚

2𝜋𝑘𝑇)3/2

𝑣2𝑒−𝑚𝑣2/2𝑘𝑇

You will find this distribution to be extremely useful in a wide range of problems

involving the speed of free particles including gases.

F. Useful Integrals

The following integral results are useful when dealing with homework problems

with Gaussian distributions like the Maxwell-Boltzmann distribution:

1. ∫ 𝑒−𝑥2𝑑𝑥 =

√𝜋

2

∞

0

2. ∫ 𝑥𝑒−𝑥2𝑑𝑥 =

1

2

∞

0

3. ∫ 𝑥2𝑒−𝑥2𝑑𝑥 =

√𝜋

4

∞

0

4. ∫ 𝑥3𝑒−𝑥2𝑑𝑥 =

1

2

∞

0

5. ∫ 𝑥4𝑒−𝑥2𝑑𝑥 =

3√𝜋

8

∞

0

6. ∫ 𝑥5𝑒−𝑥2𝑑𝑥 = 1

∞

0

IV. Density of States - g()

We often want to compare the number of particles with different energies.

It may appear that more particles will always be at the lower energy value since

the Maxwell-Boltzmann distribution falls off exponentially for increasing energy.

However, the number of particles depends not only on the probability that a single

particle can fill a single energy state but also on the number of states available!

Although the finer points of calculating energy states at the Quantum level will

have to wait till later in the course, we can understand the basic concept using the

following classical example. Consider an eraser bouncing across a table during

an earthquake. The eraser is more likely to be found at lower heights (lower

energy states) above the table than at larger distances above the table (higher

energy states). If the earthquake becomes more violent, the probability for the

eraser being at a greater height above the table increases, but it is still more likely

to be at lower heights above the table. This is just the spreading of the Maxwell-

Boltzmann distribution where the eraser is the particle and the earthquake is

analogous to our thermal energy. However, the probability of finding the eraser in

even lower energy states below the table is zero! There are no energy states

available due to the table. We could change the available energy states and the

particle probabilities by removing the table. This is analogous to band gap

engineering for electronics.

A. The density of states is a function, g(), whose value gives the number of energy

states with energy .

B. The number of particles with energy is found by multiplying the density of

states by the energy distribution function.

εεgεn f

# of Particles = (Number of states ) x (Probability of filling a state)

EXAMPLE: You are throwing bean bags at a Carnival. The probability for a bean bag to

land in a particular square, circle, or miss is 0.5, 0.25, and 0.25 respectively. What is the

ratio of the number of bags that land in a square compared to a circle?

Solution:

80.2520.54

nn

c

s

It is eight times more likely for the bags to be in a square than in a circle!