Thermodynamics: Chapter 01

September 2, 2014

§1. Basics concepts in thermodynamics

We start this course from some fundamental con-

cepts. By introducing a system that consists of many

particles and is also simply enough for us to investi-

gate the major thermal properties, the ideal gas sys-

tem, we will introduce more fundamental concepts,

laws of thermodynamics, and apply them in solving

practical thermal problems.

• Temperature:1

1. operational definition - how we measure the tem-

perature of an object;

2. theoretical definition - relation to zeroth law of

thermodynamics

The zeroth law of thermodynamics: Systems

that are in thermal equilibrium with one system,

these systems are in thermal equilibrium with

each other.

We will explore this more about temperature

Chapter 3: Its relation with the energy in the

system, entropy, absolute temperature, absolute

zero degree, etc.

3. different scales and units: pay attention to them

in applications!

Celsius (oC): water boiling point at 1 atm pressure

- 100oC, water freezing point at 1 atm pressure -

0oC

Fahrenheit (oF): water boiling point at 1 atm

pressure - 212oF , water freezing point at 1 atm

pressure - 32oF

Kelvin (K, SI unit): 0K = −273oC, the size of

1K = the size of 1oC

• Pressure: The ratio of force to the area over which

that force is distributed.

Pressure and mechanical equilibrium:

Unit: SI unit is Pascal, 1Pa = 1N/m2. Other units

see the table below.

• Equilibrium

It is a condition of a system in which competing

influences are balanced. Examples are

Equilibrium ExchangesThermal Thermal energy

Mechanical VolumeDiffusive ParticlesChemical Chemical reactions

H2O ↔ H+ +OH−

H2SO4↔ 2H+ + SO2−4

• Systems in thermal physics

Isolated system: No interaction with environment,

no matter and energy exchange.

Closed system: No matter, but has energy ex-

change.

Open system: Has matter and energy exchange.

• Energy

Kinetic energy: energy associated with motion

(and mass), of a ”particle”, of a system.

Potential energy: energy associated with interac-

tions (various fields), of a ”particle”, of a system.

Chemical energy: energy associated with EM in-

teration, of a ”particle”, of a system. Unit: SI

unit is Joule 1J = 1kg ·m2/s2

• Heat (Q): Flow of energy from one system to an-

other caused by a difference in temperature be-

tween the two systems.

Heat transfer: conduction - at microscopic level;

convection - bulk motion of medium; radiation -

electromagnetic waves, photons.

Unit: Calories, not SI unit !! 1C ≈ 4.2J

• Work (W): Any other transfer of energy (except

heat) into or out of a system.

The first law of thermodynamics: ∆U = Q+W ,

the system’s energy change (∆U) is the sum of the

changes in heat and work into or out of a system.

This is the law of energy conservation.

Aug. 28

*********

§2. The Ideal Gas: A simple thermal system that explains

a lot about thermal physics

Ideal Gas: Particles have no size; There is no interactions among

particles; Particles have elastic collisions against the wall of the

container.

This is a not-so-bad approximation for gas at low pressure/low

density and when the temperature is not so low.

In the study of ”bulk” properties:

2

.  .   .  

.  .  .  

.  

.   p  =  F/(Piston  area)  

1664 and 1676: R. Boyle and E. Mariotte discovered that ata fixed temperature T the pressure and volume satisfy therelation: pV = p0V0. Or, pV = constant!

1802: Gay-Lussac obtained V = TT0V0, at a fixed pressure p.

(He credited to the unpublished work from the 1780s by Jacques

Charles. Charles’ Law)

How p, V, T correlate to each other if we let the system change

from state (p0, V0, T0) to (p, V, T )?

(1) (p0, V0, T0) to (p, V ∗0 , T0): Fixed T

p0V0 = pV ∗0(2) (p, V ∗0 , T0) to (p, V, T ): Fixed p

V = TT0V ∗0

By canceling V ∗0 in these two equations, one has

pV

T=p0V0

T0= constant.

This constant should be proportional to the number of

particles (N) in the system, therefore we can writepV

T= Nk (1)

where k is called Boltzmann constant.

k = 1.380658× 10−23J ·K−1.

We can re-write it as

pV = nRT (2)

where n is the number of moles of gas, R is a constant that

has empirical value of R = 8.31 Jmol·K when all variables are in

SI units: Pressure in N/m2 = Pa, volume in m3, and

temperature on Kelvin,K

In-Class Exercise: A mole of molecules is Avogadro’s number

(NA) of them. Given the empirical values of R and k, what is

the value of NA?

pV = nRT

pV = NkT

nRT = NkT

NA =1.0 ·Rk

=8.31 J

mol·K1.380658× 10−23J ·K−1

NA = 6.017× 1023 mol−1

3

Example, Equation of State: Relation between state param-

eters (such as P, V, T), or f(P, V, T ) = 0.

Or, generally, f(X1, X2, ..., Xn, T ) = 0.

Example: For paramagnetism, we can write in general format

f(M,B, T,C) = 0, which in the form of Curie’s Law looks like

MT = CB

in which:

M : is the ”resulting” magnetisation, the total magnetic moment

of the material proptoµ(N↑ −N↓).

T : is absolute temperature (Kelvin).

C: is a material-specific Curie constant. (Will learn more in

Section 3.3).

B: is the magnetic field (Tesla).

4

Note: Different materials have different equation of state. Equa-

tion of state exists for a system that reaches equilibrium state.

A famous equation of state that better describes gas (and even

dense fluids!) is the van der Waals equation:(P +

an2

V 2

)(V − nb) = nRT

n is mole number. a and b are constants that depend on the

type of matter.an2

V 2 : correction to the attraction between particles.

nb: correction to the size of the particles.

Material a/(Pa ·m6 ·mol−1) b/(m3 ·mol−1)H2 0.02476 0.02661N2 0.1408 0.03913CO2 0.3639 0.04267H2O 0.5535 0.03049

We define coefficient of expansion as α = 1V

(∂V∂T

)P

(fixed pres-

sure). Calculate the coefficient of expansion for van der Waals

Gas.

If we can solve for V from the van der Waals Gas equation, it

would be great ...

Homework policy revisit:

• Due date and how to submit.

TA: Emily Dvorak, emily.dvorak@mines.sdsmt.edu

• Posting of answers.

• Questions and discussions.

5

§3. The Ideal Gas: Connection between bulk propertyand microscopic picture

§3.1 Temperature and kinetic energy of particles

Number of particles that have velocity within [~v,~v + d~v]

dN = Nf(~v)d3~v

f(~v) =1

N

dN

d3~v, it’s called the Velocity Density Distribution∫

f(~v)d3~v = 1

Assuming particles are uniformly distributed in volume V , thenumber (N) of particles that have velocity within [~v,~v + d~v] indV is

dN =N

VdV f(~v)d3~v

6

Let’s look at these particles in the volume (dV ) of the motion

of all particles in dt. dV can be written as (see Fig below),

Vz  dt   Z  

V  

Side  wall  area  =  A  

dV = Avzdt , A: area perpendicular to vz

Assuming elastic scattering on the wall

dpz = mvz − (−mvz) = 2mvz : for one particle

dFAdt = dpzdN = 2mvzdN = 2Nmv2z f(~v)d3~v

Adt

V

So, dFA = 2Nmv2z f(~v)d3~vAV , the force by all particles with veloc-

ity in [~v,~v + d~v]!!

By the definition of pressure:

p =1

A

∫dFA . Integration over all velocity. Be very careful here:

p =N

V

∫ +∞

−∞dvx

∫ +∞

−∞dvy

∫ +∞

0dvzf(~v)2mv2

z

Assuming f(~v) is independent from the direction of ~v,∫ +∞

0dvz =

1

2

∫ +∞

−∞dvz

Therefore,

pV = mN∫ +∞

−∞f(~v)v2

z d3~v (3)

The integration is the mean value of v2z !

Since the motion of gas particles is isotropic (in direction), the

mean value of the projection of velocity in all directions should

be the same. That is∫+∞−∞ f(~v)v2

z d3~v =< v2

x >=< v2y >=< v2

z >

and v2 = v2x + v2

y + v2z , we have < v2

z >= 13 < v2 >. So, Eq. (3)

can be written as

pV = mN1

3< v2 >=

2

3N < εkin > (4)

where < εkin >= 12m < v2 > is the mean kinetic energy of the

particle.

Using Ideal Gas Law, let’s take a look at the temperature:

pV = kNT

pV =2

3N < εkin >

kNT =2

3N < εkin >

T =2

3< εkin > /k

§3.2 Equipartition theorem

Let’s look at the relation between average translational kinetic

energy of a particle and the temperature of the system: From

T = 23 < εkin > /k, we get:

< εkin >= 32kT . Since < εkin >=< 1

2mv2 >= 1

2m < v2x +v2

y +v2z >,

we can have identity: 32m < v2

x >= 32kT , or for three degrees of

freedom,12m < v2

x >= 12m < v2

y >= 12m < v2

z >= 12kT .

This is a special case when the particles have only three degrees

of freedom. This result can be extended to systems (not all

thermal systems!) that have arbitrary degrees of freedom - the

Equipartition theorem: At temperature T, the average energy

of any quadratic degree of freedom is 12kT . - This can be proved

based on principles in Statistical Physics which we will learn later

in this semester.

For a system with N particles, each with f DoF, and there is NO

other non-quadratic temperature-dependent forms of energy, the

total thermal energy in the system is

Uthermal = Nf1

2kT (5)

Remarks:

• It only applies to systems in which the energy is in the form

of quadratic degree of freedom: E(q) = cq2.

• It is about ”thermal energy” of the system - those changes

with temperature, not the total energy.

• Degree of freedom: different systems require specific analy-

sis: vibration, rotation, .... The NDoF of a system may also

vary as temperature changes.

Sept. 2

*********

§3.3 Compression work

Thermal system: At macroscopic level, one cares the energy flow

into or out from the system.

There are different ways to cause energy change in a thermal

system. One of which is by doing work to/by the system.

The law of energy conservation, the first law of thermodynamics:

∆U = Q + W , the system’s energy change (∆U) is the sum of

the changes in heat Q and work W into or out of a system.

7

δx  

Piston  area  =  A  Force  =  F    

F  P  

V

Area  =    work  !  

The work done by the pushing force:

δW = Fδx, −Aδx = δV

δW = −F/AδVdW = −PdV (6)

W = −∫ V2

V1

P (V )dV (7)

Two processes:

1. Isothermal compression: temperature stays constant - slow

process / closed system.

Using Eq. (7) and (1) pV = NkT , one has

W = −∫ V2

V1

P (V )dV

= −∫ V2

V1

NkT

VdV when T is constant, we have:

W = −NkT lnV2

V1(= NkT ln

V1

V2) (8)

The first law of thermodynamics takes the following form:

∆U = Q+W

∆(Nf1

2kT ) = Q−NkT ln

V2

V1, when T is constant:

Q = NkT lnV2

V1(9)

where Q as the heat input or output depending on the work doneto or by the gas.

2. Adiabatic compression: no heat escape from the system.Examples: fast process or isolated system.

The first law of thermodynamics takes the following form:

∆U = Q+ ∆W

d(Nf1

2kT ) = Q− PdV , Q=0 for isolated system:

1

2NfkdT = −PdV For ideal gas PV = NkT (10)

1

2NfkdT = −

NkT

VdV

fdT

T= −2

dV

V(11)

Or, after the integration: (using∫ dVV = lnV )

flnT1

T2= −2ln

V1

V2(12)

This can be simplified further:

{T1

T2}f = {

V2

V1}2

{T1

T2}f/2 =

V2

V1

Tf/21 V1 = T

f/22 V2 = ... , which is

V T f/2 = constant. (13)

§3.4 Heat capacity

When we talk about the energy change in a thermal system, it

is convenient to define a quantity that measures the correlation

between heat exchange and the temperature change caused by

the heat flow.

Heat capacity: of an system is the amount of heat needed to

raise the temperature by one degree: C = Q∆T . Or, for one mass

unit as c = Qm∆T .

Looking at Q = ∆U −W ≈ [U(Tf) − U(Ti)] −∫PdV , one needs

to consider

(1) Heat capacity at constant volume CV

8

(2) Heat capacity at constant pressure CPwhich are related to partial derivatives:

CV =(∂Q

∂T

)V

=(∂U

∂T

)V

(no V change → no work) (14)

CP =(∂Q

∂T

)P

=(∂U

∂T

)P

+ P

(∂V

∂T

)P

(15)

In-class quiz:

1. Prove the heat capacity at constant volume for ideal gas is

CV = 32Nk.

2. Prove the heat capacity at constant pressure for ideal gas is

CP = CV + nR.

§3.5 The Carnot cycle: Ideal gas case

Four step of Carnot cycle:  a: Isothermal expansion at Th b: adiaba:c expansion to Tc c: isothermal compression at Tc d: adiaba:c compression back to Th  

PV diagram for an ideal  monoatomic gas  undergoing a Carnot cycle. 

Nicolas Léonard Sadi Carnot  (06/01/1796 – 08/24/1832)  

9

The Carnot cycle is very important in thermal physics because

it explains how to make a system (like engine) that can achieve

the maximum possible efficiency when it works within a given

temperature range.

Let’s see how energy changes in the cycle:

Step 1: isothermal expansion, (V1, P1, Th)→ (V2, P2, Th)

∆Q1 = −∆W1 = −(−∫ V2

V1

PdV ) =∫ V2

V1

NkThV

dV

= NkThlnV2

V1Step 2: adiabatic expansion, (V2, P2, Th)→ (V3, P3, Tc)

∆Q2 = 0,∆W2 = ∆U2 = CV (Tc − Th)

Step 3: isothermal compression, (V3, P3, Tc)→ (V4, P4, Tc)

∆Q3 = −∆W3 = NkTclnV4

V3Step 4: adiabatic compression, (V4, P4, Tc)→ (V1, P1, Th)

∆Q4 = 0,∆W4 = ∆U4 = CV (Th − Tc)

The overall energy budget is

∆Utotal = (∆Q1 + ∆W1) + ∆W2 + (∆Q3 + ∆W3) + ∆W4

∆Utotal = (0) + CV (Tc − Th) + (0) + CV (Th − Tc)∆Utotal ≡ 0

This is exactly what needed for a cycle - the state of the system

comes back to its original state!!!

For adiabatic expansion/compression, we have:

dU = dQ− PdVdU = −PdV

That is, all work done to the system is stored as thermal energy.

Using heat capacity:

CV =(∂Q∂T

)V

=(∂U∂T

)V,

we have:

CV dT = −PdV, PV = NkT → P =NkT

V

CVdT

T= −Nk

dV

Vwhich integrates to:

CV lnTc

Th= −Nkln

V3

V2

We know for Ideal Gas, CV = 32Nk. The last equation can be

re-written as:

3

2Nkln

Tc

Th= −Nkln

V3

V2(Tc

Th

)3/2

=V2

V3

So, we have the following additional relations:

Step-2:V3

V2=(ThTc

)3/2

Step-4:V1

V4=

(Tc

Th

)3/2

These give:V3

V2=V4

V1or

V1

V2=V4

V3

Since∆Q1

Th= Nkln

V2

V1(see what obtained in Step-1)

(16)

and∆Q3

Tc= Nkln

V4

V3= Nkln

V1

V2= −Nkln

V2

V1Therefore, we have:

∆Q1

Th+

∆Q3

Tc≡ 0 (17)

If we divide one big cycle into many micro-Carnot cycles, we

have∮ δQT = 0. It turns out this is true for all reversible cycles,

not just ideal gas system that undergoes the Carnot Cycle.

A cycle of an arbitrary closed path in P‐V space  can be achieved by a large number of small Carnot cycles.  

A B C D 

Isothermal lines 

Adiaba=c lines  

And more: ∆Q1Th

+ ∆Q3Tc≡ 0 is one of the most important relation

in thermal physics! The quantity ∆QT plays a very important role

in thermodynamics and statistical physics. It closely connects

with another important concept entropy and the second law of

thermodynamics.