Thermodynamics: Chapter 01

September 3, 2013

§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, energy, entropy; absolute tem-

perature; absolute zero degree; (Ideal Gas sys-

tem)

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

The zeroth law of thermodynamics: Systems

that are in thermal equilibrium with one system,

these systems are in thermal equilibrium with each

other. The theoretical basis for the concept of

”Temperature”. We will explore this more in future

(Section 3.1 in Chapter 3)

• 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. 29

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§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

.  .   .  

.  .  .  

.  

.  

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 parame-

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

Or, generally, f(X1, X2, ..., Xn, T ) = 0. Such as f(m,H, T ) = 0

for paramagnetism.

Different materials have different equation of state. Equation 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

4

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

If we can solve for V from the van der Waals Gas equation, itwould be great...

Sept. 3

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Homework policy revisit:

• Due date and how to submit.

• Posting of answers.

• Questions and discussions.

5

§3. The Ideal Gas: Connection between bulk property

and microscopic picture

§3.1 Temperature and kinetic energy of particles

Vz  dt   Z  

V  

Side  wall  area  =  A  

6

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

dN = Nf(~v)d3~v

f(~v) =1

N

dN

d3~v, velocity density distribution∫

f(~v)d3~v = 1

N of particles (with velocity in [~v,~v + d~v]) in dV is

dN = NdV

Vf(~v)d3~v , dV: volume by motion in dt

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 degreesof freedom. This result can be extended to systems (not allthermal systems!) that have arbitrary degrees of freedom - theEquipartition theorem: At temperature T, the average energyof any quadratic degree of freedom is 1

2kT . - This can be provedbased on principles in Statistical Physics which we will learn laterin this semester.

For a system with N particles, each with f DoF, and there is NOother non-quadratic temperature-dependent forms of energy, thetotal thermal energy in the system is

Uthermal = Nf1

2kT (5)

Remarks:

• It only applies to systems in which the energy is in the formof quadratic degree of freedom: E(q) = cq2.

• It is about ”thermal energy” of the system - those changeswith temperature, not the total energy.

• Degree of freedom: different systems require specific analy-sis: vibration, rotation, .... The NDoF of a system may alsovary as temperature changes.

§3.3 Compression work

Let’s deal with the energy change in a thermal system. There

are different ways to do so, one of which is by doing work to the

system:

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

VOr, after the integration: (11)

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 capacityWhen we talk about the energy change in a thermal system, itis convenient to define a quantity that measures the correlationbetween heat exchange and the temperature change caused bythe heat flow.

Heat capacity: of an system is the amount of heat needed toraise the temperature by one degree: C = Q

∆T . Or, for one mass

unit as c = Qm∆T .

Heat capacity at constant volume CV , Heat capacity at constantpressure CP , and Partial Derivative: (remember Q = ∆U −W )

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)

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In-class exercise:

• Prove the heat capacity at constant volume for ideal gas is

CV = 32Nk.

• 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)  

8

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, wehave

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

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  

The quantity ∆QT plays a very important role in thermodynamics

and statistical physics. It closely connects with another impor-

tant concept entropy and the second law of thermodynamics.