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Università di Pisa Introduction to Thermodynamics

1. Introduction. History of Thermodynamics

2. The First Law. Microscopic view. Joule

3. The Second Law. Microscopic View. Carnot

4. Thermodynamic Properties of Fluids

5. Multicomponent Systems

1. Introduction. History of Thermodynamics

2. The First Law. Microscopic view. Joule

3. The Second Law. Microscopic View. Carnot

4. Thermodynamic Properties of Fluids

5. Multicomponent Systems

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Università di PisaIntroduction

TD deals with equilibrium states. Knowing initial conditions it predicts how the system will end up, but it cannot say how long it will take.

TD was developed with steam engines, i.e. machines that convert heat into work (Watt, 1760’s).

Nature of heat: a) caloric, a conserved quantity: hot objects contain more caloric (Lavoisier, 1770’s, Laplace); b) Internal movements (B. Thompson, 1810’s).

“The production of motive power in steam engines is due to the transportation of caloric from a warm body to a cold body” (Sadi Carnot, 1824).

Carnot postulated that some caloric is lost, not being converted into mechanical work. This is the basis of the second law, which therefore predates the understanding of the first law.

Conservation of mechanical energy (i.e. kinetic plus potential) was formulated by Newton.

Joule (1940’s) demonstrated the equivalence between heat and work (1 cal. = 4.19J).

Clausius (Rudolf Gottlieb) introduced the concept of internal energy (1850) and of entropy (1862).

Boltzmann (1890’s) showed how thermodynamics can be derived from statistical mechanics (loss of information causes entropy increase).

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Università di Pisa The first law

Joule: work and heat are two forms of energy exchange.

1 calorie = 4.1868 Joule (J = N m = Kg m2 s-2)

•Work produces coherent particle movement.

•Heat produces incoherent particle movement.δQ

U

δW

0~ ;~ =+= iii uuVu

∑∑==

=+==N

i

iN

N

i

i uuuMMVEmuE1

2122

212

21

1

2

21 ~~ ;~ ;

Mechanical Energy, EK

Coherent motion

Internal Energy, U

Incoherent motionV

1~u 2

~u

3~u

4~u

Nu~

Q = 0; W=Mg ∆z = ∆U = mc ∆T; ∆T ≅ 10Kg 10 ms-2 10 m / 1 Kg 4 103 J Kg-1 K-1 ≅ 0.25K (did Joule cheat?)

MVPmuPN

i

i ==∑=

; :Note1

No” Internal Momentum”

dU = δδδδW + δδδδQ

1° law: Total energy is conserved

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Università di Pisa The Second Law (1)

EK + U1 = U2 = U3

2° law: Energy tends to disperse.

a) Distributes uniformly in space.

b) Moves from its coherent to its

incoherent forms.

V

1~u 2

~u

3~u

4~v

Nu~

1~u 2

~u

3~u

4~u

Nu~

1)

2)

time

time

1~u 2

~u

3~u

4~v

Nu~

V

These events are so improbable that they are practically

impossible (putting things in order is more difficult than the opposite)

1~u

4~u

time

Keenan’s formulation of the 2° law: an isolated system tends to reach its

state of stable equilibrium, corresponding to maximum energy dispersion.

3)

Equipartition theorem: at equilibrium each degree of freedom has the same mean energy kT/2, where

k is Boltzmann constant and T defines temperature. In ideal gas, each particle has energy 3kT/2.

a)

b)

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• Clausius: No process is possible whose

sole result is the transfer of heat from a

cooler to a hotter body (process a).

• Kelvin: No process is possible whose

sole result is the conversion of heat from

a reservoir into work (process b).

TH

TC

QC

QH

W

TH

TC

W

QC

QH

η = W / QH = 1 - QC / QH

W = QH - QC

η < 1 - TC / TH

w = QC / W = QC / (QH - QC)

w < TC / (TH – TC)

TH

TC

QC

QC

QH

W

TH

TC

W

QC

QH

Equivalence of the two formulations.

Note: the circles above indicate that we are referring to processes, in which a working fluid undergoes a

thermodynamic cycle at the end of which it is brought back to its initial conditions. In general, the second law

states that it is impossible that the sole result of a transformation is to completely convert heat into work, or

transfer heat from cold to hot.

The Second Law (2)

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Università di Pisa Problems

1. A power plant burns hydrocarbons and produces 1000 MWe, with a 40% efficiency.

a) How much does it consume? How much of that power is discharged into the cold reservoir (i.e. the sea)?

b) How many liters of hydrocarbons does it burns per hour? (Assume that the fuel produces 4 MJ/kg, with a

density of 0.9 g/cm3)

c) If the same power were to be produced in a hydroelectric power plant, operating through a 1000 m height

difference, calculate the required volumetric flux.

Answer.

a) QH = W / η = 103 MW / 0.4 = 2500 MW. QC = QH – W = 1500 MW.

b) Fuel required = 2.5 109 J/s × 3600 s/h / (4 106 J/kg × 0.9 kg/lt) = 2.5 106 lt/h = 0.66 106 gal/h. (2.5 million

liters correspond to a container 10m × 10m × 2.5m)

c) A river having a 1 m3/s flow rate would produce a power W = mg∆z=1 m3/s × 103 kg/m3 × 9.8 m/s2 × 100 m

≅ 1 MW. Therefore, producing 1000 MW requires a 1000 m3/s. (more or less, that of the Mississippi river)

2. Consider an internal combustion engine. Here the gas extracts heat from a single reservoir (i.e. the flame),

and expands, moving a piston and producing work. Does it contradict the second law?

Answer.

No. The second law states that it is impossible that the sole result of a transformation is to completely convert heat

into work, or transfer heat from cold to hot. Here at the end of the transformation the gas is not in the same

conditions as it were at the beginning and therefore the second law is not applicable.

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Università di Pisa The Second Law (3)

energy free Gibbs; =+−=−= GVdPSdTdGTSHG

P

V

A

B

C

D

1 2

entropy==⇒−==== ∫∫ ST

QdSSSdS

T

Q

T

Q

T

QAC

C

A

C

AC

ADC

H

ABC δδ

dx

pressure;/; =−==−== PAFPAdxdVPdVFdxWδ

F P A

PdVTdSWQdU −=+= δδ

enthalpy; =+=+= HVdPTdSdHPVUH

U = const. when S and V are const.

H = const. when S and P are const.

A = const. when T and V are const.

G = const. when T and P are const.

Thermal equilibrium: T = constant

Mechanical equilibrium: P = constant

energy internal=U

energy free Helmholtz; =−−=−= APdVSdTdATSUA

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Università di Pisa Thermodynamic Properties of Fluids (1)

NV = nV

u( )

V

kTNum

V

NumunP VV ==×= 2

31

61 ~~2~

nv/6 is the number of collision per unit surface and unit time. 2mv is the

momentum transferred to the wall in each collision.

= 3kT/m (equipartition theorem) So:

PV = NRT or Pv = RT where R = NAk is the gas constantNV = N NA

N = # moles

NA = Avogadro #

Ideal Gas

Real Gas / Liquid

Van der Waals:2

3

13

8

rr

rr

vv

TP −

−=2v

a

bv

RTP −

−=

volume)(reduced

pressure) (reduced

re) temperatu(reduced

C

r

C

r

C

r

v

vv

P

PP

T

TT

=

=

=

rrr

r

vTv

v

RT

PvZ

1

8

9

13

3−

−==

Law of corresponding states: Z = Z(Pr,Tr) is universal, valid for any material.

With small corrections, it works remarkably well. See plot next page.

P

v

factorility Compressib ==RT

PvZ

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Z = Pv/RT

Reduced Pressure Pr

The Compressibility Factor

Thermodynamic Properties of Fluids (2)

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Università di Pisa Thermodynamic Properties of Fluids (3)

.1

- ility compressib isothermal

.1 y expansivit volume

T

P

P

V

Vκ

T

V

Vβ

∂∂

==

∂∂

==

( )dPdTVdPP

VdT

T

VdV

TP

κβ −=

∂∂

+

∂∂

=

PTVTPSVS

VVSSVSSV

T

V

P

S

T

P

V

S

S

V

P

T

S

P

V

T

dGdAdH

S

P

V

U

SSV

U

S

U

VV

T

V

UP

S

UTPdVTdSdU

∂∂

−=

∂∂

∂∂

=

∂∂

∂∂

=

∂∂

∂∂

−=

∂∂

∂∂

−=

∂∂

∂∂

=∂∂∂

=

∂∂

∂∂

=

∂∂

⇒

∂∂

−=

∂∂

=⇒−=

; ; ;

: thefind , , , Using

. ; 2

equationsMaxwell

Using the Maxwell equations, the variation of any thermodynamic quantity for a single component system

can be expressed in terms of the variation of T and P (or V), knowing only β, κ and cV (or cP)

VV

VT

ST

T

Uc

∂∂

=

∂∂

== volumeconstant at capacity heat

PP

PT

ST

T

Hc

∂∂

=

∂∂

== pressureconstant at capacity heat ( ) ( ) ;

path)on depends (

capacity heat

dHQdUQ

Q

TQc

PV==

==

δδ

δδ

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Università di Pisa Problem

VdPT

dTcdP

T

VdT

T

cdP

P

SdT

T

SPTdS P

P

P

TP

β−=

∂∂

−=

∂∂

+

∂∂

=),( a)

a) Find dS(T,P); b) Find dS(T,V); c) Find cP – cV.

dVT

dTcdV

T

PdT

T

cdV

V

SdT

T

SVTdS V

V

V

TV κβ+=

∂∂

+=

∂∂

+

∂∂

=),( b)

( )( ) κ

β=

∂∂

∂∂−=

∂∂

∂∂

−=

∂∂

T

P

PTV PV

TV

T

V

V

P

T

P :factIn

( )κβ

κβκβ

βTV

ccdPdTVdVdVT

dTcVdP

T

dTc VPVP

2

; c) =−⇒−=+=−

Note 1: According to the equipartition principle (i.e. the second law), for a monoatomic gas U = 3/2 kT NA, so that

cV = 3/2 R (R = NAk). In general, classical physics predicts that U = 1/2 NT, where N is the number of degrees of

freedom of 1 mole of component, so that cV is a constant. Explaining why it is not so requires quantum physics.

Note 2: For an ideal gas, V = RT/P, so that β = 1/T and κ = 1/P. Consequently, cP – cV = R. Alternatively: for

an ideal gas, U = U(T); H = U + PV = U + RT = H(T), so that cV = dU/dT and cP = dH/dT, with cP = cV + R.

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Università di Pisa

∑=

+−=+=N

i

iidNPdVTdSWQdU1

µδδ µi = chemical potential of ith component

Ni = number of moles of ith component

( ) ii

N

i

iiPT NGdNdG µµ =⇒=∑=1

,µi is the molar free energy of ith component within mixture

G1 = N1g1 G2 = N2g2

G=N1µµµµ1+N2µµµµ2

Thermal equilibrium: T = constant

Mechanical equilibrium: P = constant

Chemical equilibrium: µi = constant

Gibbs phase rule. (N = # components; π = # phases; F = degrees of freedom)

Variables: T, P and xi in each phase. # variables: 2 + P(N-1).

Independent relations: µiα = µi

β =…= µiπ for each component. # relations: N (π-1).

Phase α: T, P, x1, x2, …, xN-1

Phase β: T, P, x1, x2, …, xN-1

Each phase is described by

T, P (same in each phase)

plus compositions xi=Ni/N

(N-1 independent xi)

Multicomponent Systems (1)

F = 2 - ππππ + N

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α = vap

β = liq

N = 1; π= 2; → F = 1

so: P = P(T) = Psat(T).

Small kinetic energy,

large potential energy

Large kinetic energy,

small potential energy

At equilibrium, T, P are the same. So, during phase transition

(isobaric and isothermal), G = const., so that

gα = gβ. (g=G/N).

On average, molecules in the two phases have the same energy.P

T

α = vap

β = liqCritical point ( ) ( ) ( ) ( ), , , ,s T P dT v T P dP s T P dT v T P dPα α β β− + = − +

dgα = dgβ

( )sat

dP s s h h

dT v v T v v

β α β α

β α β α

− − = = − −

β = v (vapor), α = l (liquid)

vl << vv = RT/P (ideal gas)

∆hvl = Latent heat of

vaporization(Clausius-Clapeyron)

Valid at low P

2

1

RT

h

dT

dP

P

lvsat∆=

)/1(

ln

Td

PdRh

satlv −=∆

1/T

lnPsat

Single Component, Two Phase Systems

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Università di Pisa

T

P

Q

T

P, x1,g1 P, x2, g2

T

P, g

Ideal gas

( ) ( ) const.) ( ;lnln 2211 =∆−=−∆=∆+−=∆ T,P,vsTTshgxxxxRs

VVxxx ii / ;121 ==+

( ) 221122112211 lnln µµ xxxxxxRTgxgxg +=+++=

( )iiiii fxRTxRTg lnln =+=µ

Mixing of ideal mixtures

i

f

v

v

v

vv

vR

v

dv

T

RTPdv

TT

W

T

Qs

f

i

f

i

ln1

===−==∆ ∫∫

( ) ( )PRTdvdPdgT

ln== PffRTg =≡ :gas ideal (fugacity) ln

( ) ( )TPfPff sat=⇒≠ :LiquidVap, T, Psat, f = Psat

Liq, T, Psat, f = Psat

Multicomponent Systems (2)

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Raoul law: valid for ideal mixtures.

in particular, valid for ideal liquid mixtures, i.e. benzene /

toluene, composed of species having similar molecules

In general, Raoult law is valid when xi →→→→ 1.

Multicomponent Systems (3)

( )TPxPyp sat

iiii ==

Vap T, P, yi

Liq T, P, xi

Henry’s law: the solubility of a gas in a liquid at a particular temperature is proportional to the

pressure of that gas above the liquid:

constant law sHenry' == iiii HxHp

In particular, Hi = Pisat for ideal mixtures.

In general, Henry’s law is valid when xi→→→→ 0.

Vapor – Liquid Equilibrium

liq

ii

vap

ii

liq

i

vap

i fxRTfyRT lnln =⇒= µµ