Thermo & Stat Mech - Spring 2006 Class 9 1 Thermodynamics and Statistical Mechanics Change of Phase.

Thermo & Stat Mech - Spring 2006 Class 9

1

Thermodynamics and Statistical Mechanics

Change of Phase


2

Thermodynamic Potentials

We know that for an isolated system, S ≥ 0. Therefore, any processes in an isolated system can only increase entropy, and the system will be in equilibrium when it reaches maximum entropy.

But what of a system that is not isolated?


3

Helmholtz Function

Suppose a system is in contact with a reservoir at temperature T. The system undergoes a process, and Q is transferred from reservoir to system. S0 is the entropy change of the reservoir, and S is the entropy change of the system.

S0 + S ≥ 0


4

Work

T

QS 0 so, S0 + S ≥ 0 becomes

fiffii

ifif

FFTSUTSUW

TSTSUUW

STUWQUW

QSTST

Q

)()(

)()(

or ,0


5

Helmholtz Function

W ≤ – F (Constant T)

If work is zero for the process,

F ≤ 0, or Ff ≤ Fi

System tends to go to lowest F.

At stable equilibrium when dF = 0


6

Gibbs Function

If contact with the reservoir keeps both temperature and pressure constant, the system goes to the lowest value of the Gibbs function. As before, TS ≥ Q, but in addition, W = PV. Then, Q = U + PV, or

U + PV – Q = 0

U + PV – TS ≤ 0


7

Gibbs Function

U + PV – TS ≤ 0

(Uf + PVf – TSf ) – (Ui + PVi – TSi ) ≤ 0

G ≤ 0

Gf ≤ Gi

Constant T and P.

Stable equilibrium when dG = 0


8

Gibbs Function

If non-mechanical work is done by the system, at constant T and P, then as with F,

Wnm ≤ G


9

Phase Transition

A phase transition, as from a liquid to a vapor, usually takes place at constant temperature and pressure. Therefore the system will go to the state of lowest Gibbs function. Let us see how the specific Gibbs function changes with temperature.

gis the vapor and g is the liquid.


10

Gibbs Function at Transition


11

Gibbs Function

dG = – SdT + VdP G(T, P)

VTP

TP

P

GV

T

GS

dPP

GdT

T

GdG

and


12

Transition


13

Transition

ggGG

gngnG

gngnG

nnnn

so 21

222

111

2211


14

Real Substance


15

Transition

heatLatent )(

)()(

23

ssTvv

ss

dT

dP

dPvvdTss

dPvdTsdPvdTs

gdgd

gg


16

Clausius-Clapeyron Equation

liquid-Solid )(

vapor-Solid )(

vapor-Liquid )(

12

12

13

13

23

23

vvTdT

dP

vvTdT

dP

vvTdT

dP


17

Enthalpy and Latent Heat

du = đq – Pdv

At transition, u2 – u1 = l12 – P(v2 – v1)

l12 = (u2 + Pv2) – (u1 + P v1)

l12 = h2 – h1


18

Enthalpy and Latent Heat

ion)(subllimat vapor tosolid

ion)(vaporizat vapor toliquid

(fusion) liquid tosolid

13

23

12

hh

hh

hh


19


20

Regelation


21

Problem

Consider a sealed steel container completely filled with water at 0ºC and pressure of one atmosphere. Lower the temperature to – 1ºC. What happens? Water starts to freeze, but tries to expand. That raises pressure, so freezing point is lowered. How much?


22

Freezing Problem

K 1

K 273

/kgm 1005.9

J/kg 1034.3

)(

so )(

35

512

1212

12

T

T

vv

vvT

TP

vvTdT

dP


23

Freezing Problem

atm 135

atm 134Pa/atm 101.01

Pa 1035.15

7

P

P

How much freezes?Call the fraction that freezes x.


24

Freezing Problem

vdPvdTvv

dPP

vdT

T

vvv

vv

vvx

vxvxv

if

TPif

ff

fi

ffi

)1(


25

Freezing Problem

/kgm 10

/kgm 1005.9

Pa 1035.1 Pa 1012

K 1 K 1067

33

35

71-11

1-6

v

vv

P

T

vv

PvTv

vv

vvx

ff

ffff

fi


26

Freezing Answer

x = 0.017 = 1.7%

Not much!

Thermo & Stat Mech - Spring 2006 Class 9 1 Thermodynamics and Statistical Mechanics Change of Phase.

Documents

thermo stat mech spring

p slide

transition slide

regelation slide

f f f i system

latent heat slide

freezing problem slide

real substance slide