Thermodynamics of Water - 1

Take notes!

1. Quick review from 121A

To solve any thermo problem for dry air…

Consider whether the Gas Law alone will help!

* *

*

/ , universal gas constant, molecular weight of the gas ( or ???)

number of moles of gas

d

d

d

p R T

R R m Rm m M

pV nR T

n

p R T

review …

If that’s not enough…

Consider whether the First Law of Thermodynamics will help (and maybe the Gas Law)

heat (energy) supplied (extracted) (inexact)internal energy changework done on/by system (inexact)

expansion/contraction work for an ideal gas

So:

v

v p

q du w

qduw

w pddu c dT

q c dT pd c dT dp

review …

And if that’s not enough…

Consider whether the Second Law of Thermodynamics will help

Involves entropy, :

"the entropy of a an isolated system undergoing an irreversible process must increase" - CAR p. 263

"entropy is a property which tells us in which direction a process will go"

s

qdsT

example: top of p. 259...beaker of boiking water...

review …

We look at various special cases:

1) Isothermal processes (dT = 0)2) Isobaric processes (dp = 0)3) Isosteric processes (d = 0)

4) Adiabatic processes (d = 0)

TD Diagrams

• For an ideal gas, we have three unknowns:p T

• But the Eqn. of State allows us to reduce to two:

p, or p,T or ,T

TD Diagrams

• We often represent processes on diagrams with axes (p,) or (p,T) or (,T)

• The (p,) diagram is often used• CAR p. 234 Fig. VIII-8

TD Diagrams

• Shows a cyclic process

• Area enclosed = work done during process

• Area = pd = w

TD Diagrams

• CAR p. 248-249 Figs. VIII-12-14

• Show isothermal, isosteric, isobaric processes

TD Diagrams

• Tsonis p. 39 (handout)

For a p-diagram:

• Isotherms are “equilateral hyperbolas”• Adiabats are too, but are steeper (Fig. a)

TD Diagrams

Note…

• The 2nd Law is often derived via analysis of the Carnot Cycle – a cycle involving two adiabatic processes and two isothermal processes (p. 260, Fig. VIII-19).

TD Diagrams

For a pT-diagram:

• Isochores (constant ) are straight lines (Fig. b)

• Adiabats are “equilateral hyperbolas”

TD Diagrams

In pT-space:

• Fig. d• Adiabats are “equilateral hyperbolas”

WATER!!!

Water

Water is hugely important and interesting!

We will look at:

• Vapor alone (briefly…it’s a gas)• Liquid alone (briefly)• Ice alone (briefly)

Water

We will look at:

• The coexistence of water in two states.

Mainly…• Vapor & liquid• Liquid and ice

Water

Concepts…

• Vapor pressure (e)

• Saturation vapor pressure (es)

– Actually … esw = SVP over water

• Latent heats

Water

And …

• The Clausius-Clapeyron Equation

– Gives the relationship for es(T)

Water

Consider pure water …

• CAR p. 274 Figs. IX-1-3 show the pT- surface for water.

• Find the three phases: solid, liquid, vapor/gas

Water

• Note the projections/slices in the pT-plane (Fig. 2) and the p-plane (3).

• Note the triple line/triple point of water: the temperature and pressure at which all three phases coexist (Tt, pt).

Water

• In the pT-plane there is the triple point.

• pt = 6.11 mb

• Tt = 0C = 273 K

• Note the critical point where (vapor) = (liquid)!

Water

• pc = 220,598 mb (yikes!)

• Tt = 647 K

• T > Tt means we have a gas

• T < Tt means we have a vapor

Water vapor alone…

• Water vapor is an ideal gas and obeys its Equation of State:

“p=RT” - which we write as:

where e = vapor pressure(and everything else should be obvious!)

ev=RvT

Water vapor alone…

• At saturation, e es – the saturation vapor pressure (SVP) - in which case:

esv=RvT

*

*

mol. wt. of dry air

mol. wt. of water vapor

1 with 0.622

v v

dv v d

d

dv v

d v

dv

v d

vv d

d

e R T

me R T m

m

m Re T mm m

m Re Tm m

me R T

m

Liquid or ice …

• For liquid water:

• w = 10-3 m3/kg density

• For ice:

• w = 1.091 x 10-3 m3/kg density

Specific heats…

Vapor Cpv 1846

Vapor Cvv 1384.5

Cpv - Cpv = Rv

Liquid Cw 4187

Solid ci 2106

2. Phase changes

• Look at CAR p. 275 Fig. IX-3…

• Imagine starting with vapor @ To in a piston• Compress the vapor (p , ) point “b” at

which vapor is saturated (e es)• Further reduction in volume liquid appears

– the two phases coexist

Phase changes

• Continued reduction in volume occurs @ constant pressure and constant temperature (line “bc”) but not constant volume (since the vapor-liquid mix is NOT an ideal gas)

• At “c”, all water is in liquid form• Now we need much larger pressure increases

to get volume changes

Phase changes

• In going from “b” to “c”, the vapor is compressed and work is done on the vapor.

• Heat is liberated during the process … latent heat (latent heat of condensation here)

• In this case in the atmosphere, the surrounding “air” would be warmed by this – not the water substance.

Phase changes

• Whenever water goes to a state of reduced molecular energy, latent heat is released

• Thus, latent heat is heat released (or absorbed) during the process:

L = Qp

Phase changes

• Remember enthalpy?

• Specific enthalpy is: h = u + p• With: dh = cpdT called sensible heat

Phase changes

• h = u + p• dh = du + pd + dp• dh = du + pd since isobaric!• dh = q from the 1st Law!

• Thus: L = Qp l = qp = dh

• Latent heat enthalpy change

Latent heats…

Vaporization (vapor-liquid)

lv or lwv2.5 x 106 J/kg

Fusion (liquid-ice)

li or liw3.34 x 105 J/kg

Sublimation(vapor-ice)

li or liv2.5 x 106 J/kg

Latent heats

• Latent heats are “reversible”

– Example: lwv = - lvw

• Also they are “additive”

– Example: liv = liw + lwi

Clausius-Clapeyron

• Unsaturated water vapor: ev=RvT

• Saturated water vapor: es = es =(T)

• We’d like to know how es varies with T

Clausius-Clapeyron

• Following CAR, we use Gibbs (free) energy to derive a relationship.

• p.268… g = u – Ts + p• s = entropy

Clausius-Clapeyron• So: dg = du –Tds – sdT + pd + dp

• Total work is: w = Tds – du (VII-82)

• By substitution: wtot = – dg – sdT + pd + dp

(VII-100)

• Here, pd = expansion work

• And, – dg – sdT + dp = other work done during in the process

Clausius-Clapeyron

• Importantly, for an isothermal, isobaric process:

dg = 0

• Gibbs free energy is unchanged

Clausius-Clapeyron

• In a phase change (say vapor liquid), temperature & pressure are unchanged.

• See Fig. IX-1

• Hence: dg = 0

• Or: g1 = g2 (e.g., gvapor = gliquid)

Clausius-Clapeyron

• From Fig. IX-4

• At one (p,T):g1 = g2

• At another (p,t) = (p+dp,T+dT)(g+dg)1 = (g+dg)2

• Thus:dg1 = dg2

Clausius-Clapeyron• Again in a phase change (say vapor liquid), the only work done is

expansion work.

• So from slide 42, wtot = pd

• And VII-100 gives:– dg – sdT + pd + dp = pd

– dg – sdT + dp = 0

dg = dp – sdT in a phase change

Clausius-Clapeyron

dg = dp – sdT

• Going back to having two values (p,T) and (p+dp,T=dT) – remember we are varying T to determine the variation of es with T – we have:

• dg1 = dg2

• 1dp – s1dT = 2dp – s2dT

Clausius-Clapeyron

• 1dp – s1dT = 2dp – s2dT

• (1 - 2)dp = (s1 – s2)dT

• dp = (s1 – s2) dT (1 - 2)

Clausius-Clapeyron

• From slide 37:

T(s1 – s2) = l = qp = dh

• dp = l dT T(1 - 2) Clausius-

clapeyron equation

Clausius-Clapeyron

• Special case: vaporization• Then: p es

• And thus:

( )sw wv

v w

de ldT T

Clausius-Clapeyron

• If we know how lwv varies with T, we can integrate!

• For vapor liquid:

• We know: w (slide 30)

• We know: v = RvT/esw (slide 28)

Clausius-Clapeyron

• And thus:

2sw sw wv

v

de e ldT T R

Clausius-Clapeyron

• If we assume lwv is constant, we can solve:

• esw = (es)t @ triple point

1 1exp wvsw s t

v t

le eR T T

Clausius-Clapeyron

• CAR also derives expressions for the other processes.