thermodynamics of water - 1
DESCRIPTION
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!. review …. If that’s not enough… Consider whether the First Law of Thermodynamics will help (and maybe the Gas Law). review …. - PowerPoint PPT PresentationTRANSCRIPT
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Thermodynamics of Water - 1
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Take notes!
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1. Quick review from 121A
To solve any thermo problem for dry air…
Consider whether the Gas Law alone will help!
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* *
*
/ , 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
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review …
If that’s not enough…
Consider whether the First Law of Thermodynamics will help (and maybe the Gas Law)
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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
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review …
And if that’s not enough…
Consider whether the Second Law of Thermodynamics will help
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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...
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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)
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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
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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
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TD Diagrams
• Shows a cyclic process
• Area enclosed = work done during process
• Area = pd = w
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TD Diagrams
• CAR p. 248-249 Figs. VIII-12-14
• Show isothermal, isosteric, isobaric processes
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TD Diagrams
• Tsonis p. 39 (handout)
For a p-diagram:
• Isotherms are “equilateral hyperbolas”• Adiabats are too, but are steeper (Fig. a)
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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).
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TD Diagrams
For a pT-diagram:
• Isochores (constant ) are straight lines (Fig. b)
• Adiabats are “equilateral hyperbolas”
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TD Diagrams
In pT-space:
• Fig. d• Adiabats are “equilateral hyperbolas”
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WATER!!!
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Water
Water is hugely important and interesting!
We will look at:
• Vapor alone (briefly…it’s a gas)• Liquid alone (briefly)• Ice alone (briefly)
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Water
We will look at:
• The coexistence of water in two states.
Mainly…• Vapor & liquid• Liquid and ice
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Water
Concepts…
• Vapor pressure (e)
• Saturation vapor pressure (es)
– Actually … esw = SVP over water
• Latent heats
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Water
And …
• The Clausius-Clapeyron Equation
– Gives the relationship for es(T)
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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
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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).
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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)!
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Water
• pc = 220,598 mb (yikes!)
• Tt = 647 K
• T > Tt means we have a gas
• T < Tt means we have a vapor
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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
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Water vapor alone…
• At saturation, e es – the saturation vapor pressure (SVP) - in which case:
esv=RvT
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*
*
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
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Liquid or ice …
• For liquid water:
• w = 10-3 m3/kg density
• For ice:
• w = 1.091 x 10-3 m3/kg density
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Specific heats…
Vapor Cpv 1846
Vapor Cvv 1384.5
Cpv - Cpv = Rv
Liquid Cw 4187
Solid ci 2106
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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
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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
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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.
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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
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Phase changes
• Remember enthalpy?
• Specific enthalpy is: h = u + p• With: dh = cpdT called sensible heat
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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
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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
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Latent heats
• Latent heats are “reversible”
– Example: lwv = - lvw
• Also they are “additive”
– Example: liv = liw + lwi
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Clausius-Clapeyron
• Unsaturated water vapor: ev=RvT
• Saturated water vapor: es = es =(T)
• We’d like to know how es varies with T
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Clausius-Clapeyron
• Following CAR, we use Gibbs (free) energy to derive a relationship.
• p.268… g = u – Ts + p• s = entropy
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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
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Clausius-Clapeyron
• Importantly, for an isothermal, isobaric process:
dg = 0
• Gibbs free energy is unchanged
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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)
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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
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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
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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
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Clausius-Clapeyron
• 1dp – s1dT = 2dp – s2dT
• (1 - 2)dp = (s1 – s2)dT
• dp = (s1 – s2) dT (1 - 2)
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Clausius-Clapeyron
• From slide 37:
T(s1 – s2) = l = qp = dh
• dp = l dT T(1 - 2) Clausius-
clapeyron equation
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Clausius-Clapeyron
• Special case: vaporization• Then: p es
• And thus:
( )sw wv
v w
de ldT T
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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)
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Clausius-Clapeyron
• And thus:
2sw sw wv
v
de e ldT T R
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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
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Clausius-Clapeyron
• CAR also derives expressions for the other processes.