Clausius-Clapeyron Equation
Cloud drops first form when the vaporization equilibrium point is reached(i.e., the air parcel becomes saturated)
Here we develop an equation that describes how the vaporization/condensation equilibrium point changes as a function of pressure and temperature
T
C
T (ºC)
p (mb)
3741000
6.11
1013
221000
Liquid
Vapor
Solid
Who are these people?
Clausius-Clapeyron Equation
Benoit Paul Emile Clapeyron1799-1864
French Engineer / Physicist
Expanded on Carnot’s work
Rudolf Clausius1822-1888German
Mathematician / Physicist
“Discovered” the Second LawIntroduced the concept of entropy
Basic Idea:
• Provides the mathematical relationship(i.e., the equation) that describes anyequilibrium state of water as a functionof temperature and pressure.
• Accounts for phase changes at eachequilibrium state (each temperature)
Clausius-Clapeyron Equation
T
C
T (ºC)
p (mb)
3741000
6.11
1013
221000
Liquid
Vapor
Solid
V
P(mb)
Vapor
Liquid
Liquidand
Vapor
T
esw
Sections of the P-V and P-T diagrams forwhich the Clausius-Clapeyron equation is derived in the following slides
Mathematical Derivation:
Assumption: Our system consists of liquid water in equilibrium withwater vapor (at saturation)
• We will return to the Carnot Cycle…
Clausius-Clapeyron Equation
Temperature
T2 T1
esw1
esw2
Satu
ratio
n va
por p
ress
ure
A, D
B, C
Volume
T2
T1esw1
esw2
Satu
ratio
n va
por p
ress
ure
A D
B C
Isothermal processAdiabatic process
Mathematical Derivation:
• Recall for the Carnot Cycle:
• If we re-arrange and substitute:
Clausius-Clapeyron Equation
21NET QQW
1
21
1
21
TTT
QQQ
where: Q1 > 0 and Q2 < 0
21
NET
1
1
T-TW
TQ
Volume
T2
T1esw1
esw2
Satu
ratio
n va
por p
ress
ure
A D
B C
Isothermal processAdiabatic process
WNET
Q1
Q2
Volume
T2
T1esw1
esw2Sa
tura
tion
vapo
r pre
ssur
eA D
B C
Isothermal processAdiabatic process
WNET
Q1
Q2
Mathematical Derivation:
Recall:
• During phase changes, Q = L
• Since we are specifically workingwith vaporization in this example,
• Also, let:
Clausius-Clapeyron Equation
21
NET
1
1
T-TW
TQ
v1 LQ
TT1
dTTT 21
Mathematical Derivation:
Recall:
• The net work is equivalent to thearea enclosed by the cycle:
• The change in pressure is:
• The change in volume of our system ateach temperature (T1 and T2) is:
where: αv = specific volume of vaporαw = specific volume of liquiddm = total mass converted from
vapor to liquid
Clausius-Clapeyron Equation
dmααdV wv
sw2sw1sw eede
21
NET
1
1
T-TW
TQ
dpdVWNET
Volume
T2
T1esw1
esw2Sa
tura
tion
vapo
r pre
ssur
eA D
B C
Isothermal processAdiabatic process
WNET
Q1
Q2
Mathematical Derivation:
• We then make all the substitutions into our Carnot Cycle equation:
• We can re-arrange and use the definition of specific latent heat ofvaporization (lv = Lv /dm) to obtain:
Clausius-Clapeyron Equation for the equilibrium vapor pressurewith respect to liquid water
Clausius-Clapeyron Equation
21
NET
1
1
T-TW
TQ
dTdedmαα
TL swwvv
wv
vsw
ααTdTde
l
Temperature
T2 T1
esw1
esw2
Satu
ratio
n va
por p
ress
ure
A, D
B, C
General Form:
• Relates the equilibrium pressure between two phases to the temperatureof the heterogeneous system
where: T = Temperature of the system l = Latent heat for given phase change
dps = Change in system pressure at saturationdT = Change in system temperatureΔα = Change in specific volumes between
the two phases
Clausius-Clapeyron Equation
TΔdTdps l
T
C
T (ºC)
p (mb)
3741000
6.11
1013
221000
Liquid
Vapor
Solid
Equilibrium States for Water(function of temperature and pressure)
Application: Saturation vapor pressure for a given temperature
Starting with:
Assume: [valid in the atmosphere]
and using: [Ideal gas law for the water vapor]
We get:
If we integrate this from some reference point (e.g. the triple point: es0, T0) to some arbitrary point (esw, T) along the curve assuming lv is constant:
Clausius-Clapeyron Equation
wv αα
TRαe vvsw
2v
v
sw
sw
TdT
Rede l
wv
vsw
ααTdTde
l
T
T 2v
ve
esw
sw
0
sw
s0 TdT
Rede l
Application: Saturation vapor pressure for a given temperature
After integration we obtain:
After some algebra and substitution for es0 = 6.11 mb and T0 = 273.15 K we get:
Clausius-Clapeyron Equation
T
T 2v
ve
esw
sw
0
sw
s0 TdT
Rede l
T1
T1
Reeln
0v
v
s0
sw l
T(K)1
273.151
Rexp11.6(mb)e
v
vsw
l
Application: Saturation vapor pressure for a given temperature
A more accurate form of the above equation can be obtained when we do notassume lv is constant (recall lv is a function of temperature). See your book forthe derivation of this more accurate form:
Clausius-Clapeyron Equation
T(K)1
273.151
Rexp11.6(mb)e
v
vsw
l
)(ln09.5
)(680849.53exp11.6(mb)esw KT
KT
Application: Saturation vapor pressure for a given temperature
What is the saturation vapor pressure with respect to water at 25ºC?
T = 298.15 K
esw = 32 mb
What is the saturation vapor pressure with respect to water at 100ºC?
T = 373.15 K Boiling point
esw = 1005 mb
Clausius-Clapeyron Equation
)(ln09.5
)(680849.53exp11.6(mb)esw KT
KT
Application: Boiling Point of Water
At typical atmospheric conditions near the boiling point:
T = 100ºC = 373 Klv = 2.26 ×106 J kg-1
αv = 1.673 m3 kg-1
αw = 0.00104 m3 kg-1
This equation describes the change in boiling point temperature (T) as a functionof atmospheric pressure when the saturated with respect to water (esw)
Clausius-Clapeyron Equation
wv
vsw
ααTdTde
l
1sw Kmb36.21dT
de
Application: Boiling Point of Water
What would the boiling point temperature be on the top of Mount Mitchell if the air pressure was 750mb?
• From the previous slide we know the boiling pointat ~1005 mb is 100ºC
• Let this be our reference point:
Tref = 100ºC = 373.15 Kesw-ref = 1005 mb
• Let esw and T represent thevalues on Mt. Mitchell:
esw = 750 mb
T = 366.11 KT = 93ºC (boiling point temperature on Mt. Mitchell)
Clausius-Clapeyron Equation
1
ref
refswsw Kmb36.21TTee
refrefsw T
eT
36.21esw
1sw Kmb36.21dT
de
Equilibrium with respect to Ice:
• We will know examine the equilibriumvapor pressure for a heterogeneoussystem containing vapor and ice
Clausius-Clapeyron Equation
T
C
T (ºC)
p (mb)
3741000
6.11
1013
221000
Liquid
Vapor
Solid
C
V
P(mb)
Vapor
Solid
Liquid
T
6.11 T
ABesi
Equilibrium with respect to Ice:
• Return to our “general form” of theClausius-Clapeyron equation
• Make the appropriate substitution forthe two phases (vapor and ice)
Clausius-Clapeyron Equation for the equilibrium vaporpressure with respect to ice
Clausius-Clapeyron Equation
T
C
T (ºC)
p (mb)
3741000
6.11
1013
221000
Liquid
Vapor
Solid
TdTdes l
iv
ssi
ααTdTde
l
Application: Saturation vapor pressure of ice for a given temperature
Following the same logic as before, we can derive the following equation forsaturation with respect to ice
A more accurate form of the above equation can be obtained when we do notassume ls is constant (recall ls is a function of temperature). See your book forthe derivation of this more accurate form:
Clausius-Clapeyron Equation
T(K)1
273.151
Rexp11.6(mb)e
v
ssi
l
)(ln555.0
)(629316.26exp11.6(mb)esi KT
KT
Application: Melting Point of Water
• Return to the “general form” of the Clausius-Clapeyron equation and make theappropriate substitutions for our two phases (liquid water and ice)
At typical atmospheric conditions near the melting point:
T = 0ºC = 273 Kls = 0.334 ×106 J kg-1
αw = 1.00013 × 10-3 m3 kg-1
αi = 1.0907 × 10-3 m3 kg-1
This equation describes the change in melting point temperature (T) as a functionof pressure when liquid water is saturated with respect to ice (pwi)
Clausius-Clapeyron Equation
iw
fwi
ααTdTdp
l
1wi Kmb135,038dT
dp