Phase Changes
Courtesy www.lab-initio.com
The BIG Idea
• Students know energy is released
when a material condenses or
freezes and is absorbed when a
material evaporates or melts.
constant
Temperature remains __________
during a phase change.
Water phase
changes
Phase Changes
Energy must be absorbed to change from a
solid to a liquid; heat of fusion or enthalpy of
fusion, DHfus
Energy must be absorbed to change from a
liquid to a gas; heat of vaporization or
enthalpy of vaporization, DHvap
Phase Changes
For water, the DHfus = 6.02 kJ/mol
the DHvap = 40.7 kJ/mol
Why is the DHvap much higher than DHfus?
Example #1
How much energy is needed to convert 25g of ice
at -15C to 200C steam?
Plan: Follow the heating curve to determine where the
energy goes at each step in the process
Can you identify the FIVE steps in the process?
Example #1 Continued
Step 1: heat ice from -15C to 0C
Cice = 2.03J/gC
q = mCDT
q = 25g (2.03J/gC)(15C) = 761.25J = .761kJ
Step 2: phase change (s) (l)
25g (1mol/18.02g) (6.02kJ/mol) = 8.35kJ
Step 3: heat water from 0C to 100C
Cwater = 4.18J/gC
q=mCDT
q=25g (4.18J/gC)(100C)= 10450J = 10.4kJ
Example #1 Continued
Step #4: phase change (l) (g)
25g (1mol/18.02g)( 40.7kJ/mol) = 56.5kJ
Step #5: heat steam from 100C to 200C
q=mCDT
q = 25g (2.0J/gC)(100C) = 5000J = 5kJ
Csteam = 2.0J/gC
Total heat required:
0.761kJ + 8.35kJ + 10.4kJ + 56.5kJ + 5kJ = 81kJ
Now let’s try one:
The molar heat of fusion for benzene, C6H6, is 9.92kJ/mol. The
molar heat of vaporization is 30.7kJ/mol.
Calculate the heat required to melt 8.25g of benzene at its
normal melting point.
1.05kJ
Calculate the heat required to vaporize 8.25g of benzene at its
normal boiling point.
3.24kJ
Why is the heat of vaporization more than 3 times the heat of
fusion?
Another Example
You have a 75g sample of water at 125C. What phase(s) are
present when 215kJ of energy is removed from the sample?
Hint: Identify the steps on the cooling curve that may occur:

Energy to cool from 125C to 100C

Energy to phase change from a vapor to a liquid

Energy to cool from 100C to 0C

Energy to phase change from a liquid to a solid

Energy to cool the solid
A mixture of H2O(l) and H2O(s) will be present
Vapor Pressure
Pressure of the vapor present when equilibrium is achieved
between the rate of vaporization and the rate of condensation.
At the boiling point, the Patm = Pvapor
As the vapor pressure on a pot of water is reduced, the energy
needed to boil that water is also reduced.
Pressure Cooker: By increasing the vapor pressure, additional
energy is needed for the water to boil, therefore the water can
boil at temperatures above 100C.
Effect of Pressure on Boiling Point
Water
Phase Diagram
 Represents phases as a function of temperature and
pressure.
 Critical temperature: temperature above which the
vapor can not be liquefied.
 Critical pressure: pressure required to liquefy AT the
critical temperature.
 Critical point: critical temperature and pressure (for
water, Tc = 374°C and 218 atm).
Phase changes by Name
Carbon dioxide
Phase
Diagram
for Carbon
dioxide
Carbon
Phase
Diagram
for Carbon
Vapor Pressure vs. Temperature
As the temperature increases, a greater number of molecules
have sufficient kinetic energy to convert from the liquid to the
vapor phase.
There is a nonlinear relationship between the vapor pressure of
a liquid and temperature.
Vapor Pressure vs. Temperature
The Clausius – Clapeyron Equation
• A mathematical expression which relates the variation of
vapor pressure to temperature
• ln P = (-DHvap/RT) + C where C is a constant
• IMPORTANCE:
• When the ln P is plotted vs (1/T) you create a line where the
slope is equal to the –DHvap/R
• Which means you can calculate the heat of vaporization from
the slope of the line.
• R = 8.314 J/Kmol
Convert all Temps to Kelvin
Example #1
Calculate the heat of vaporization of a compound that has a
vapor pressure of 560 torr at 25C and a normal boiling point of
45C.
Hint: normal BP occurs at 1atm (760 torr)
Variation of the C-C Equation is used:
ln (P1/P2) = (DHvap/R )(1/T2 – 1/T1)
ln(560torr/760torr) = (DH/8.314J/Kmol) (1/318 – 1/298)
12033J, 12kJ = DH
Example #2
In Breckenridge, CO, the typical atmospheric pressure is 520 torr.
What is the boiling point of water (DH=40.7kJ/mol) in
Breckenridge?
Hint: 760torr BP=100C / 373K
ln(520torr/760torr)= (40,700J mol-1 /8.314)(1/373 - 1/T)
-0.379 = 4896(1/373 - 1/T)
-7.75 E -5 = (1/373 - 1/T)
1/T = 2.76 E -3
T = 363K
Example #3
The temperature inside a pressure cooker is 115C. Calculate the
vapor pressure of water inside the pressure cooker.
ln (P/1atm) = (40,700kJ mol-1 / 8.314)(1/373 – 1/388)
lnP = 0.51
P = e.51 = 1.7atm
What would be the temperature inside the pressure cooker if
the vapor pressure of water was 3.50atm?
ln(3.50/1.00) = (40.7kJ mol-1 / 8.314)(1/373 – 1/T)
413K = T
Example #4
The normal boiling point for acetone is 56.5C . At an elevation of
5600ft, the atmospheric pressure is 630 torr. What would be the
boiling point of acetone at this elevation? (DHvap = 32.0kJ/mol).
ln(760/630) = (32kJ mol-1/8.314)(1/T – 1/330)
325K = T
What would be the vapor pressure of acetone at 25C at this
elevation?
ln(630/P) = (32kJ mol-1/8.314)(1/298 – 1/325)
P = 221 torr