Experiment #2
VAPOR PRESSURE OF WATER
What are we doing in
this experiment?
1.Determine the vapor pressure of water
at different temperatures.
2. Determine the heat of vaporization of
water.
Review of Gas laws
1
Boyle’s Law: P  ( at cons tan t T and n)
V
Charle’s Law: V
Avogadros’s
Law:
 T (at cons tan t P and n)
V  n(at cons tan t P and T )
PV  nT
IDEAL GAS LAW PV  nRT
IDEAL GAS LAW
PV  nRT
Units of P: atmospheres
Units of V: Liters
Units of T: kelvin
Units of R: L.atm
mol .k
Dalton’s law of partial pressure
A+B+C
Ptotal= PA + PB + PC
Ptotal
nA RT   nB RT   nC RT 





 V   V   V 
Relation between mole fraction and
partial pressure of a gas
A+B
Ptotal= PA + PB
nA RT 

nB RT 

PA  

PB  

 V 
 V 
PAV 

PBV 

nA  

nB  

RT


 RT 
 nA 
Mole fraction of A   A  

 nA  nB 
 nA 
Mole fraction of A   A  

 nA  nB 
  PAV  
 
 
 A    RT  
  PAV  PBV  


  RT RT  
V  


 PA 
 
P
P
RT


A

A  
 A
 V  P  P   PA  PB Ptotal

A
B 
 RT

PA
A 
PA  Ptotal   A
Ptotal
What is
VAPOR PRESSURE OF WATER?
It is the pressure (technically, partial pressure)
exerted by the water molecules in the vapor
phase (gas, water vapor) above the surface
of a liquid at equilibrium at that temperature.
Vapor phase
or
Water vapor
Pvap
Liquid phase
Heat
Define Pressure
It is defined as force applied per unit area
Force ( N )
Pr essure 
Area (m 2 )
Units of pressure:
1N
2
 1Pascal
m
5
1 atm  1.0110 Pascal
1 atm  760 mm Hg  760 torr
1 atm  29.92 inches Hg
1 atm  14.7 psi
1 inch  2.54 cm
1 cm  10 mm
Why does water not vaporize
by itself quickly?
Atmosphere
Vapor phase
or
Water vapor
Liquid phase
Strong inter molecular hydrogen bonds
H
O--H-O-H - - O
H
H
H
When does a liquid boil?
Vapor pressure of the liquid = surrounding
atmospheric pressure
What is boiling point of a liquid ?
It is the temperature at which,
Vapor pressure of the liquid = surrounding
atmospheric pressure
Normal boiling point: If Patm= 1.0 atm
Composition of Dry Air
Gas
N2
O2
Ar
CO2
Rest
% by volume
78.09
20.94
0.93
0.03
0.01
What does it mean if a liquid
has high vapor pressure?
1. More molecules are found in the
vapor phase.
2. Does not require a lot of energy to
vaporize the liquid.
3. Weak inter molecular attractive forces
holding the molecules together in their
liquid state.
A comparison
Open system
Heat
Closed system
Heat
Rate of vaporization cannot equal Rate of vaporization can equal
rate of condensation
rate of condensation
Dynamic equilibrium not
Dynamic equilibrium
reachable
reachable
A comparison
Open system
Heat
As the temperature is increased,
We can send more and more
molecules in to
Vapor phase
Closed system
Heat
There can only be a fixed
number of molecules that
can be in
the vapor phase
A comparison
Open system
Heat
The maximum vapor pressure
that can be reached here is
atmoshpheric pressure
Closed system
Heat
The maximum vapor pressure
that can be reached here is
more than the atmoshpheric
pressure
What is
VAPOR PRESSURE OF WATER?
It is the pressure (technically, partial pressure)
exerted by the water molecules in the vapor
phase (gas, water vapor) above the surface
of a liquid at equilibrium at that temperature.
Vapor phase
or
Water vapor
Pvap
Liquid phase
Heat
EXPERIMENTAL SETUP
Surface of water
should cover test
tube.
Magnetic stir-bar
(position
carefully)
Stirrer/hot
plate
Air bubble at the top
of a inverted test tube
Cut the scale out from the
manual
0
1
2
3
4
5
6
7
8
Note:
This scale is unit
less and does not
correspond to
inches or
centimeters.
Paste the scale on a 100 mm
test tube as shown, with the
scale facing inside of the tube
EXPERIMENTAL SETUP
Surface of water
should cover test
tube.
Magnetic stir-bar
(position
carefully)
Stirrer/hot
plate
Air bubble at the top
of a inverted test tube
What are we going to do?
1. Measure the volume of the bubble at
different temperatures
Charle’s Law
Low temperature Higher temperature
As temperature increase, the volume increases (at constant P,n)
What are we going to do?
2. Calculate the pressure contribution from N2 and O2
in the bubble using the ideal gas equation.
3. Calculate the vapor pressure of water.
Ptotal= PN2 + PO2+ Pwvap
PN2 and PO2 are much greater than Pwvap
Ptotal= PNO+ Pwvap
But,
Ptotal inside the bubble = Patmosphere
So, we have
Patmosphere= PNO+ Pwvap
Pwvap= Patmosphere- PNO
How to convert the change in
the number of divisions to
change in volume?
Have to find out the relationship
between the number of divisions
to number of mL
Vi
Vo
S=4.5
Burrette
Filled with
water
VN
Vs 
V  V 
N
O
N
V  Vu  S  Vs   0.2
Unscaled
Volume, Vu=vo-vi
DATA
Water
Temp
(oC)
T*
Scale
Value
S*
Water
Temp (K)
TK
Bobble
Vol.
(mL)
VB
Press.
N2 &
O2
(atm)
PNO
Water
Vap P
(atm)
PW Vap
Ln PW
Vap
Please use 3 sig. figs., and scientific notation
1
/
T
K
HEAT OF VAPORIZATION
DEFINITION
It is defined as the heat required to vaporize
(converting a liquid to gas) one mole of a
substance at constant pressure and temperature.
The heat of vaporization is expressed in kJ/mol. `
vap
HEAT OF VAPORIZATION
Clausius-Clapeyron equation

  H Vap H 2O  1
 
 C

R

 T



y

Ln PWater Vapor
m
 H Vap
slope  m 
R
R=8.314 J/K mol
 H Vap
m R 
 R   H Vap
R
x

b
intercept
LnPwvap (unitless)
y
X1
1/Tk, (k-1)
LnPwvap= m(1/T) + Z
x2
Equation of the best-fit line:
Y1
x1
Y=mX + z
Best-fit line
Y2
x
m = slope =
2
 HxVap
slope  m 
R
R=8.314 J/K mol
 H Vap
m R 
 R   H Vap
R
Y2  Y1
X 2  X1