Liquids
11.3 – 11.6
Viscosity of Liquids
• Viscosity - resistance of a liquid to flow molecules slide over one another.
•Viscosity depends on the strength of IMF:
stronger IMF means higher viscosity.
• Viscosity decreases with an increase in
temperature. (think syrup!)
– Higher temp means more
energy to overcome IMFs
Surface Tension
• Bulk molecules in the liquid are
equally attracted to all of their
neighbors.
• Surface molecules are only
attracted inward towards the bulk
molecules.
• Surface molecules are packed
more closely than bulk molecules.
• This causes the liquid to behave as
if it had a “skin”.
• Surface tension - amount of energy required to
increase the surface area of a liquid by a unit
amount.(like m2 or in2)
• Stronger IMFs cause higher surface tension.
• Higher temperature - lower surface tension
• Cohesive forces - IMF that bind molecules
to one another.
• Adhesive forces - IMF that bind molecules
to a surface (container).
water and
mercury
MOST substances cohesive > adhesive
adhesive > cohesive
6
Capillary action
• The rise of liquids up very narrow tubes.
• The liquid climbs until adhesive / cohesive forces
are balanced by gravity.
Phase Changes and Energy
Sublimation: solid  gas Hsub > 0 (endothermic)
.
Melting or fusion: solid  liquid Hfus > 0 (endothermic)
Vaporization: liquid  gas
Deposition (crystalization): gas
Hvap > 0 (endothermic)
 solid
Condensation: gas  liquid
Freezing: liquid  solid
Hdep < 0 (exothermic)
Hcon < 0 (exothermic)
Hfre < 0 (exothermic)
Harder with weak IMFs
Easier with strong IMFs
Exothermic
Solid--------Liquid--------Gas
Easier with weak IMFs
Harder with strong IMFs
Endothermic
Heating Curves
• Plot of temperature change versus heat
added is a heating curve.
• During a phase change, temp. remains
constant even though energy is being added
(plateau on heating curve)
• Added energy overcomes IMF rather than
cause temperature change.
Heating Curve
for water
Notice the plateaus
where temperature
levels off even
though energy is
still being added.
Notice that there
are 5 different
parts to the “curve”
11
• To calculate the amount of heat
absorbed/released from a heating curve:
– Use q=mΔTcp for increasing/decreasing lines
– Use q=mΔH for flat lines
• Add q values for each line segment to find
the total q for the process.
ENERGY CONVERSIONS
Sample Problem #7
Calculate the energy required to convert 25.0 g
of ice at -13°C to steam at 125°C.
This problem needs 5 steps
step 1 ice at -13°C to ice at 0°C (raising T to mp)
q = (mass) (ΔT) (S.H.) = (25.0 g) (13°C) (2.09 J/g °C) = 680 J
step 2 ice at 0°C to liquid water at 0°C (melting)
q = (mass) (heat of fusion) = (25.0 g) (334 J/g) = 8350 J
step 3 water at 0°C to water at 100°C (raising T to bp)
q = (mass) (ΔT) (S.H.) = (25.0 g) (100°C) (4.184 J/g °C) = 10500 J
step 4 water at 100°C to steam at 100°C (boiling)
q = (mass) (heat of vaporization) = (25.0 g) (2260 J/g) = 56500 J
step 5 steam at 100°C to steam at 125°C (raising to final T)
q = (mass) (ΔT) (S.H.)= (25.0 g) (25°C) (1.84 J/g °C) = 1200 J
final answer = 680 J + 8350 J + 10500 J + 56500 J + 1200 J
13
= 77200 J
• Critical temperature is the highest
temperature at which a substance can exist
as a liquid - above Tc molecules have too
much energy to be a liquid
• Critical pressure is the pressure required for
liquefaction at this critical temperature.
– The greater the IMF, the higher the critical
temperature.
Vapor Pressure
• Some molecules on the surface of a liquid
have enough energy to escape the attraction of
the bulk liquid and move into the gas phase.
• As the number of molecules in the gas phase
increases, some of the gas phase molecules strike
the surface and return to the liquid.
• After some time the pressure of the gas will be
constant.
– A dynamic equilibrium has been established
• Vapor pressure is the pressure exerted when the
vapor is in dynamic equilibrium with its liquid
Vapor Pressure
17
Volatility
• Liquids with lower IMFs tend to have high
vapor pressures
– Liquids that evaporate easily are said to be
volatile. (have low IMF)
Vapor Pressure and Temperature
• Increasing temperature will increase vapor
pressure exponentially
– Higher temps mean the molecules have more energy
to overcome the IMFs that hold them in the liquid
phase.
– increasing temperature will increase evaporation and
will produce more vapor molecules, so more pressure
– increasing temperature will increase the pressure of any gas
(or vapor) (Charles’ law)
20
Vapor Pressure and Boiling Point
• Boiling occurs when the vapor pressure equals the
pressure above the liquid
• The normal BP is the temperature when the
vapor pressure is equals standard pressure (1 atm)
1 atm
22
Vapor Pressure and Boiling Point
• Decreasing air pressure (going up in altitude) lowers the temperature at which
a liquid will boil – food does not cook as fast (high altitude cooking
directions on boxed foods)
23
Two ways to get a liquid to boil:
(1) increase temperature
(2) decrease pressure above liquid
• Clausius-Clapeyron Equation (Relationship
between vapor pressure and temperature)
-H vap
ln P =
+C
RT
• more often written for a gas (or vapor) at 2 different P
and T as:
ln P2 = ln P1 +
 H vap   1


 R  T

 1
1
- 
T2 
25
-H vap
ln P =
+C
RT
y
= mx
+ b
Graphing 1/T versus ln P
ln P
The slope of this line =
- ΔHvap / R
The slope can be used to
determine ΔHvap
ΔHvap = - (slope)(R)
1/T
26
Use of Clausius-Clapeyron Equation to find ΔH
Ether has Pvap = 400 mm Hg at 17.9oC and a normal BP of 34.6oC. What is the
heat of vaporization, Hvap , for ether in kJ/mol?
ln P2 = ln P1 +
 ΔH vap  1


 R   T1


P1 = 400 mm Hg T1 =17.9oC = 290.9 K
P2 = 760 mm Hg T2 = 34.6oC =307.6 K
R = 8.3145 J/mol·K
ln 760 = ln 400 +
Hvap =
 6.633



1
- 
T2 
* Normal BP is at standard pressure
 ΔH vap   1


 8.3145   290.9



1

307.6 
- 5.991  8.3145 J/mol  K 

 = 28600
1
1
290.9 K 307.6 K
J/mol = 28.6 kJ/mol
27
Use of Clausius-Clapeyron Equation to find a pressure
The vapor pressure of ethanol at 34.7oC is 100.0 mm Hg. The heat of
vaporization is 38.6 kJ/mole. What is the vapor pressure of ethanol in mm Hg at
65.0oC?
ln P2 = ln P1 +
 ΔH vap  1


 R   T1



1
- 
T2 
P1 = 100.0 mm Hg
T1 = 34.7oC = 307.7 K
P2 = ??
T2 = 65.0oC = 338.0 K
Hvap = 38.6 kJ/mol = 38600 J/mol
R = 8.3145 J/mol·K
Thus:
ln P2 = ln 100.0 +
 38600 J/mol 
1


8.3145
J/mol

K

  307.7 K

1
338.0 K 
ln P2 = 5.9577
P2 = e5.9577 = 386.7 mm Hg
28
Phase Diagrams
• plot of pressure vs. temperature summarizing all equilibria
between phases.
• Phase diagrams tell us which phase will exist at a given T and P
Features of a phase diagram:
Vapor-pressure curve: generally as T increases, Pvap increases.
Critical point: critical temperature and pressure for the gas.
Normal melting point: melting point at 1 atm (101 kPa)
Normal boiling point: boiling point at 1 atm (101 kPa)
Triple point: T and P at which all three phases are in equilibrium.
Any T and P combination on a curve represent 2 phases
Any T and P combination not on a curve represents a single phase.
Generic Phase Diagram
Liquid
Melting
Freezing
Solid
Vaporization
Critical point
Triple point
Sublimation
Deposition
Condensation
Gas (vapor)
Notice that the solid-liquid line for water has a negative slope. This is
because solid ice is less dense than liquid water.
What phase or phases are present
a) at any point on curve YX?
b) at any point on curve AD?
c) at any point on curve AB?
d) at point A or X?
solid and vapor
liquid and vapor
solid and liquid
solid, liquid and vapor
Looking at the water diagram on the left, what effect would each of the
following have on a sample of water at any point on the AD curve.
a) increasing T at constant P
turns into vapor
b) decreasing P at constant T?
turns into vapor
c) decreasing T at constant P?
turns into liquid then solid
d) increasing P at constant T?
turns into liquid
Using the two diagrams above:
a) points A and X are called
b) point B is called
c) point C is called
d) point Y is called
e) points D and Z are called
triple point
normal melting/freezing point
normal boiling/condensing point
normal sublimation/deposition point
critical point
Good explanation of phase diagrams:
http://www.chemguide.co.uk/physical/phasee
qia/phasediags.html