COOLING CURVE
COOLING OF THE GAS SLOPE
q = mC∆T
COOLING OF THE LIQUID SLOPE
q = mC∆T
CONDENSATION/VAPORIZATION
EQUILIBRIUM PLATEAU.
(G)  (L)
COOLING OF THE
SOLID SLOPE
q = mC∆T
q = ∆HVAPm
FREEZING/MELTING EQUILIBRIUM
PLATEAU. (L)  (S)
q=∆HFUSIONm
EVAPORATION
-Attractions
BROKEN as gas
heats.
MELTING
-STRONG
ATTACTIONS
WEAKENED.
LIQUID PHASE
HEATING OF
LIQUID
-”STICKY”
attractions
-POTENTIAL
ENERGY
CHANGE.
-Assumes shape
of container but
not the volume,
-energy is gainedendothermic.
-Not compressible.
-fluidity, viscosity.
-EVAPORATION
occurs at constant
temp, POTENTIAL
ENERGY
CHANGE.
GAS PHASE
HEATING OF
GAS
-No attractions
-Assumes both
shape and volume
of container,
compressible.
-can be heated
infinitely.
GAS PHASE
COOLING OF
GAS
-No attractions
-Assumes both
shape and volume
of container,
compressible.
-can be heated
infinitely.
CONDENSATION
-Attractions form
as gas cools
causing
condensation
-Condensations
occurs at constant
temp, POTENTIAL
ENERGY
CHANGE.
LIQUID PHASE
COOLING OF
LIQUID
-”STICKY”
attractions
-Assumes shape
of container,
-Not compressible.
-fluidity, viscosity.
FREEZING
-STRONG
ATTACTIONS
FORMED
-POTENTIAL
ENERGY
CHANGE.
-energy is lostexothermic.
SOLID
-Strong
attraction
s. lowest
energy
COOLING CURVE
(a) – first
appearance of
liquid
(c) – first
appearance of
solid
(b) – condensation
nearly complete
(d) – freezing
complete.
• THE GAS PHASE
• No attractions (ideal gas). Real gases do have
attractions – proof is they condense.
• Move in random straight line motion, collide with
each other and container wall.
• Assume shape and volume of container.
• Compressible – volume is indirect with pressure and
direct with temp. Variable density.
• At the boiling point, the gas either condenses
(cooling curve) or evaporates (heating curve.)
• At the higher energy plateau (potential energy) there
is an equilibrium of gas and liquid at the same
temperature at the same time. And condensation and
vaporization ALWAYS occur together.
• THE LIQUID PHASE
• Particles have attractions which are not as strong as
in the solid.
• Assumes shape but NOT the volume of the container.
Not compressible with uniform density.
• Particles have some freedom of motion that causes
fluidity, however they are not independent as in the
gas phase.
• Viscosity or “thickness” is a function of attraction
strength and temp.
• Some of the particles have enough energy to break
attractions and escape into the gas phase below the
boiling temp. These newly vaporized particles are
called VAPOR and exert vapor pressure. The vapor
pressure is a direct function of pressure.
• MELTING/FREEZING EQUILIBRIUM
• The solid and liquid exist at equilibrium at
the same time at the same temperature
(freezing point/melting point).
• Melting (fusion) and freezing occur together
ALWAYS in an equilibrium. Freezing is faster
in a cooling curve, melting is faster in a
heating curve.
• Attraction strengthen as a substance freezes,
attractions weaken as a substance melts.
• THE SOLID STATE.
• Assumes neither the shape or volume of its
container.
• The strongest attractions of any phase.
• No momentum, however particles can vibrate
in place.
• Solids are not compressible and have
uniform density.
U DO IT NOW!!!
1)CALCULATE THE q WHEN 100.0 G OF WATER
GAS IS COOLED FROM 120.0 oC to 100.0 oC.
2)CALCULATE THE HEAT REMOVED WHEN 100
g OF WATER IS CONDENSED AT 100oC.
3)CALCULATE THE HEAT REMOVED WHEN
100g OF WATER LIQUID IS COOLED FROM
100.0 TO 0.0 oC.
4) CALCULATE THE HEAT REMOVED WHEN
100G OF WATER FREEZES,
5) CALCULATE THE HEAT REMOVED (q) WHEN
100G OF ICE IS COOLED FROM 0.0 C TO -20
C.
• 1)CALCULATE THE q WHEN 100.0 G OF WATER GAS
IS COOLED FROM 120.0 oC to 100.0 oC.
q = mC∆T
Q = 100.0g * 4.18 J * -20oC = - 8360
goC
∆T is negative
when T decreases,
as is q
• 2)CALCULATE THE HEAT REMOVED WHEN 100 g
OF WATER IS CONDENSED AT 100oC.
q = m-Hvap
Q = 100.0g * -2260.0 J = -226000.0 J
g
Condensation is
exothermic, has a
negative sign. It is
the reverse of
evaporation.
Condensation is
EXOTHERMIC, q is
negative
• 3)CALCULATE THE HEAT REMOVED WHEN 100g OF
WATER LIQUID IS COOLED FROM 100.0 TO 0.0 oC.
q = mC∆T
q = 100.0g * 4.18 J * -100oC = -41800 J = 41.800kJ
goC
• 4) CALCULATE THE HEAT REMOVED WHEN 100G
OF WATER FREEZES,
q = mHfusion
q = 100.0g * -334.0 J = -33400.0 J = -33.400 kJ
g
FREEZING Q = ∆Fusion = -
MELTING
Q=+
∆Fusion = +
CONDENSATION
Q=∆vaporization = -
VAPORIZATION
Q=+
∆vaporization= +
EXOTHERMIC:
Energy EXITS, is
released
ENDOTHERMIC
Energy ENTERS,
is absorbed
EXOTHERMIC:
Energy EXITS, is
released
ENDOTHERMIC:
Energy ENTERS,
is absorbed
The heat is
equal in number
but opposite in
sign to heat of
fusion
Heat of
fusion is
positive
The heat is
equal in number
but opposite in
sign to heat of
vaporization
Heat of
vaporization
is positive