1. LIQUIDS, SOLIDS, SOLUTIONS
AND PHASE CHANGES
Copyright Sautter 2003

2. SOLIDS, LIQUIDS AND PHASE CHANGES
What is the difference between a solid, a liquid and a gas?
ENERGY STATE ! Solids are the lowest energy state, liquids are intermediate and gases are the highest energy state of matter.
The energy differences which separate each state are called the Heats of Phase Change. The Heat of Phase Change is the energy required to change the state of a substance without changing its temperature.
The Heats of Phase Change are defined by the phase change which occurs, for example, Heat of Melting (often called fusion) is the heat needed to change a solid into a liquid at its melting point. The Heat of Vaporization is the heat needed to change a liquid into a gas at its boiling point.

3. SOLIDS, LIQUIDS AND PHASE CHANGES
When solids, liquids and gases are heated or cooled and do not undergo phase change, the heated added or removed from the substance can be calculated using the equation: q = m x c x T
The letter q represents heat is calories or joules. The mass of the substance in grams is m. The letter c stands for “specific heat” in calories per gram degree Celsius or joules per gram degree Celsius. Specific heat indicated how easy or difficult it is to change the temperature of a substance. Large specific heats mean that it requires a large amount of heat addition or removal to change temperature. Small values mean that the substance changes temperature readily.
Delta T (T) is the Celsius temperature change of the substance

4. CALCULATING HEAT QUANTITIES (NO PHASE CHANGE OCCURS)
TEMPERATURE
CHANGE IN 0C
q = M x C x T
SPECIFIC HEAT
IN JOULES / G x 0C
OR CALORIES / G x 0C
HEAT IN JOULES
OR CALORIES
MASS IN
GRAMS

5. SOLIDS, LIQUIDS AND PHASE CHANGES
Problem: What is the temperature of 15.0 grams of iron when 250 joules of heat are removed from it?
(specific of iron = j / g x 0C)
Solution: q = m x c x T, q = 250 joules
m = 15.0 grams, c = j / g x 0C
T = q / (m x c) = 250 / (15.0 x 0.450) = C
Since heat is removed from the sample the temperature is lowered by 37.0 degrees, therefore the answer is – C

6. SOLIDS, LIQUIDS AND PHASE CHANGES
Problem: A 100 gram sample of aluminum at 100 0C is placed in 200 grams of water at 25 0C. What is the final temperature of the system? (c for Al = j / g 0C and c for water = 4.18 j / g 0C)
Solution: Heat lost = Heat gained
(Conservation of Energy)
Al loses heat to the water since the Al is at the higher initial temperature.
mAl x cAl x TAl = mw x cw Tw,
T = temperature of system at equilibrium
100g(0.900 j /g0C)(1000C – T) = 200g(4.18 j /g0C)(T-250C)
9000 – 90T = 836T – 20900
926T = 29900, T = C

7. SOLIDS, LIQUIDS AND PHASE CHANGES
When a substance undergoes phase change (change in physical state) heat is added or removed from the material however the temperature of the material remains unchanged. All energy that is released or absorbed during the phase change is used to change the state of the substance. None is used to alter the temperature.
Only after the phase change is complete can the temperature of the system rise or fall. (Note the temperature plateaus on the next graph)
The heat of phase change for different processes is described by the physical change which is occurring for example: Heat of Melting (sometimes called fusion), Heat of Vaporization, Heat of Freezing, etc.

8. Heating Curve for Water (-10 to 110 0C)
All
steam
Boiling
starts T E M P R A U 0C
110
100
-10
Water & steam
Boiling
completed
Melting
starts
All water
Ice & water
Melting
completed
All ice
Time of Heating

9. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
During a phase change the temperature of a substance remains constant. Based on that observation, q = m c T cannot be used to find heat gained or lost since T = 0.
In order to find heat values during phase change we must multiply the heat of phase change by the quantity of substance involved. Heat of phase change is the quantity of heat which is gained or lost when a substance changes state without a temperature change.
The heat of phase change may be expressed in calories or joules per gram of substance or calories or joules per mole of substance.
Q = Heat of Phase Change x Amount of Substance

10. DURING HEATING & COOLING
HEATS OF PHASE CHANGE
DURING HEATING & COOLING T E M P R A U
Heat of Vaporization
Heat of Condensation
Heat of Melting
Heat of Freezing
Time of Heating

11. CALCULATING HEAT GAINED HEAT OF MELTING (FUSION)
OR LOST DURING A PHASE
CHANGE
AMOUNT OF SUBSTANCE
CHANGING PHASE
(GRAMS OR MOLES)
Heat = Heat of Phase Change X Amount
HEAT OF MELTING (FUSION)
HEAT OF FREEZING
HEAT OF VAPORIZATION
HEAT OF CONDENSATION
HEAT OF SUBLIMATION
(ALL IN JOULES / GRAM,
CALORIES/ GRAM
OR PER MOLE
HEAT REQUIRED
FOR PHASE
CHANGE

12. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
Problem: How much heat is needed to melt 200 grams of ice at its melt point (0 0C) ? The heat of fusion for ice is joules per mole.
Solution: Since a phase change is occurring without a temperature change:
qmelting = Heat of Melting x moles of ice melted
Moles = grams / grams per mole, moles = 200 / 18 = 11.1
qmelting = 6019 joules / mole x 11.1 moles = 66,878 joules or 66.9 kilojoules.

13. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
Problem: How much heat is needed change 100 grams of ice at –10 0C to steam at 110 0C. (c for ice = 2.06 j / g 0C, c for water = 4.18 j / g 0C, c for steam = 2.02 j / g 0C, Heat of Fusion for ice = 6.02 Kj / mole, Heat of Vaporization = 40.7 Kj / mole)
Solution: qheating ice = m x c x T
qheating ice = 100 x 2.06 x (0 – (-10)) = 2060 j
qmelting = Heat of Melting x moles of ice melted
qmelting = (6020 j / mole)(100 / 18 ) mole = 3.34 x 104 j
qheating water = 100 x 4.18 x (100 – 0) = j
qvaporization = (40.7 x 103 j / mole)(100 / 18) mole = 2.26 x 105 j
qheating steam = 100 x 2.02 x (110 – 100) = 2020 j
qtotal = x x = 2.67 x 105 j

14. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
IMPORTANT POINTS IN CALCULATING HEAT QUANTITIES:
(1) The specific heat of a substance is different for the same substance in a different physical state. For example, as we have seen, the specific heat of ice (2.06 j /g0C), that for water (4.18 j / g0C) and steam (2.02 j / g0C) are different.
(2) Heats of phase change are different for different changes involving the same material. For example Heat of Fusion for ice is 6.02 Kj / mole while Heat of Vaporization for water is 40.7 Jj / mole. Note that the Heat of Vaporization for a substance is always noticeably larger than its Heat of Fusion.
(3) Heat of phase change for a substance in the heating process are equal but opposite in sign as compared for the reverse phase change in the cooling process. For example, the Heat of Melting for ice is 6.02 Kj / mole while the Heat of Freezing for water is – 6.02 Kj / mole.

15. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE
Phase changes occur as:
Melting or Fusion – solid to liquid (endothermic)
Freezing or Solidification – liquid to solid (exothermic)
Vaporization – liquid to gas (endothermic)
Condensation – gas to liquid (exothermic)
Sublimation – solid to gas (endothermic)
Deposition – gas to solid (exothermic)
All changes of physical state are affected by both temperatures and pressures. Gas state is favored by low pressures and high temperatures both allowing for free, random motion. Solids are favored by low temperatures and high pressures both contributing to an ordered low energy state. A phase diagram shows the relationship between temperature, pressure and physical state under a variety of conditions.

16. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE
After a certain temperature (depending on the substance) no amount of pressure can cause a gas to form a liquid. This temperature is called the critical temperature of the substance. The pressure which the gas exhibits at this temperature is called the critical pressure.This temperature – pressure point is called the critical point.
At a specific temperature and pressure all three phases of a substance can coexist (solid, liquid and gas). This point is called the triple point. At this point a very slight change in temperature and / or pressure can cause a solid to melt, a liquid to vaporize, a solid to sublime, a liquid to freeze, etc.

17. PHASE DIAGRAM a g b c d f e P a = freezing R E S b = melting U
c = vaporization b
LIQUID
d = condensation c
SOLID
e= deposition
f = sublimation d
g = critical point
GAS f e
TRIPLE
POINT
TEMPERATURE

18. PHASE DIAGRAM (for water)U b
Notice that moving
along the line
a-b, as pressure
rises, the solid
melts without a
temperature increase
LIQUID
SOLID
This behavior is
typical for
water but not
most substances
GAS a
TEMPERATURE

19. PHASE DIAGRAM (for most substances)
Notice that moving
along the line
a-b, as pressure
rises, the solid
Will not melt
Increasing pressure
without a temperature
change will not
melt the substance
LIQUID
SOLID
GAS
This behavior is
typical for
most substances a
TEMPERATURE

20. VAPOR PRESSURE OF LIQUIDS
The measure of the tendency of a liquid to form a gas is indicated by its vapor pressure. Liquids which evaporate readily have high vapor pressures. The tendency of a liquid to enter the gaseous state varies based on several factors:
(1) the intermolecular forces which exist between the liquid molecules. Factors such as polar character, London Forces and hydrogen bonding (all of which have been discussed in other programs) affect vaporization. As the strength of intermolecular forces increases the vapor pressure of the liquid decreases.
(2) the Heat of Vaporization of a substance. Materials with large Hvap have low vapor pressures.
(3) the temperature of the liquid. As temperature rises the liquid molecules increase their kinetic energies and the tendency to form a gas increases.

21. VAPOR PRESSURE OF LIQUID A IS GREATER THAN LIQUID B BECAUSE OF:
WEAK INTERMOLECULAR BONDS
LOWER MOLECULAR WEIGHT AND / OR
HIGHER TEMPERATURE
LOWER Hvaporization
LIQUID A
HIGH VAPOR PRESSURE
RAPID VAPORIZATION
LIQUID B
LOW VAPOR PRESSURE
SLOW VAPORIZATION

22. RELATING VAPOR PRESSURE, TEMPERATURE &  HVAPORIZATION
Vapor pressure of liquids is directly related to the temperature of the liquid and inversely related to the Heat of Vaporization of that liquid. As temperature rises, the vapor pressure rises. As the Heat of Vaporization becomes larger the tendency to evaporate decreases as shown by a reduced vapor pressure.
The temperature at which the vapor pressure of a liquid reaches that of the surroundings (generally atmospheric pressure) is called the boiling point of the liquid. Liquids, therefore, can be boiled by increasing the temperature or reducing the pressure on the liquid. (Boiling refers to the spontaneous vaporization of the liquid at all point within the liquid not just at the surface of the liquid which is evaporation.)

23. KINETIC ENERGY DISTRIBUTION CURVE FOR A LIQUIDM B E R O F L S
TEMPERATURE = T2
TEMPERATURE = T1
HEAT OF VAPORIZATION
MOLECULES WITH
SUFFICIENT ENERGY
TO VAPORIZE AT T2
MOLECULES WITH
SUFFICIENT ENERGY
TO VAPORIZE AT T1
KINETIC ENERGY
AS TEMPERATURE , VAPORIZATION RATE

24. VAPOR PRESSURE VS TEMPERATURE
FOR VARIOUS LIQUIDS V A P O R E S U
C2H5OH
H2O
DIETHYL ETHER
1.0
NORMAL
BOILING
OCCURS
WHEN
VAPOR
PRESSURE
EQUALS 1 ATM
TEMPERATURE

25. RELATING VAPOR PRESSURE, TEMPERATURE &  HVAPORIZATION*
The vapor pressure of a liquid can be calculated at a particular temperature, if its Heat of Vaporization is known, using the Clausius-Clapeyron Equation.
In its basic form the equation is:
P = A x e -(H / RT)
Or in expanded form:
ln P2 – ln P1 = ( Hvap/ R) ((1 / T1) – (1 / T2))
Where: P1 = vapor pressure at T1 in Kelvin
P2 = vapor pressure at T2 in Kelvin
 Hvap = Heat of Vaporization in joules
R = 8.31 joules / mole Kelvin

26. THE CLAUSIUS – CLAPEYRON EQUATION*
ln P ln P = Hvap ) ( _ 2 1 _ _ _ _ R T T 2 1
R = 8.31 JOULES / MOLE K
HVAP IN JOULES
T = KELVIN TEMPERATURE
PRESSURE IN TORRS OR ATM

27. RELATING VAPOR PRESSURE, TEMPERATURE &  HVAPORIZATION*
Problem: Hexane, C6H14, has a heat of vaporization of 30.1 Kj / mole. At 25 0C its vapor pressure is 148 torrs. What is the normal boiling point of hexane?
Solution:
ln P2 – ln P1 = ( Hvap/ R) ((1 / T1) – (1 / T2))
P2 = 760 torr (normal air pressure) recall the definition of boiling previously mentioned
P1 = 148 torr, T1 = ( ) = 298 K
 Hvap = 30.1 x 103 joules / mole, T2 = ?
ln(760) – ln(148) = (30.1 x 103 / 8.31)((1/298) –(1/T2))
1.64 = 3622( – (1/ T2)), (1.64 / 3622) = – (1/T2)
(1/T2) = – , T2 = 344 K = (344 – 273) = 71 0C

28. COLLIGATIVE PROPERTIES OF SOLUTIONS
Colligative means “collective”. Recall that “The Gas Laws” were a set of equations that applied to all gases generally irrelevant of their individual nature. Boyle’s Law, Charles Law, Gay-Lussac’s Law, etc. were applied to hydrogen, oxygen, neon, etc. equally well. (In actuality all gases were assume to exhibit ideal behavior which is not entirely true.)
In the case of solutions we can also discuss the behavior of solutions in general without regard to the individual nature of the particular solutes and solvents used. Again we will assume ideal behavior which is not always entirely true.

29. Vapor Pressure of Solutions*
A solution is a solute dissolved in a solvent. The vaporization of a solution differs from that of the solvent from which it is made. The differences are based on the molar composition of the solution and the nature of the solute and solvent used.
The vapor pressure of a solution is related to its molar composition by Raoult’s Law. It tells us that the vapor pressure is directly related to the mole fraction of the solvent(s). Mole fraction refers to the number of moles of a particular component out of the total moles present.
Mole fraction (X) = (moles of substance / total moles)
Raoult’s Law:
PA = (mole fraction of A) x (vapor pressure of A)
And if more than one liquid components are present:
PSOLUTION = PA + PB + PC + …

30. Vapor Pressure of Solutions*
Problem: What is the vapor pressure of an aqueous solution containing 250 grams of water and 25.0 grams of sucrose (C12H22O11) at 35 0C? The vapor pressure of water at 35 0C is 41.2 torr.
Solution: Find the mole fraction of the water.
Mole fraction (X) = (moles of substance / total moles)
Xwater = (250 / 18) / ((250/18) + (25/342)) = / 13.95
Xwater = 0.996
Pwater = (mole fraction of water) x (vapor pressure of water)
Pwater = x 41.2 = 41.0 torrs
Note – the vapor pressure of a solution is always less than that of the pure solvent from which it is made.

31. Vapor Pressure of Solutions*
When a solution consists of two or more liquid components, the vapor pressure of the solution requires that the vapor pressure of each component be determined using Raoult’s Law and then added together.
Problem: Find the vapor pressure of a solution consisting of 29 grams of acetone (C3H6O) and 9.0 grams of water at 20 0C.(v.p. or acetone = 162 torrs and v.p. of water = 17.5 torrs)
Solution: Find the moles fractions of each
Mole fraction (X) = (moles of substance / total moles)
Moles of acetone = 29 / 58 = moles
Moles of water = 9.0 / 18.0 = moles
PA = (mole fraction of A) x (vapor pressure of A)
Xacetone = / ( ) = 0.500, V.P. = 162 x = 81.0 t
Xwater = / ( ) = 0.500, V.P. = 17.5 x = 8.75 t
PSOLUTION = PA + PB + PC + …
Psolution = 81.0 torrs torrs = 89.8 torrs

32. Freezing point Depression & Boiling point Elevation of Solutions
Solutions always have lower freezing points and higher boiling points than the solvents from which they are made.
The exact freezing point depressions and boiling point elevations can be calculated for solutions using the equation:
T = k x m x i
k is called the freezing point depression constant or the boiling point elevation constant depending upon the calculation
m is the molality of the solution (not molarity !)
molality = moles of solute / kilograms of solvent
i is called the Van’t Hoff factor = the number of particles (ions or molecules) in solution formed by each dissolved solute component

33. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
Problem: What is the freezing point of a solution made with 20.0 grams of sucrose (C12H22O11) in 100 grams of water? (kfreezing = C / m)
Solution: T = k x m x i
molality = moles of solute / kilograms of solvent
Moles = 20.0 g / 342 grams per mole =
Molality = / kg of water = m
T = C/m (0.585 m) 1 = C
i = 1 because sucrose does not make ions. One molecule dissolves to make only 1 dissolved molecule!
The normal freezing point of water is 0 0C and it is lowered by C
F.P. solution = normal F.P. - T
F.P. solution = 0 0C – C = C

34. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
Problem: Find the molecular mass of a substance if 5.65 g when dissolved in 110 g of benzene gives the solution a freezing point of C. (kf for benzene is C/m and normal f.p. is C)
Solution:
If we find molality we can then calculate molar mass
T = k x m x i, benzene does not form ions, i = 1
(5.450C – 4.390C) = 5.070C/m x m x 1, molality = m
molality = moles of solute / kilograms of solvent
0.209 m = moles of solute / kg, moles = moles
Moles = grams / molar mass, mole = 5.65 g / molar mass
Molar mass = 5.65 g / moles = 246 grams per mole

35. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
Problem: Find the boiling point of a molal solution of calcium chloride. Assume the CaCl2 dissociates 100%. The Kb for water is C/m.
Solution:T = k x m x i
CaCl2(s)  Ca+2(aq) Cl-(aq), three particles (ions) are formed from each CaCl2 dissolved, i = 3.
T = C/m (0.0500) x 3 = C
The normal boiling point for water = C
B.P. solution = normal B.P. + T =
B.P. solution = C

36. Osmotic Pressure *
Osmosis is defined as the diffusion (moving) of water across a semi-permeable membrane from an area of higher concentration to an area of lower concentration.
A semi-permeable membrane is one that a allows some materials to pass through while blocking others. As water moves through a membrane at unequal rates, a higher concentration and pressure builds up on the side of the membrane with the larger quantity of water.
Osmotic pressure can be calculated using the equation:
 = M x R x T,  is used to represent osmotic pressure
M = Molarity (not molality)
R = atm liters / moles K
T = Kelvin temperature
This equation is derived from: PV = nRT, P = (n/V) RT
Since moles / liters = molarity, Pressure = M x R x T

37. * Osmotic Pressure Semi permeable membrane
water
sugar
More
water
moves
left
and
pressure
rises in
the left
chamber
Semi permeable membrane
Allows water to pass through
but not sugar molecules

38. *
Osmotic Pressure
Problem: Find the osmotic pressure in the system shown on the previous slide if the sugar concentration is M and the temperature is 25 0C.
Solution:  = M x R x T
 = M x ( atm liter / mol K) x 298 K
 = atm or 0.24 x 760 = 18 torrs

39. Final Note
All solutions used in the preceding problems are assumed to exhibit ideal behavior meaning that they conform to the laws and equations discussed and that no solute particle interaction occurs.
In reality, no solution is ideal just as no gas was ideal when the gas laws where discussed.
Like gases, some solutions are more ideal than others. Recall that gases of low molecular weight which there non polar, at high temperatures and low pressures were most ideal. Solutions with solutes which are non polar and non ionic and which are very dilute approach ideal behavior most closely.
Like the gas laws, the relationships for solutions give good approximations for many systems.

40. SUMMARY (1) q = m x c x T (Heat to warm or cool a material)
(2) Heat lost = Heat gained (Conservation of Energy)
(3) Q = Heat of Phase Change x Amount of Substance
(Heat lost of gained during a phase change)
(4) ln P2 – ln P1 = ( Hvap/ R) ((1 / T1) – (1 / T2))
(Clausius-Clapeyron Equation for vapor pressures)
(5) PA = (mole fraction of A) x (vapor pressure of A)
(6) PSOLUTION = PA + PB + PC + …
(Equations (5) & (6) are Raoult’s Law)
(7) T = k x m x i (Changing F.P. or B.P. for solutions)
(8)  = M x R x T (Osmotic Pressure)

41. The End