Copyright Sautter 2003. SOLIDS, LIQUIDS AND PHASE CHANGES What is the difference between a solid, a liquid and a gas? ENERGY STATE ! Solids are the lowest.

Copyright Sautter 2003

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.

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

HEAT IN JOULES OR CALORIES

MASS INGRAMS

SPECIFIC HEATIN JOULES / G x 0C

OR CALORIES / G x 0C

TEMPERATURECHANGE IN 0C

CALCULATING HEAT QUANTITIES(NO PHASE CHANGE OCCURS)

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 = 0.450 j / g x 0C)• Solution: q = m x c x T, q = 250 joules

m = 15.0 grams, c = 0.450 j / g x 0C

T = q / (m x c) = 250 / (15.0 x 0.450) = 37.0 0C

Since heat is removed from the sample the temperature is lowered by 37.0 degrees, therefore the answer is – 37.0 0C

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 = 0.900 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 = 32.3 0C

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.

Time of Heating

TEMPERATURE

0C

110100

0-10

All ice

Meltingstarts

Meltingcompleted

Ice & water

All water

Boiling starts

Water & steam

Boiling completed

All steam

Heating Curve for Water (-10 to 110 0C)

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

Heat of Melting

Heat of Vaporization

Heat of Condensation

Heat of Freezing

TEMPERATURE

Time of Heating

HEATS OF PHASE CHANGEDURING HEATING & COOLING

HEAT REQUIREDFOR PHASE

CHANGE

HEAT OF MELTING (FUSION)HEAT OF FREEZING

HEAT OF VAPORIZATIONHEAT OF CONDENSATIONHEAT OF SUBLIMATION(ALL IN JOULES / GRAM,

CALORIES/ GRAMOR PER MOLE

AMOUNT OF SUBSTANCECHANGING PHASE

(GRAMS OR MOLES)

CALCULATING HEAT GAINEDOR LOST DURING A PHASE

CHANGE

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 6019 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.

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) = 4180 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 = 2060 + 3.34 x104 + 4018 + 2.26 x 105 + 2020 = 2.67 x 105 j

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.

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.

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.

PRESSURE

TEMPERATURE

GAS

SOLID

LIQUID

PHASE DIAGRAM

TRIPLEPOINT

a

b

c

d

ef

ga = freezing

b = melting

c = vaporization

d = condensation

e= deposition

f = sublimation

g = critical point

PRESSURE

GAS

SOLID

LIQUID

PHASE DIAGRAM(for water)

a

b Notice that movingalong the line

a-b, as pressurerises, the solidmelts without a

temperature increase

TEMPERATURE

This behavior istypical for

water but notmost substances

PRESSURE

GAS

SOLID

LIQUID

PHASE DIAGRAM(for most substances)

a

b Notice that movingalong the line

a-b, as pressurerises, the solidWill not melt

Increasing pressure without a temperature

change will notmelt the substance

TEMPERATURE

This behavior istypical for

most substances

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.

LIQUID AHIGH VAPOR PRESSURERAPID VAPORIZATION

LIQUID BLOW VAPOR PRESSURESLOW VAPORIZATION

VAPOR PRESSURE OF LIQUID A IS GREATER THAN LIQUID B BECAUSE OF:

(1) WEAK INTERMOLECULAR BONDS(2) LOWER MOLECULAR WEIGHT AND / OR

(3) HIGHER TEMPERATURE(4) LOWER Hvaporization

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.)

KINETIC ENERGY DISTRIBUTION CURVEFOR A LIQUID

TEMPERATURE = T1

TEMPERATURE = T2

KINETIC ENERGY

NUMBER

OF

MOLES

HEAT OF VAPORIZATION

MOLECULES WITHSUFFICIENT ENERGYTO VAPORIZE AT T1

AS TEMPERATURE , VAPORIZATION RATE

MOLECULES WITHSUFFICIENT ENERGYTO VAPORIZE AT T2

VAPOR PRESSURE VS TEMPERATUREFOR VARIOUS LIQUIDS

VAPOR

PRESSURE

TEMPERATURE0 10 20 30 40 50 60 70 80 90 100

H2OC2H5OHDIETHYL ETHER

NORMALBOILINGOCCURSWHEN VAPOR

PRESSUREEQUALS 1 ATM

1.0

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

R = 8.31 JOULES / MOLE KHVAP IN JOULES

T = KELVIN TEMPERATUREPRESSURE IN TORRS OR ATM

THE CLAUSIUS – CLAPEYRON EQUATION

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 = (25 + 273) = 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(0.00336 – (1/ T2)), (1.64 / 3622) = 0.00336 – (1/T2)

• (1/T2) = 0.00336 – 0.000453 , T2 = 344 K = (344 – 273) = 71 0C

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.

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 + …

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)) = 13.89 / 13.95• Xwater = 0.996• Pwater = (mole fraction of water) x (vapor pressure of water)• Pwater = 0.996 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.

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 = 0.500 moles• Moles of water = 9.0 / 18.0 = 0.500 moles• PA = (mole fraction of A) x (vapor pressure of A)• Xacetone = 0.500 / (0.500 + 0.500) = 0.500, V.P. = 162 x 0.500 = 81.0 t• Xwater = 0.500 / (0.500 + 0.500) = 0.500, V.P. = 17.5 x 0.500 = 8.75 t• PSOLUTION = PA + PB + PC + …• Psolution = 81.0 torrs + 8.75 torrs = 89.8 torrs

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

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 = 1.86 0C / m)

• Solution: T = k x m x i• molality = moles of solute / kilograms of solvent• Moles = 20.0 g / 342 grams per mole = 0.0585• Molality = 0.0585 / 0.100 kg of water = 0.585 m T = 1.86 0C/m (0.585 m) 1 = 1.09 0C• 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 1.09 0C • F.P. solution = normal F.P. - T • F.P. solution = 0 0C – 1.09 0C = -1.09 0C

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 4.39 0C. (kf for benzene is 5.07 0C/m and normal f.p. is 5.45 0C)

• 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 = 0.209 m

• molality = moles of solute / kilograms of solvent

• 0.209 m = moles of solute / 0.110 kg, moles = 0.0230 moles

• Moles = grams / molar mass, 0.0230 mole = 5.65 g / molar mass

• Molar mass = 5.65 g / 0.0230 moles = 246 grams per mole

Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions

• Problem: Find the boiling point of a 0.0500 molal solution of calcium chloride. Assume the CaCl2 dissociates 100%. The Kb for water is 0.512 0C/m.

• Solution:T = k x m x i

• CaCl2(s) Ca+2(aq) + 2 Cl-

(aq), three particles (ions) are formed from each CaCl2 dissolved, i = 3.

T = 0.512 0C/m (0.0500) x 3 = 0.0768 0C• The normal boiling point for water = 100.000 0C

• B.P. solution = normal B.P. + T = 100.000 + 0.0768

• B.P. solution = 100.0768 0C

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 = 0.0821 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

Osmotic Pressurewatersugar

Semi permeable membraneAllows water to pass through

but not sugar molecules

Morewater moves

leftand

pressurerises inthe left

chamber

Osmotic Pressure

• Problem: Find the osmotic pressure in the system shown on the previous slide if the sugar concentration is 0.0010 M and the temperature is 25 0C.

• Solution: = M x R x T = 0.0010 M x (0.0821 atm liter / mol K) x 298 K = 0.024 atm or 0.24 x 760 = 18 torrs

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.

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)