prentice hall © 2003chapter 11 the forces holding solids and liquids together are called...

• Sublimation: solid gas.• Vaporization: liquid gas.• Melting or fusion: solid liquid.• Deposition: gas solid.• Condensation: gas liquid.• Freezing: liquid solid.

Phase ChangesPhase Changes

Phase ChangesPhase Changes

• ENERGY ASSOCIATED WITH HEATING CURVES

Topics

• Vapor Pressure

• Normal Boiling Point

• Normal Freezing

• Specific Heat

• Enthalpy (Heat) of Vaporization

• Enthalpy (Heat) of Fusion

Vapor Pressure

• THE PRESSURE OF A VAPOR IN EQUILIBRIUM WITH ITS LIQUID (OR ITS SOLID)

NORMAL BOILING POINT & FREEZING POINTS

• NORMAL BOILING PT. - THE TEMPERATURE @WHICH VAPOR PRESSURE = 1 atm

• NORMAL FREEZING PT. – THE TEMPERATURE @ WHICH THE VAPOR PRESSURE OF THE SOLID AND THE LIQUID ARE THE SAME

Heat Capacity aka Specific Heat (C)

• Specific Heat (C) = the amount of energy required to raise the temperature of 1 gram of substance 1 degree celcius

Specific Heat (C) aka Heat Capacity

• Units for: specific heat (C) = J/g-oC

where J = joulesoC = temperature in oC

g = mass in grams

Specific Heat (C) Values(aka Heat Capacity)

• Example: Water

• LIQUID: CLiq = 4.18 J/ (oC . g)

• LIQUID: Csol = 2.09 J/ (oC . g)

• LIQUID: Cgas = 1.84 J/ (oC . g)

Use of Specific Heat

• q = mCT

• q = gm substance x specific heat x T

• where:

• M = mass of substance in grams

• q = amount of heat (energy)

• C = specific heat

• And T = change in temperature

Enthalpy of Vaporizationaka heat of vaporization (Hvap)

• Is the amount of heat needed to convert a liquid to a vapor at its normal boiling point

Enthalpy of Fusion aka heat of fusion (Hfus)

• Is the amount of heat needed to convert a solid to a liquid at its normal melting (freezing) point

Units for Hvap, Hfus and heat(q)

Heat of fusion Hvap = kJ/mol

Heat of vaporization Hfus = kJ/mol

Heat (q) = Joules

Therefore:

• To come up with Joules which is the unit of heat, if:

H is given, then:qvap = Hvap x moles

and

qfus = Hfus x moles

(2) Specific heat (C) is given, then: q = mCT

Sample Problem

• Calculate the enthalpy change upon converting 1 mole of ice at -25 oC to steam at 125 oC under a constant pressure of 1 atm? The specific heats are of ice, water and steam 2.09 J/g-K for ice, 4.18 J/g-K for water and 1.84 J/g-K for steam. For water, Hfus= 6.01 kJ/mol, and Hvap = 40.67kJ/mol.

• Note: The total enthalpy change is the sum of the changes of the individual steps.

Take Home Quiz

• Do Problem 11.34, page 444

• Due Wednesday at the beginning of class, Dec. 1

• Must show work for full credit

• Individual work not required

HEATING CURVES

• ENERGY ASSOCIATED WITH HEATING CURVES

• During a phase change, adding heat causes no temperature change.

Critical Temperature and Pressure• Gases liquefied by increasing pressure at some

temperature.• Critical temperature: the minimum temperature for

liquefaction of a gas using pressure.• Critical pressure: pressure required for liquefaction.

Phase ChangesPhase Changes

• Phase diagram: plot of pressure vs. Temperature summarizing all equilibria between phases.

• Given a temperature and pressure, phase diagrams tell us which phase will exist.

• Any temperature and pressure combination not on a curve represents a single phase.

Phase DiagramsPhase Diagrams

Phase DiagramsPhase Diagrams

The Phase Diagrams of H2O and CO2

Phase DiagramsPhase Diagrams