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.

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).

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water andMOST substances

adhesive > cohesive

mercurycohesive > adhesive

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 Hvap > 0 (endothermic)

Deposition (crystalization): gas solid Hdep < 0 (exothermic)

Condensation: gas liquid Hcon < 0 (exothermic)

Freezing: liquid solid Hfre < 0 (exothermic)

Solid--------Liquid--------Gas

Easier with weak IMFsHarder with strong IMFs

Endothermic

Harder with weak IMFsEasier with strong IMFs

Exothermic

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.

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Heating Curvefor water

Notice the plateauswhere temperaturelevels off eventhough energy isstill being added.

Notice that thereare 5 differentparts to the “curve”

• 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.

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ENERGY CONVERSIONSSample 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

= 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

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Vapor Pressure

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)

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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)

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1 atm

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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)

Two ways to get a liquid to boil:

(1) increase temperature

(2) decrease pressure above liquid

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• Clausius-Clapeyron Equation (Relationship between vapor pressure and temperature)

• more often written for a gas (or vapor) at 2 different P and T as:

vap- Hln P = + C

RT

vap2 1

1 2

H 1 1ln P = ln P + - R T T

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Graphing 1/T versus ln P

The slope of this line = - ΔHvap / R

The slope can be used to determine ΔHvap

ΔHvap = - (slope)(R)

vap- Hln P = + C

RT

ln P

1 / T

y = m x + b

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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?

P1 = 400 mm Hg T1 =17.9oC = 290.9 KP2 = 760 mm Hg T2 = 34.6oC =307.6 K * Normal BP is at standard pressureR = 8.3145 J/mol·K

vap2 1

1 2

ΔH 1 1ln P = ln P + - R T T

vap

6.633 - 5.991 8.3145 J/mol KH = = = 28.6 kJ/mol

1 1 - 290.9 K 307.6 K

28600 J/mol

Use of Clausius-Clapeyron Equation to find ΔH

vapΔH 1 1ln 760 = ln 400 + - 8.3145 290.9 307.6

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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?

P1 = 100.0 mm Hg T1 = 34.7oC = 307.7 K P2 = ?? T2 = 65.0oC = 338.0 KHvap = 38.6 kJ/mol = 38600 J/molR = 8.3145 J/mol·K

Thus:

ln P2 = 5.9577

P2 = e5.9577 = 386.7 mm Hg

vap2 1

1 2

ΔH 1 1ln P = ln P + - R T T

238600 J/mol 1 1ln P = ln 100.0 + -

8.3145 J/mol K 307.7 K 338.0 K

Use of Clausius-Clapeyron Equation to find a pressure

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 PFeatures 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.

Solid

Liquid

Gas (vapor)

Sublimation

Deposition

Melting

Freezing

Vaporization

Condensation

Critical pointTriple point

Generic Phase Diagram

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

Notice that the solid-liquid line for water has a negative slope. This is because solid ice is less dense than liquid water.

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

b) decreasing P at constant T?

c) decreasing T at constant P?

d) increasing P at constant T?

turns into vapor

turns into vapor

turns into liquid then solid

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/phaseeqia/phasediags.html