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The Clausius-Clapeyron Equation

Applications

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Areas in which Clausius-Clapeyronapplies

1. Apply the Clausius-Clapeyron equation toestimate the vapor pressure at anytemperature.

2. Estimate the heat of phase transitionfrom the vapor pressures measured attwo temperatures.

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Approximation to simplify the equation1. The change in volume that accompanies

evaporation or sublimation is assumed to be equal

to the volume of the vapor produced. This is agood assumption at moderate or low pressures. Forexample, water is about 1000 times more densethan its vapor around room temperature andpressure, so the volume change for complete

evaporation of water equals the volume of vaporproduced to about three significant figures.2. The enthalpy of vaporization (or sublimation) is

assumed to be constant over the temperaturerange of interest. This is never really true, but

changes in these H's are very small at low andmoderate pressures. As one approaches the criticalpoint, this assumption will fail completely.

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3.The vapor is assumed to be an ideal gas.Again, this is a good assumption atmoderate pressures for most substances.

4.The external pressure doesn't affect thevapor pressure. There is a slightdependence on external pressure.

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The vaporization curves of most liquids have similarshape. The vapour pressure steadily increase as thetemperature increases. A good approach is to find a

mathematical model for the pressure increase as afunction of temperature. Experiments showed that thepressure P, enthalpy of vaporization, , andtemperature Tare related,

If P1and P2are the pressures at two temperatures T1and T2, the equation has the form:

The Clausius-Clapeyron equation allows us to estimatethe vapor pressure at another temperature, if thevapor pressure is known at some temperature, and if theenthalpy of vaporization is known.

122

1 11lnTTnR

H

P

Por

H

tconsAandKJ mol.where R (nRT

HAP -- tan31458exp 11

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Gas Constant in Various Units

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Example 1 The vapor pressure of water is 1.0 atm at 373 K, and the

enthalpy of vaporization is 40.7 kJ mol-1. Estimate the

vapor pressure at temperature 363 K and 383 Krespectively. Solution

Using the Clausius-Clapeyron equation, we have: P at 363 = 1.0 exp (- (40700/8.3145)(1/363 - 1/373)

= 0.697 atmP at 383 = 1.0 exp (- (40700/8.3145)(1/383 - 1/373)= 1.409 atm

Note that the increase in vapor pressure from 363 K to373 K is 0.303 atm, but the increase from 373 to 383 K is0.409 atm. The increase in vapor pressure is not a linearprocess.

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The Clausius-Clapeyron equation applies to anyphase transition. The following example shows itsapplication in estimating the heat of sublimation.

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Example 2 The vapor pressures of ice at 268 and 273 are 2.965 and

4.560 torr respectively. Estimate the heat of sublimation

of ice. Solution

The enthalpy of sublimation is Hsub. Use a piece of paperand derive the Clausius-Clapeyron equation so that youcan get the form:

Hsub= Rln (P268 / P273) (1/268 - 1/273)

= 8.3145*ln(2.965/4.560) / (1/268 - 1/273)

= 52370 J mol-1 Note that the heat of sublimation is the sum of heat

of melting and the heat of vaporization.

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Examples to solve

Example 1:Water has a vapor pressureof 24 mmHg at 25oC and a heat ofvaporization of 40.7 kJ/mol. What is the

vapor pressure of water at 67oC?AnswerP2= 182 mmHg

Example 2: An unknown liquid has avapor pressure of 88mmHg at 45oC and39 mmHg at 25oC. What is its heat ofvaporization? AnswerHvap= 32.0 kJ

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The heat of vaporization for water is 40.7 kJ mol-1.

Calculate vapor pressureat 300 K.P T/ K

P2 300760 373expect U to know

R = 8.314 J mol-1

P2 - Hvap 1 1

ln ----- = ---------- ( ---- - ---- )

P1 R T 2 T1

P2 -40700 J mol-1 1 1

ln ------- = ------------------- ( ---- - ----- ) = - 3.914

760 8.31 J mol-1 300 373

P2 / 760 = e-3.914= 0.020

P2 = 0.020*760 mmHg = 15.2 mmHg

when P1=800, b.p.=?

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Practice Problems

Use a calculator to evaluate the vapor pressure of water at 272,

273, and 360 K. (Use data from lecture material)H/ R= 41000/8.3142=4931

P at 272 K = 748 Pa; P at 273 K = 799 Pa, P at 360K = 62846 Pa

P at 373 = 101.3 k Pa

Use a spreadsheet to plot the vapor pressure of ice for temperature

between 253 and 275 K.H/R = 47000/8.3142 = 5653;

Facts about H2O;

m.p. = 272 K

b.p. = 373 K

Hsub= Hf+ Hv

T P /Pa250 91

253 119

255 141

273 610

274 658

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Solve by graph

The dissociation pressure of calcium carbonate atvarious temperatures are tabulated below.Calculate the mean heat of the reaction.

Temperatureo

C

650 700 750 800Pressuremm 8.2 25.3 68.0 168.0

Slope should be around (-818)

Answer 37,700 cal per mole

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What are the heat of vaporization (in kJ mol-1)

and the normal boiling point of a liquid that

has a vapor pressure of 254 mm Hg at 25oC

and a vapor pressure of 648 mm Hg at 45oC ?

(R = 8.314 J mol

-1

K

-1

)

HOME WORK ONE WEEK

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Assignment

1) The vapour pressure of dichloromethane at 24.1 oC is 53.3 kPaand its enthalpy of vaporization is 28.7 kJ mol-1. Estimate thetemperature at which its vapour pressure is 70.0 kPa.

Ans= 304 K @ 31oC

2) The vapour pressure of a liquid in the temperature range 200K to260K was found to fit the expression ln P = (16.255-2501.8)/T.Calculate the enthalpy of vaporization of the liquid.

Ans= 20.7 kJ mol-1

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T (oC) P (kPa)

0 1.92

20 6.38

40 17.7

50 27.7

70 62.3

80 89.3

90 124.9

100 170.9

3) The vapour pressure, P, of nitric acid varies with temperature as

follows:

What are (a) the enthalpy of vaporization of nitric acid and (b) the

normal boiling point?

Ans = a) Tb = 357 K b) H = 38 kJ mol-1