4,7 b solman

CHEM 3411, Fall 2010Solution Set 4

In this solution set, an underline is used to show the last significant digit of numbers. For instance in

x = 2.51693

the 2,5,1, and 6 are all significant. Digits to the right of the underlined digit, the 9 & 3 in the example, are notsignificant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values incalculations reduces rounding errors and ensures we get the same answer regardless of the order of arithmetic steps.Numbers without underlines (including final answers) are shown with the proper number of sig figs.

1 Exercise 4.1b pg 153

Question

How many phases are present at each of the points marked in Fig. 4.23b?

Solution

• a. 1 - point is entirely inside a single phase region (not on any boundaries).

• b. 3 - point on a boundary where 3 phases meet.

• c. 3 - (same explanation as b)

• d. 2 - point occurs on boundary between two phases (on a single line).

1



Given

Iron is heated from Ti = 25◦C to Tf = 100◦C. Over this temperature range S⊖m = 53 JK−1 mol−1.

In terms of given variables, this is written:

Ti = 25◦C

Tf = 100◦C

S⊖m = 53 JK−1 mol−1

Find

By how much does its chemical potential change?

Strategy

First, temperatures are converted to Kelvin.

Ti = 100 + 273◦C = 373K

Tf = 1000 + 273◦C = 1273K

We can use text book Equation 4.2 (pg 143) to relate temperature changes to changes in chemical potential.(∂µ

∂T

)p

= −S⊖m

which gives the differential

dµ = −S⊖mdT

Integrating dµ over the temperature range gives the change in chemical potential.

∆µ = −∫ Tf

Ti

S⊖mdT

= −S⊖m (Tf − Ti)

= −53J

Kmol(1273K− 373K)

= −47700J

mol

= −47.7kJ

mol

where the integral has been easily solved, as we’ve assumed S⊖m is constant over this temperature range.

2


Solution

∆µ = −50kJ

mol

3



Given

The vapour pressure of a liquid between 15◦C and 35◦C fits the expression

log(p/ torr) = 8.750− 1625/(T/K)

Find

Calculate . . .

• (a) the enthalpy of vaporization

• (b) the normal boiling point of the liquid

Strategy

We’ll start with Equation 4.11 (pg 148).

d ln p

dT=

∆vapH⊖

RT 2

Through rearrangement we solve for ∆vapH⊖.

∆vapH⊖ = RT 2 d ln p

dT

To find d ln pdT , the given expression for log(p/ torr) can be converted to an expression for ln(p/ torr) using the change

of base formula.

logb x =logk x

logk b

and this gives the equation

ln(p/ torr) = ln(10) (7.960− 1625/ (T/K))

Differentiating this expression by T gives

ln(dp)

dT=

ln(10)× 1625K

T 2

(The Torr units were discarded as the units of pressure would only lead to constant shift in the expression for ln(p)and this constant is lost on differentiation.)

Lastly we can substitute this expression for ln(dp)dT into our earlier expression for ∆vapH

⊖.

4


∆vapH⊖ = R��T 2 ln(10)× 1625K

��T 2

= R× ln(10)× 1625K

= 8.314J

��Kmol× ln(10)× 1625��K

= 31108.5J

mol

= 31.1085kJ

mol

The normal boiling point can be found by solving the given expression p(T ) for the temperature at which p(T ) =patmosphere where the atmospheric pressure patmosphere = 1.00 atm = 760 torr.

log(760��torr/��torr) = 8.750− 1625/(T/K)

T =1625K

8.750− log 760

= 276.9K

Solution

• (a) ∆vapH⊖ = 31.11 kJ

mol

• (b) T = 280K

5



Given

On a cold, dry morning after a frost, the temperature was T = −5◦C and the partial pressure of water in theatmosphere fell to pH2O = 0.30 kPa.

In terms of given variables, this is written:

T = −5◦C

p = 0.30 kPa

Find

• (a) Will the frost sublime?

• (b) What partial pressure of water pH2O would ensure that the frost remained?

Strategy

In this exercise, the second question answers the first question, in that once we know the partial pressure of waterpH2O needed to ensure the frost remains, we know any pH2O below this will lead to frost sublimation.

The partial pressure needed to prevent sublimation is found by determing the solid-vapor pressure for water at thistemperature; this is the pressure at this temperature on the coexistance curve for ice and water vapor. We knowthat if the atmosphere has a lower partial pressure of water than the solid-vapor pressure, then the water vapourwill be favored over the solid and the ice will sublime. At higher partial pressures the solid phase is favored.

We can find the solid-vapor pressure using Equation 4.12 (pg 149) from our text book

p = p⋆e−χ χ =∆subH

⊖

R

(1

T− 1

T ⋆

)where we’ve replaced ∆vapH

⊖ in the original expression with ∆subH⊖ as we’re concerned with the sublimation

coexistence point instead of the vaporization point.

The sublimation enthalpy ∆vapH⊖ is found from the vaporization enthalpy and the fusion enthalpy, ∆vapH

⊖ and∆fusH

⊖ respectively.

∆subH⊖ = ∆fusH

⊖ +∆vapH⊖

= 6.008 kJmol−1 + 44.016 kJmol−1

= 50.024 kJmol−1

= 5.0024× 104 Jmol−1

Using Equation 4.12 we can calculate the pressure p at temperature T when we know the reference pressure p⋆ at thereference temperature T ⋆. As we’re solving for a p(T ) on the solid/gas coexistence curve, we’ll need the reference toalso fall on this curve. Therefore we’ll use the the triple point of water as our reference giving the following values:

T ⋆ = 273.16K

6


p⋆ = 0.61173 kPa

This allows us to find the solid-vapor pressure p on the solid/gas coexistence curve for the temperature T = −5◦C =268K.

χ =∆subH

⊖

R

(1

T− 1

T ⋆

)=

5.0024× 104�J��mol−1

8.314�J��K−1��mol−1

(1

268��K− 1

273.16��K

)= 0.42409

p = p⋆e−χ

= 0.61173 kPa× e−0.42409

= 0.40029 kPa

Now that we know the solid-vapor pressure p = 0.40029 kPa at the given temperature, we know the ice will sublimeinto any gas system with a partial pressure of water pH2O < 0.40029 kPa. This gives us the answer to part b aspH2O = 0.40029 kPa.

Additionally, can determine that the ice will sublime into the atmosphere as the partial pressure of water is pH2O =0.30 kPa.

Solution

• (a) Yes.

• (b) pH2O = 0.40029 kPa

7



Question

What fraction of the enthalpy of vaporization ∆vapH of ethanol is spent on expanding its vapour?

Strategy

We can use the definition of enthalpy, H = U + pV (Equation 2.18 pg 56) to decompose the enthalpy of vaporization∆vapH into an internal energy component ∆vapU and an expansion work component ∆vap(pV ).

∆vapH = ∆vapU +∆vap(pV )

Additionally, we can assume the pressure is constant and the liquid volume Vliq is negligible relative to the gas volumesuch that

∆vap(pV ) = p(Vgas − Vliq) ≈ pVgas

Next, we can use the ideal gas law pV = nRT to relate pVgas to RT for a molar quantity of gas by assuming theethanol vapor is ideal. This gives

∆vap(pV ) = RT

And thereby the ratio of expansion work ∆vap(pV ) to expansion enthalpy ∆vapH is

∆vap(pV )

∆vapH=

RT

∆vapH

Using text book table 2.3 we find that ethanol vaporizes at T = 352K and its expansion enthalpy is 43.5 kJmol−1.Substituting these values gives the ratio value

∆vap(pV )

∆vapH=

RT

∆vapH

=8.3145�J��K−1��

mol−1 × 352��K

43.5��kJ��mol−1 × 1000�J

1��kJ= 0.0673

= 6.73%

Solution

∆vap(pV )

∆vapH= 6.73%

8


6 Problem 4.4 pg 154

Question

Calculate the difference in slope of the chemical potential against temperature on either side of (a) the normal freezingpoint of water and and (b) the normal boiling point of water. (c) By how much does the chemical potential of watersuper cooled to −5.0◦C exceed that of ice at that temperature.

Strategy

To solve parts (a) and (b) we’ll use Equation 4.13 (part 2 pg 150) that relates how the temperature derivative (slope)of chemical potential, ∂µ

∂T , changes across a coexistence curve.(∂µ (β)

∂T

)p

−(∂µ (α)

∂T

)p

= −Sm (β) + Sm (α) = −∆trsS = −∆trsH

Ttrs

where α and β denote the two phases.

For part (a) we’re interested in the difference between the liquid l and solid s phases, the fusion transition ∆fusH⊖,

which gives the following expression.(∂µ (l)

∂T

)p

−(∂µ (s)

∂T

)p

= −Sm (l) + Sm (s) = −∆fusS = −∆fusH

Tf

Substituting in the enthalpy and temperature of water’s fusion transition, ∆fusH = 6.008 kJmol−1 and Tf = 273.15Krespectively as found in Table 2.3 of our text book, gives

(∂µ (l)

∂T

)p

−(∂µ (s)

∂T

)p

= −∆fusH

Tf

= −6.008 kJmol−1

273.15K

= −0.021995 kJmol−1 K−1

= −21.995 Jmol−1 K−1

Likewise for part (b) we’re interested in the difference between the gas g and the liquid l phases which is thevaporization transition ∆vapH.(

∂µ (g)

∂T

)p

−(∂µ (l)

∂T

)p

= −Sm (g) + Sm (l) = −∆vapS = −∆vapH

Tv

Substituting in ∆vapH = 40.656 kJmol−1 and Tv = 373.15K from Table 2.3 gives

(∂µ (g)

∂T

)p

−(∂µ (l)

∂T

)p

= −∆vapH

Tv

= −40.656 kJmol−1

373.15K

= −0.108953 kJmol−1 K−1

= −108.953 Jmol−1 K−1

9


For part (c) we’re interested in calculating the difference in chemical potential of liquid water at −5◦C, µ(l,−5◦C)and the chemical potential of solid water at the same temperature µ(s,−5◦C) and we’ll call this quantity x.

x = µ(l,−5◦C)− µ(s,−5◦C)

To calculate x we’ll take advantage of the normal freezing point of water, which implies that at T = 0◦C solid waterand liquid water have the same chemical potential: µ(l, 0◦C) = µ(s, 0◦C)

Therefore we can subtract µ(l, 0◦C)− µ(s, 0◦C) (as this quantity is 0) from the expression we’re working to solve.

x = µ(l,−5◦C)− µ(s,−5◦C)

= µ(l,−5◦C)− µ(s,−5◦C)− [µ (l, 0◦C)− µ (s, 0◦C)]

= [µ (l,−5◦C)− µ (l, 0◦C)]− [µ (s,−5◦C)− µ (s, 0◦C)]

= ∆µ(l)−∆µ(s)

In the third equation above we’ve rearranged the righthand side so as to place the difference in chemical potentials atthe two temperatures in brackets for each phase. We call this quantity ∆µ(α) where ∆µ(α) = µ (α,−5◦C)−µ (α, 0◦C)and α specifies either the liquid (l) or solid (s) phase.

We can calculate ∆µ(l) and ∆µ(s) using Equation 4.2 (pg 143) which gives the change in chemical potential withtemperature. (

∂µ

∂T

)p

= −Sm

Hence the change in chemical potential for the α phase can be calculated as

∆(α) =

∫ Tf

Ti

(∂µ

∂T

)p

dT

=

∫ Tf

Ti

SmdT

= Sm (Tf − Ti)

where Sm is assumed temperature independent.

Using Tf − Ti = ∆T = −5K and the standard entropies of liquid water Sm(l) and solid water Sm(s) we get thefollowing expressions for chemical potential changes

∆(l) = Sm(l)∆T

∆(s) = Sm(s)∆T

Substituting these expression into our equation for x gives

x = ∆µ(l)−∆µ(s)

= [Sm (l)− Sm (s)]∆T

The above term in brackets, the difference in entropy of liquid and solid water, is just the opposite of the entropy offusion −∆fusS which we know from part (a).

10


x = [Sm (l)− Sm (s)]∆T

= −∆fusS∆T

= −(+21.995 Jmol−1��K−1

)×−5��K

= 109 Jmol−1

We are reminded that x is just the difference in chemical potentials that we were solving.

µ(l,−5◦C)− µ(s,−5◦C) = x = 109 Jmol−1

This positive difference in chemical potential between liquid and solid phases implies higher free energy for the liquidphase relative to the solid phase (ice) and explains why ice is favorable at this temperature.

Solution

• (a) (∂µ (l)

∂T

)p

−(∂µ (s)

∂T

)p

= −22.00 Jmol−1K−1

• (b) (∂µ (g)

∂T

)p

−(∂µ (l)

∂T

)p

= −108.95 Jmol−1 K−1

• (c)µ(l,−5◦C)− µ(s,−5◦C) = x = 100 Jmol−1

11

4,7 b solman

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