CHEM 3411, Fall 2010
Solution 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 not
significant and would be rounded off at the end of calculations. Carrying these extra digits for intermediate values in
calculations 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
CHEM 3411, Fall 2010
Solution Set 4
2
Exercise 4.5b pg 153
Given
⊖
Iron is heated from Ti = 25◦ C to Tf = 100◦ C. Over this temperature range Sm
= 53 J K−1 mol−1.
In terms of given variables, this is written:
Ti = 25◦ C
Tf = 100◦ C
⊖
Sm
= 53 J K−1 mol−1
Find
By how much does its chemical potential change?
Strategy
First, temperatures are converted to Kelvin.
Ti = 100 + 273◦ C = 373 K
Tf = 1000 + 273◦ C = 1273 K
We can use text book Equation 4.2 (pg 143) to relate temperature changes to changes in chemical potential.
(
∂µ
∂T
)
⊖
= −Sm
p
which gives the differential
⊖
dµ = −Sm
dT
Integrating dµ over the temperature range gives the change in chemical potential.
∫
∆µ = −
Tf
⊖
Sm
dT
Ti
⊖
= −Sm
(Tf − Ti )
J
= −53
(1273 K − 373 K)
K mol
J
= −47700
mol
kJ
= −47.7
mol
⊖
where the integral has been easily solved, as we’ve assumed Sm
is constant over this temperature range.
2
CHEM 3411, Fall 2010
Solution Set 4
Solution
∆µ = −50
kJ
mol
3
CHEM 3411, Fall 2010
Solution Set 4
3
Exercise 4.11b pg 153
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
∆vap H ⊖
=
dT
RT 2
Through rearrangement we solve for ∆vap H ⊖ .
∆vap H ⊖ = RT 2
d ln p
dT
ln p
To find ddT
, 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)
ln(10) × 1625 K
=
dT
T2
(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 ∆vap H ⊖ .
4
CHEM 3411, Fall 2010
Solution Set 4
∆vap H ⊖
ln(10) × 1625 K
2
T
R × ln(10) × 1625 K
J
8.314
× ln(10) × 1625
K
K mol
J
31108.5
mol
kJ
31.1085
mol
2
= R
T
=
=
=
=
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.
= 8.750 − 1625/(T / K)
log(760
torr/
torr)
T
1625 K
8.750 − log 760
= 276.9 K
=
Solution
kJ
• (a) ∆vap H ⊖ = 31.11 mol
• (b) T = 280 K
5
CHEM 3411, Fall 2010
Solution Set 4
4
Exercise 4.14b pg 153
Given
On a cold, dry morning after a frost, the temperature was T = −5◦ C and the partial pressure of water in the
atmosphere fell to pH2 O = 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 pH2 O 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 water
pH2 O needed to ensure the frost remains, we know any pH2 O 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 this
temperature; this is the pressure at this temperature on the coexistance curve for ice and water vapor. We know
that if the atmosphere has a lower partial pressure of water than the solid-vapor pressure, then the water vapour
will 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
⋆ −χ
p=p e
∆sub H ⊖
χ=
R
(
1
1
− ⋆
T
T
)
where we’ve replaced ∆vap H ⊖ in the original expression with ∆sub H ⊖ as we’re concerned with the sublimation
coexistence point instead of the vaporization point.
The sublimation enthalpy ∆vap H ⊖ is found from the vaporization enthalpy and the fusion enthalpy, ∆vap H ⊖ and
∆fus H ⊖ respectively.
∆sub H ⊖
=
∆fus H ⊖ + ∆vap H ⊖
= 6.008 kJ mol−1 + 44.016 kJ mol−1
= 50.024 kJ mol−1
= 5.0024 × 104 J mol−1
Using Equation 4.12 we can calculate the pressure p at temperature T when we know the reference pressure p⋆ at the
reference temperature T ⋆ . As we’re solving for a p(T ) on the solid/gas coexistence curve, we’ll need the reference to
also fall on this curve. Therefore we’ll use the the triple point of water as our reference giving the following values:
T ⋆ = 273.16 K
6
CHEM 3411, Fall 2010
Solution Set 4
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 =
268 K.
χ
)
1
1
=
− ⋆
T
T
)
(
−1
1
5.0024 × 104
J
mol
1
=
−
−1
−1
268
K 273.16
K
mol
8.314
J
K
= 0.42409
∆sub H ⊖
R
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 sublime
into any gas system with a partial pressure of water pH2 O < 0.40029 kPa. This gives us the answer to part b as
pH2 O = 0.40029 kPa.
Additionally, can determine that the ice will sublime into the atmosphere as the partial pressure of water is pH2 O =
0.30 kPa.
Solution
• (a) Yes.
• (b) pH2 O = 0.40029 kPa
7
CHEM 3411, Fall 2010
Solution Set 4
5
Exercise 4.17b pg 154
Question
What fraction of the enthalpy of vaporization ∆vap H 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
∆vap H into an internal energy component ∆vap U and an expansion work component ∆vap (pV ).
∆vap H = ∆vap U + ∆vap (pV )
Additionally, we can assume the pressure is constant and the liquid volume Vliq is negligible relative to the gas volume
such 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 the
ethanol vapor is ideal. This gives
∆vap (pV ) = RT
And thereby the ratio of expansion work ∆vap (pV ) to expansion enthalpy ∆vap H is
∆vap (pV )
RT
=
∆vap H
∆vap H
Using text book table 2.3 we find that ethanol vaporizes at T = 352 K and its expansion enthalpy is 43.5 kJ mol−1 .
Substituting these values gives the ratio value
∆vap (pV )
∆vap H
=
RT
∆vap H
−1
−1
8.3145
J
mol
× 352
K
K
−1
1000
J
43.5
mol × 1 kJ
kJ
= 0.0673
= 6.73%
=
Solution
∆vap (pV )
= 6.73%
∆vap H
8
CHEM 3411, Fall 2010
Solution Set 4
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 freezing
point of water and and (b) the normal boiling point of water. (c) By how much does the chemical potential of water
super 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)
∂µ
of chemical potential, ∂T
, changes across a coexistence curve.
(
∂µ (β)
∂T
(
)
−
p
∂µ (α)
∂T
)
= −Sm (β) + Sm (α) = −∆trs S = −
p
∆trs H
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 ∆fus H ⊖ ,
which gives the following expression.
(
∂µ (l)
∂T
)
(
−
p
∂µ (s)
∂T
)
= −Sm (l) + Sm (s) = −∆fus S = −
p
∆fus H
Tf
Substituting in the enthalpy and temperature of water’s fusion transition, ∆fus H = 6.008 kJ mol−1 and Tf = 273.15 K
respectively as found in Table 2.3 of our text book, gives
(
∂µ (l)
∂T
(
)
−
p
∂µ (s)
∂T
)
= −
p
∆fus H
Tf
6.008 kJ mol−1
273.15 K
= −0.021995 kJ mol−1 K−1
= −21.995 J mol−1 K−1
= −
Likewise for part (b) we’re interested in the difference between the gas g and the liquid l phases which is the
vaporization transition ∆vap H.
(
∂µ (g)
∂T
)
(
−
p
∂µ (l)
∂T
)
= −Sm (g) + Sm (l) = −∆vap S = −
p
∆vap H
Tv
Substituting in ∆vap H = 40.656 kJ mol−1 and Tv = 373.15 K from Table 2.3 gives
(
∂µ (g)
∂T
)
(
−
p
∂µ (l)
∂T
)
= −
p
∆vap H
Tv
40.656 kJ mol−1
373.15 K
= −0.108953 kJ mol−1 K−1
= −
= −108.953 J mol−1 K−1
9
CHEM 3411, Fall 2010
Solution Set 4
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 water
and 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 at
the 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 with
temperature.
(
∂µ
∂T
)
= −Sm
p
Hence the change in chemical potential for the α phase can be calculated as
∫
Tf
∆(α) =
Ti
∫
(
∂µ
∂T
)
dT
p
Tf
=
Sm dT
Ti
=
Sm (Tf − Ti )
where Sm is assumed temperature independent.
Using Tf − Ti = ∆T = −5 K and the standard entropies of liquid water Sm (l) and solid water Sm (s) we get the
following 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 of
fusion −∆fus S which we know from part (a).
10
CHEM 3411, Fall 2010
Solution Set 4
x = [Sm (l) − Sm (s)] ∆T
= −∆fus S∆T
(
)
× −5 −1
= − +21.995 J mol−1
K
K
109 J mol−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 J mol−1
This positive difference in chemical potential between liquid and solid phases implies higher free energy for the liquid
phase relative to the solid phase (ice) and explains why ice is favorable at this temperature.
Solution
• (a)
• (b)
• (c)
(
(
∂µ (l)
∂T
∂µ (g)
∂T
(
)
−
p
(
)
−
p
∂µ (s)
∂T
∂µ (l)
∂T
)
= −22.00 J mol−1 K−1
p
)
= −108.95 J mol−1 K−1
p
µ(l, −5◦ C) − µ(s, −5◦ C) = x = 100 J mol−1
11