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Kinetics and Phase Transformation / Problem Set 1 Solution / Due Feb. 6, 2014
Problem 1
The temperature dependence of the vapor pressure of solid sulfur dioxide can be
approximated by the relation log (P ) = 10.5916 − 1871.2/T and that of liquid sulfur
dioxide by log (P ) = 8.3186 − 1425.7/T . Pressure P is in Torr, and temperature T is
in Kelvin. Estimate the temperature and pressure of the triple point of sulfur dioxide.
From Wikipedia, vapor pressure is defined as the pressure exerted by a
vapor in thermodynamic equilibrium with its condensed phases at a given
temperature in a closed system. Therefore, the given expressions of vapor
pressure are actually the equations for phase boundaries. The triple point
then can be determined by solving the intersection of these two phase
boundaries:
log (P ) = 10.5916 − 1871.2/T
log (P ) = 8.3186 − 1425.7/T
Let the right-hand parts equal to each other:
10.5916 − 1871.2/Ttri = 8.3186 − 1425.7/Ttri ⇒ Ttri = 196.02 K
Plug in Ttri into one of the given relation and solve for Ptri :
log (Ptri ) = 10.5916 − 1871.2/196.02 ⇒ Ptri = 11.11 Torr
Note: Since the Clausius-Clapeyron equation of vapor-condensed phases
can be approximated in exponential form, so Ptri = 2.85 Torr if you used
natural logarithm.
by Qingjie Li / Instructor: Prof. Erlebacher / Department of MSE @ JHU
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Kinetics and Phase Transformation / Problem Set 1 Solution / Due Feb. 6, 2014
Problem 2
Calculate the difference in slope of the chemical potential against temperature on
either side of (a) the normal freezing point of water and (b) the normal boiling point
of water. (c) By how much does the chemical potential of water supercooled to −5.0◦ C
exceed that of ice at that temperature?
From thermodynamics,
chemical potential is defined as molar Gibbs free
∂G
energy:µi = ∂n
, where ni 6= nj . Therefore,
i
T,P,nj
dµ = dGm = −Sm dT + Vm dP ⇒
∂µ
∂T
= −Sm .
P
From this, we see that the slope of the chemical potential against temperature at constant pressure is the negative molar entropy.
(a) At freezing point of water, the difference in slopes of chemical potential against temperature on either side is:
l
s
s→l
∂µ
∂µ
∆Hm
l
s
s→l
.
−
= −Sm
+ Sm
= −∆Sm
=−
∂T P
∂T P
Tf
s→l
s→l
The last step comes from the fact ∆Gs→l
m = ∆Hm − ∆Sm Tf = 0. We know
s→l
∆Hm
= 6.01 kJ/mol, Tf = 273.15 K, so
l
s
s→l
∂µ
∂µ
∆Hm
6.01 kJ/mol
= −22.0 JK−1 mol−1 .
−
=−
=−
∂T P
∂T P
Tf
273.15 K
(b) Similarly, at boiling point,
v
l
l→v
∂µ
∂µ
∆Hm
40.6 kJ/mol
−
=−
=−
= −109.0 JK−1 mol−1 .
∂T P
∂T P
Tv
373.15 K
(c) The difference in chemical potential of ice and supercooled liquid water
at −5◦ C is:
l
s
µl (−5◦ C) − µs (−5◦ C) = µl (0◦ C) + Sm
∆T − (µs (0◦ C) + Sm
∆T ) .
At freezing point, µl (0◦ C) = µs (0◦ C), therefore
s→l
∆µ(−5◦ C) = ∆Sm
∆T = −22.0 JK−1 mol−1 × (−5) K = 110 Jmol−1 .
by Qingjie Li / Instructor: Prof. Erlebacher / Department of MSE @ JHU
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Kinetics and Phase Transformation / Problem Set 1 Solution / Due Feb. 6, 2014
Problem 3
The change in enthalpy is given by dH = Cp dT + V dP . The Clausius-Clapeyron
Equation relates dP and dT at equilibrium, and so in combination the two equations can be used to find how the enthalpy changes along a two-phase boundary
as the temperature changes and the two phases remain in equilibrium. Show that
d (∆H/T ) = ∆Cp d (ln T ).
Suppose infinitesimal change of temperature dT and pressure dP along a
two-phase boundary, the enthalpy changes in each phase(let the two phases
be α and β, respectively)are:
dH α = Cpα dT + V α dP,
dH β = Cpβ dT + V β dP.
dH α − dH β = d H α − H β = d (∆H) .
dH α − dH β = Cpα − Cpβ dT + V α − V β dP = ∆Cp dT + ∆V dP.
⇒ d (∆H) = ∆Cp dT + ∆V dP
(1)
From Clausius-Clapeyron equation,
dP =
∆H
dT.
T ∆V
Plug dP into equation (1),
d (∆H) = ∆Cp dT + ∆V
∆H
∆H
dT = ∆Cp dT +
dT.
T ∆V
T
(2)
Rearrange equation (2),
d (∆H) −
∆H
dT = ∆Cp dT.
T
(3)
Multiply 1/T on both sides of equation (3),
d (∆H) T − ∆HdT
dT
= ∆Cp .
2
T
T
∆H
⇒d
= ∆Cp d (ln T ) .
T
by Qingjie Li / Instructor: Prof. Erlebacher / Department of MSE @ JHU
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Kinetics and Phase Transformation / Problem Set 1 Solution / Due Feb. 6, 2014
Problem 4
Methane is the primary component of natural gas. The thermophysical properties
of methane (J. Phys. Chem. Ref. Data 18, 583 (1989)) includes the following data
describing the liquid-vapor phase boundary:
(a)Plot the liquid-vapor phase boundary; (b) Estimate the standard boiling point of
methane; (c) Compute the standard enthalpy of vaporization of methane, given that
the molar volumes of the liquid and vapor at the standard boiling point are 3.8 × 10−2
and 8.89 dm3 mol−1 , respectively.
(a) The liquid-vapor phase boundary is shown as below:
(b) The standard boiling point is the temperature at which the liquid is in
equilibrium with the standard pressure of 1 bar (0.1 Mpa). Interpolation
of the the plotted points gives:
Tb − 110 K
112 K − 110 K
=
⇒ Tb = 111.5 K.
0.104 MPa − 0.088 MPa
0.1 MPa − 0.088 MPa
(c) From Clausius-Clapeyron equation,
dP
∆vap H
dP
=
⇒ ∆vap H = (T ∆vap V )
.
dT
T ∆vap V
dT
The slope can be obtained by fitting the points nearest the boiling point:
dP/dT = 8.14 × 10−3 MPaK−1 . So,
∆vap H = 112 K × (8.89 − 0.0380) × 10−3 m3 mol−1 × 8.14 kPaK−1 = 8.07 kJ/mol.
by Qingjie Li / Instructor: Prof. Erlebacher / Department of MSE @ JHU
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