MATSCI 204
THERMODYNAMICS AND PHASE EQUILIBRIA
Winter 2013
Problem Set #3
Due: Friday February 1st
Problem 1:
(25 points)
1 % $V (
The isentropic compressibility is defined as " S = # ' * . Express βS as a function of
V & $P ) S
CP, α and β (isothermal compressibility). By using the expression linking CP and CV that
C
you can find in the notes, show that " S = " V .
CP
!
!
!
!
Answer
1 % $V (
" S = # ' * therefore,!we need to find V(P,S). We will use the procedure shown in
V & $P ) S
class:
# "V &
# "V &
C
dV = % ( dP + % ( dS and dS = P dT " #VdP . Hence:
$ "P ' S
$ "S ' P
T
+# "V &
.
# "V &
C # "V &
dV = -% ( ) *V % ( 0dP + P % ( dT . We identify the coefficients from V(P,T)
$ "S ' P /
T $ "S ' P
,$ "P ' S
obtained in class:
!
#
&
#
C P "V
"V & )VT
% ( = )V * % ( =
$ "S ' P C P
T $ "S ' P
2
!
# "V &
# "V &
# "V & T (*V )
) +V
% ( ) *V % ( = )+V , % ( =
$ "P ' S
$ "S ' P
$ "P ' S
CP
and we obtain:
2
(
% C # TV$ 2 (
1 % T ($V )
CV
" S = # ''
# "V ** = "' P
*="
V & CP
CP
CP
&
)
)
!
Problem 2:
(30 points)
!
Consider the S/L line of the phase diagram of water. In this problem, you’ll derive why it
always looks like a line rather than a curve.
1
1- Calculate
dP/dT
of
the
S/L
equilibrium
line
at
1
atm.
2- Calculate dP/dT of the S/L equilibrium line at 100 atm. For simplicity, assume
that the melting temperature at both pressures is very similar (i.e. the line is nearly
vertical) and that the compressibility of both phases is 0.
Physical data:
Latent heat of melting: Lm=5.9 kJ/mol.
Thermal expansion coefficient of water near 0°C: α= -0.68x10-4 K-1
Molar volume of water near 0°C: vw=18 cm3/mol
Thermal expansion coefficient of ice near 0°C: α=5x10-5 K-1
Molar volume of ice near 0°C: vw=20 cm3/mol
Answer
1 dP/dT=Δs/Δv and Δs=Lm/T therefore
dP
Lm
5900
=
=
= "10.8 MPa /K
dT T (v L " v S ) 273 # 2 $10"6
(notice negative sign).
!
!
2 If you assume that β~0 and ΔT~0 then Δv~0 upon compression but Δs may depend on
!
P.
# "S &
# "V &
% ( = )% ( = )*V (Maxwell relation).
$ "P 'T
$ "T ' P
Thus, for the water phase we can write
100
100
# "S &
S100atm = S1atm + ) % ( dP = S1atm * ) +V dP = S1atm * +V,P
$ '
1 "P T
1
The same can be written for the ice phase, therefore
"S100 = "S1 # "P ($ wVw # $ iVi ) . We find ΔS100=21.59 J.K-1.mol-1 while ΔS1=21.61 J.K1
.mol-1, a difference of ~0.1%. Therefore, the slope changes by less than 0.1% and it is a
good approximation to consider it constant.
!
Problem 3:
(40 points)
By solving this problem you’ll determine whether the vapor pressure of a solid is the
same whether it evaporates in a vacuum or in the atmosphere (you should already know
the answer to this question anyway…).
See the phase diagram of Fe below.
1- Rank all the phases in order of increasing entropy. (5 points)
2- At P=10-4 atm., draw a sketch of G vs. T of all the phases. Label as quantitatively
as possible the intersection points. (10 points)
3- We set the temperature at 1000°C and conduct the following imaginary
experiment. A piece of Fe (in the γ phase) is placed in an enclosed volume that
2
has been completely evacuated. Describe what happens and explain where you
would read the vapor pressure of Fe at equilibrium. (10 points)
4- Now, imagine that the piece of iron (still in the vacuum and at 1000°C) is
squeezed by a clamp up to a pressure P=1 atm. Calculate the ratio of the new
equilibrium vapor pressure of Fe to the one estimated in (3). Assume that Fe
vapor is an ideal gas (10 points)
5- Based on the answer of questions 3 and 4, what would have happened to the vapor
pressure of Fe if rather than squeezing it with a clamp you had opened to
atmosphere the evacuated container, let air in and then closed it? (5 points)
Physical data of Fe at 1 atm. (ppm=parts-per-million=10-6):
VFe=7.4 cm3/mol, α=66 ppm/K, β=1.1 ppm/atm, CP=24.27+8.28x10-3T J.K-1mol-1
Answer
1- α−γ−δ-liquid-vapor (higher S phases appear at higher T)
# "g &
2- % ( = )s
$ "T '
!
3
3- At P=0, the equilibrium phase is vapor so Fe will start evaporating until it reaches
the equilibrium pressure, which is read at the intersection between the γ-v
equilibrium curve and the vertical line at T=1000°C. PFe~10-9.5 atm.
4- By squeezing the Fe with a clamp to P=1 atm., we increase the chemical potential
of the solid: dµs=vdP.
The vapor will be in equilibrium if dµv=dµs and dµv=RTdln(P) (ideal gas).
In order to calculate the chemical potential increase of the solid, we need to know
v(P).
The definition of compressibility gives: d ln(v ) = "#dP therefore
v ( P ) = v P0 . exp["# ( P " P0 )] where P0 is atmospheric pressure.
P0
Thus
" dµ
PV
!
P0
S
=
P0
v P0 .
" v (P )dP =v !" exp[#$ (P # P )]dP = $
P0 .
PV
0
PV
PV
exp[#$ ( P # P0 )]
P0
P0
Plugging in the numbers, we get
" dµ
S
= 0.75 J /mol .
PV
! If the vapor pressure in equilibrium with the solid at pressure P0 is labeled Pv’, we
P0
* 1 P0
# P' &
P'
have: " dµS = RT ln% v ( ) v = exp,
d
µ
" S //.
P
P
RT
,
$
'
!
v
v
PV
PV
+
.
P'v
" 1+ 71#10$6 thus the pressure increases by
Plugging the numbers we get
Pv
! ~70 ppm.
By opening the container to atmosphere, the same effect as in #4 is obtained: the vapor
!
pressure increases by a negligible
amount. Therefore, it is a good approximation to
assume that the vapor pressure above a solid is the same whether it evaporates in a
vacuum (the situation expressed in a unary phase diagram) or in a 1 atm. of total pressure.
4
Problem 4:
(25 points)
We have seen that if I control P and T, in a spontaneous transformation gsys must
decrease. In this problem, you’ll understand explicitly how that’s related to the the
entropy of the Universe (which must of course increase in a spontaneous transformation).
We will show that if water solidifies at T<Tm (a spontaneous process), ΔSUNIVERSE>0. Our
initial system is thus liquid water in contact with a thermal bath at T<Tm.
1- Calculate ΔSsys of 1g of water during the water->ice transformation at T=-2°C (10
points)
2- Calculate ΔSsurroundings during the same transformation at T=-2°C (10 points)
3- Verify that ΔSUNIVERSE>0 (5 points)
Data:
Cp (water)
4.184 J.g-1.K-1
Cp (ice)
2.05 J.g-1.K-1
Lm (ice) at T = Tm
333.55 kJ.kg-1
Assume the above properties are independent of temperature
Answer
1- At T=Tm, ΔSm=ΔHm/Tm because Δgm=0. ΔHm=Lm, we obtain ΔSm(0°C)=1.222 J.g-1.K-1
At T<Tm, I can’t equate ΔSm=ΔHm/Tm because we are not equilibrium. However, I can
271 water
$ 271'
C
-1 -1
water
water
" S0°C
= # P = CPwater ln&
calculate S"2°C
) =-0.03 J.g .K
%
(
T
273
273
271 ice
$ 271'
C
-1 -1
ice
ice
" S0°C
= # P = CPice ln&
Similarly, S"2°C
) ==-0.015 J.g .K
% 273 (
273 T
271 ice
271 water
!Hence S ice " S water = CP " CP " $S 273 =-1.208 J.g-1.K-1
#
# T
"2°C
"2°C
m
273 T
273
! you prefer closed formulas, "S 271 = "H m + "C ln#% 271&( = )"S 271
(If
m
P
solidification )
$ 273 '
Tm
! 2- The surroundings are just a conventional heat bath and since ΔHm does not depend on
"H m
271
temperature, we have "Ssurroundings
=1.231 J.g-1.K-1
=
T
!
$1 1 '
$ 271'
-1 -1
271
3- ΔSsys+ΔSsurroundings= "SUNIVERSE
= Lm & # ) # "CP ln&
) =0.023 J.g .K >0.
% 273 (
% T Tm (
!
Problem 5: !
(30 points)
5
Historically, phase diagrams were obtained from calorimetry measurements. This
problem is meant to show you how much information can be obtained from these
experiments.
A unary system only exhibits 3 phases (liquid, solid and gas). We put 1 mole of the solid
at low temperature in a calorimeter that delivers a constant heating rate of 1 kJ/minute.
The heat transfer is assumed to be reversible. At three different pressures we obtain the
three following curves for T(time) (they are offset vertically for clarity). At P=10 atm., no
second plateau is ever observed.
1- Draw the most general unary phase diagram in the (P,T) plane compatible with
these measurements. Use what you know about the shape of these phase diagrams
and make it as quantitative as possible. (8 points)
2- What is the pressure at the triple-point and what would the heating curve look like
at the pressure of the triple point? (8 points)
3- Find the temperature where the pressure of the gas is equal to 0.1 atm. (6 points)
4- We get a more accurate measurement of the plateau temperature of the P=10 atm:
T=249.98K. (8 points)
a. Does the liquid or the solid have a larger density?
b. Can you estimate vL-vS?
Answer
1- The plateaus correspond to phase transformations. Since we start with the solid,
the first transformation can be S->L or S->G. However at the lower pressure we
6
have 2 transformations at 2 different temperatures therefore the simplest version
is:
2- S/L line is nearly vertical therefore to a good approximation TTP~250K. In order
to calculate the pressure, we use the Clausius Clapeyron equation on the L/G line.
The heating curve gives us the heat of vaporization at P=1 atm. and T=310K:
ΔHvap=6 kJ. As a result:
)H vap " 1
"P %
1 %
ln$ TP ' = (
(
$
' from which we get PTP=0.57 atm.
# 1 &
R # TTP 310 &
The heating curve has a single phase transformation and the heat of the
transformation is roughly the sum of the heat of melting and that of vaporization.
!
!
3- Now we use the Clausius Clapeyron on the sublimation line. We get
ΔHsubl=ΔHmelt+ΔHvap since H is a state function. We get ΔHmelt from the heating
curve (3 kJ).
Since
the
triple
point
is
on
the
S/G
line:
01
* "P % " R % 1 T0.1atm. = +ln$ TP ' ( $
. =178.3 K.
'+
, # 0.1 & # )H subl. & TTP /
4- The melting temperature is decreased as the pressure is increased therefore the
material must increase its volume upon freezing (like water).
*1
dP "H
"H $ dP '
-7
3
=
# "v =
Using the Clausius equation
& ) =2.6x10 m /mol
dT T"v
T % dT (
(remember that 1 atm. is 101325 Pa, which is the SI unit of pressure).
!
7