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3.012 PS 3
THERMO MODEL SOLUTIONS
Issued: 10.03.05
Not due
3.012
Fall 05
THERMODYNAMICS
1. Why is it hard to make measurements at constant volume? We have discussed in class
that experimentally, Cv is difficult to measure due to thermal expansion. Let’s quantify this
difficulty: Suppose you have 1 mole of iron that has a volume of 7.31 cm3 at 293 K.
Determine the pressure that would have to be applied after this material is heated to 298 K
(only 5 degrees warmer!) to compress it to the volume it had at 293K- thus maintaining
constant volume.
Data for Fe:
α = 6.3x10-5 K-1
κ = 1.10x10-6 atm-1
at 298K
We start with the formula for the thermal expansion coefficient, in order to calculate the amount of
expansion that occurs when the temperature is raised to 298K:
"=
"dT =
!
dV
V
TH
VH
# "dT =
TL
1 $ #V '
& )
V % #T ( P
#
VL
dV
V
" (TH $ TL ) = "%T = ln
VH
VL
Thus the volume after heating is:
!
V H = VL e"#T
Next, we need to determine the pressure required to compress the material back to VL:
1 % $V (!
" =# ' *
V & $P )T
1
dP = #
dV
"V
V
1 L dV 1 V H 1 VL e-+T -+T
+P = # ,
= ln
= ln
=
= 286.3atm
" V H V " VL "
VL
"
3.012 PS 3 !
1 of 7
10/6/05
…thus the difficulty in making constant volume measurements. Even this small re-compression
requires quite a significant applied pressure.
2. Thermal properties of gases. Calculate the thermal expansion coefficient and isothermal
compressibility of an ideal gas.
1 % $V (
' *
V & $T ) P
nRT
V=
P
% $V (
nR
' * =
& $T ) P P
1 % nR ( % P (% nR ( 1
" = ' * ='
*' * =
V & P ) & nRT )& P ) T
"#
!
1 & %V )
" #$ ( +
V ' %P *T
nRT
V=
P
& %V )
nRT
( + =$ 2
' %P *T
P
1 & nRT ) & P )& nRT ) 1
, " = $ ($ 2 + = (
+(
+=
V ' P * ' nRT *' P 2 * P
3. Cooking with a sealed pot. Engel and Reid problem P3.5, p. 59.
!
We make use of the expression derived in class relating temperature/volume changes to pressure
changes in a material:
dP =
"
1
dT $
dV
#
V#
Taking the a and k as approximately constant during this process, we can integrate the first term
directly:
!
' 2.04 %10&4 K &1 *
"
dT
=
333K & 298K ) = 156bar
)
$#
&5
&1 ,(
4.59
%10
bar
(
+
298
333
To integrate the second term, we need to estimate the volume change allowed during the expansion.
Since the thermal expansion coefficient of the vessel is smaller than that of water, the expansion of
the vessel will !
be the limiting factor. We can calculate the volume change allowed by the vessel:
3.012 PS 3
2 of 7
10/6/05
" vessel =
1 $ #V '
& )
V % #T ( P
4. Irreversibility of a free expansion. Recall in lecture 2 that we discussed irreversible
processes, and gave several
! examples- one of these example irreversible processes was the
expansion of a gas to fill a vacuum. All irreversible processes only occur in one direction
spontaneously because it is in this direction that the entropy of the universe is increased, in
accord with the second law. Let’s show this for the irreversible expansion: consider the
diagram below. An ideal gas is initially contained in one half of a chamber with adiabatic
walls, the volume of the half of the chamber containing the gas is Vo. The partition between
the right and left halves is suddenly removed, and the gas expands isothermally and
adiabatically to fill the entire space (Vf = 2Vo). Show that this process fulfills the requirements
of the second law for spontaneity by calculating the entropy change in the gas, the
surroundings, and the universe, and show that for the process to run in reverse (spontaneous
collapse of the gas back to one half of the chamber) will violate the second law.
dSuniverse = dSsystem + dSsurr = dSsystem =
dU P
+ dV
T
T
Because the temperature is held constant and for an ideal gas U = U(T), the internal energy is
constant:
!
" nRT %
$
'
nR
# V &
dSuniverse = 0 +
dV =
dV
T
V
V
()Suniverse = nRln f = nRln2 > 0
Vi
Because the entropy change in the universe is positive, the process is spontaneous. For the
process to run in reverse, the entropy change is negative:
!
3.012 PS 3
3 of 7
10/6/05
"Suniverse = nRln
Vf
1
= nRln = #nRln2 < 0
Vi
2
!
5. Let’s revisit the liquid bismuth adiabatic cooling problem we examined on the last problem
set. The physical situation is described again below, and the physical data for the system is
also given. From your calculations on the last problem set, you showed that the alumina
crucible and bismuth equilibrate at a final temperature of 433 K. Demonstrate that this
process (cooling of the bismuth, warming of the alumina crucible) obeys the second law of
thermodynamics, by calculating the total entropy change of the universe for the process.
Twenty kg of liquid bismuth at 600 K is introduced into a 10 kg alumina (Al2O3) crucible (initial
temperature 298K), filling the crucible to the top; the crucible and bismuth are then surrounded
by adiabatic walls (illustrated below) and the system is allowed to equilibrate. At equilibrium,
according to the zeroth law, the temperatures of the bismuth and alumina crucible must be equal.
Use the following thermodynamic data:
C palu min a,solid = 106.6 + 0.0178T
C pBi,solid = 18.8 + 0.023T
!
J
K " mole
C pBi,liquid = 20 + 0.00615T
!
J
K " mole
J
K " mole
!
Tmalu min a = 2327K
!
!
TmBi = 544K
"H mBi = 10,900
J
mole
!
To show the process is spontaneous, we must calculate the entropy change in the universe:
"Suniverse = "Ssystem + "Ssurr = "Ssystem + 0
"Suniverse = "SBi + "SAl2O3
3.012 PS 3
!
4 of 7
10/6/05
We know the entropy change in the surrounding is zero because the only energy transfer occurring
here is heat transfer, and the samples are surrounded by an adiabatic boundary- thus no heat is
moving into the surroundings to change their entropy.
Tf
"SAl2O3
433
C pAl2O3
(106.6 + 0.0178T)
= #n
dT = # (98.08)
dT
T
T
Ti
298
J
K
s
544
C pl
n"H m 433 C p
J
"SBi = # n
dT $
+ #n
dT = $2.79
T
T
T
K
600
544
J
"Suniverse = "SBi + "SAl2O3 = 1.35
K
"SAl2O3 = 4.14
!
! change of the universe is positive, the process is predicted to be spontaneous.
Since the entropy
!
6. Confirming your intuition using the second law. Using your ability to calculate entropy
and enthalpy changes at phase transitions and the data given below for nickel, consider the
following possible processes and confirm your intuition by a calculation showing whether the
process is spontaneous or not according to the second law. The key to this problem is to
correctly identify what heat is transferring between the system and the surroundings in these
processes, and to carefully calculate the entropy changes for the system. Also critical: pay
attention to the number of significant figures you carry in your calculation.
a. One mole of solid nickel at 1716 K isothermally transforms completely to liquid at this
temperature. The surroundings are a heat bath at 1716 K, and the process occurs at
constant pressure.
b. One mole of solid (superheated) nickel at 1736 K isothermally transforms completely
to liquid at this temperature. The surroundings are a heat bath at 1736 K, and the
process occurs at constant pressure.
C ps = 16.49 + 0.0187T
C pL = 38.91
!
!
J
K " mole
J
K " mole
"H m = 17.47
Tm = 1726K
!
kJ
mole
!
a. The transformation we are concerned with is shown schematically below on
qualitative plots of the entropy and enthalpy of the system near the temperature of
interest. Our process is isothermal and isobaric. To test for spontaneity, we need, as
in the above problems, to calculate the entropy change in the universe:
3.012 PS 3
5 of 7
10/6/05
"Suniverse = "Ssystem + "Ssurr
The entropy change in the system will be given by the entropy change along the dashed line from A
to B on the entropy plot below. We calculate the length of this line segment as illustrated graphically
in the plot:
!
"Ssystem = "Sm # "S1 + "S2
"Sm =
"H m
J
= 10.12
Tm
K
1726
C pl
(38.91)
1726
J
"S1 = $ n dT = $ (1)
dT = 38.91ln
= 0.226
T
T
1716
K
1716
1716
1726
1726
C ps
(16.49 + 0.0187T)
1726
J
"S2 = $ n dT = $ (1)
dT = 16.49ln
+ 0.0187(1726 #1716) = 0.0958 + 0.187 = 0.283
T
T
1716
K
1716
1716
1726
%"Ssystem = 10.18
J
K
Because the surroundings are at constant temperature, and we assume no work is occurring, the
entropy change in the surroundings derives completely from the heat transfer from the system to the
surroundings:
!
"Ssurr =
#qsys
=
#"H sys
T
(1716K)
"H sys = "H m # "H1 + "H 2 = 17,470 # 389.1+ 486.7 = 17.57kJ
1726
"H1 =
1726
$ nC pl dT =
1716
$ (1)(38.91)dT = 389.1J
1716
1726
"H 2 =
s
p
$ nC dT = 16.49(1726 #1716) +
1716
%"Ssurr = #10.23
0.0187
(1726 2 #1716 2 ) = 164.9 + 321.8 = 486.7J
2
J
K
Finally, the total entropy change of the universe is:
!
"Suniverse = "Ssystem + "Ssurr = 10.18 #10.23 = #0.05
J
K
Since the entropy change is less than zero, this process will not occur.
!
3.012 PS 3
6 of 7
10/6/05
b. Now, we follow the exact same scheme to predict the spontaneity of melting a
superheated solid nickel sample: We must show that the entropy change of the
universe in this process is positive:
"Suniverse = "Ssystem + "Ssurr
1736
"Ssystem = "Sm #
$n
1726
"Ssurr =
1736
C ps
Cl
J
dT + $ n p dT = 10.12 # .282 + 0.225 = 10.06
T
T
K
1726
#qsys
#"H sys
=
(1736K) (1736K)
1736
qsys = "H m #
$ nC psdT +
1726
1736
l
p
$ nC dT = 17.37kJ
1726
J
%"Ssurr = #10.01
K
J
%"Suniverse = 0.05
K
…which is a spontaneous process.
!
3.012 PS 3
7 of 7
10/6/05