Chemistry 4460
Physical Chemistry II
Problem Set #3
Spring 2018
1.
One mole of an ideal monatomic perfect gas at 10 atm and 200 K is to be transformed to 0.15 atm
and 900 K. Calculate ∆S1, ∆S2, and the total entropy change for the system which occurs along
each of the following two-step reversible paths:
a.
b.
c.
d.
2.
Isobaric expansion and isochoric temperature drop.
Isobaric expansion and adiabatic temperature drop.
Isothermal expansion and isochoric temperature increase.
Isothermal expansion and isobaric temperature increase.
The third law states that the entropies of crystalline substances are zero at absolute zero. This
should apply to different crystalline forms of the same substance. The following heat capacity
relationships have been determined for rhombic and monoclinic forms of sulfur between 15 and
370 K:
Cp
R
Cp
R
2
5
0.08 2.17 x10 T 6.2 x10 T
1.96 x10 2 T
4.8 x10 5 T 2
forms
2
8
6.9 x10 T
4.7 x10 8 T
(rhombic)
tobezero
3 assuming
t
sy So
5368
3
(monoclinic)
tf
HT
GIt f qdT Soisplugin18K
Both
Calculate the third-law entropies for each form at 368.54 K (the rhombic → monoclinic transition
temperature). You will need to apply the Debye approximation to calculate the entropy content
of each phase at 15 K. What is the difference in the total entropy content of the two phases at
368.54 K? Does your answer support the idea that both have zero entropies at 0 K? (Note that the
uncertainty in the heat capacities is on the order of 0.05R).
between
if twovalues lie
indistinguishable
i supportszero entropies
3.
R
Derive an expression for the efficiency of a Stirling engine. Remember that the work done by the
engine is equal to the total heat extracted from the hot reservoir minus the heat dumped into the
cold reservoir. As opposed to the Carnot engine, heat either enters or leaves the system in all four
steps.
ok
4.
5.
Consider a perfect gas contained in a cylinder and separated by a frictionless adiabatic piston into
two sections A and B. All changes in B are isothermal, but the temperature in A is allowed to vary.
There is 2.00 mol of the gas in each section. Initially TA=TB=300 K. VA=VB=2.00 L. Heat is supplied
to section A, and the piston moves reversibly until the final volume in section B is 1.00 L.
a.
Calculate ∆SA and ∆SB
b.
Calculate ∆AB and ∆GB. (It turns out that these two parameters are indeterminate in
section A and cannot be calculated.)
c.
Calculate the total ∆S for both the system and the surroundings.
Given that:
A U TS
derive the following relationships:
dA
A
V
PdV
a.
A
T
P
T
P
T
6.
SdT
V
S
V
T
Using available tabulated data, calculate the entropy change for the following reaction at
298 K and 750 K. You may assume that the heat capacities are constant with respect to
temperature.
SO2(g) + ½O2(g) → SO3(g)
b.
S
V
Calculate ∆Grxn at the same two temperatures.
7.
Consider CO2(g) at 320 K. The compression factor at this temperature can be expressed by the
following pressure-dependent function:
Z ( P,320K )
Where:
A1
A2 P A3 P 2
A4 P 3
A5 P 4
A1 = 1.000
A2 = -4.007x10-3 atm-1
A3 = -1.16x10-5 atm-2
A4 = 5.5x10-8 atm-3
A5 = -1.65x10-9 atm-4
Graph the fugacity as a function of P.
8.
The protein lysozyme unfolds at a transition temperature of 75.5°C and the standard enthalpy of
transition is 509 kJ/mol. Calculate the entropy of unfolding of lysozyme at 25.0°C, given that the
constant pressure heat capacity of the unfolded form of the protein is 6.28 J K-1 mol-1 greater
than the folded form. Both heat capacities can be assumed to be temperature independent.
Hint: Imagine the process occurring in three steps: (1) heating the folded protein to the
transition temperature, (2) unfolding at the transition temperature, and (3) cooling the unfolded
protein back to 25.0°C.
9.
Suppose that an internal combustion engine runs on octane, for which the enthalpy of
combustion is -5512 kJ/mol and take the mass of 1 gal of fuel as 3 kg. What is the maximum
height, neglecting all forms of friction, to which a car of mass 1000 kg can be driven on 1.00 gal
of fuel given that the engine cylinder temperature is 2000°C and the exit temperature is 800°C.
You may assume for this problem that the engine operates reversibly.
AnnaWilson
10Atm 200k
imollo082ltmoY.iml2o
v
PVHRT
0.15atm900k
Tz Pfp
1.64L
fatmFYY.ff.bz
matgqy
6ooO l0at Vz
l900k 4926L
oat.TT
lmollo0821fn
0.15ohm
a Isobaricexpansion andisochorio temp drop
Ds
bnf9 31.3FmoyK
5 28314 0174
dT
DSz
31218.3144mgK
1
en 4880 23.74m01K
Astor55.0
b
1
mob
isobaric expansion and adiabatic temperaturedrop
DS 31.34mAK
D82 0 adiabat
DStot3134m01K
c IsothermalExpansion
and isochoriotemp increase
BS n Ren FT
1moe 8.3144mAK en 4,926
647
7
052 28.314 01K en 92
47.5
18.8Hmolk
DStof 66.34m01 K
d
isothermal Expansion
and isobarictemp
increase
DS 47.54m01 K
ASI E 8.3144mAK en 788 3134mg k
DS.pt 78.8Knot'k
01k
PP l
DWiz
ing
He
I
b
V
Vz
PIVdN hRTze.hr IT
lsochori
DWz3
DWy
0 isothermal
DW34
NRTienfYz
n
8
tCvTztTil7o
DQ Ql2tQ4FnRtzen
DQiz DWiz o
DQzz Cv Tt Tz 0
bQ34 DWzyco
DQ4 Cultz71170
DW W.ztwzye hrftzTDlnf.TT
p
lT2
HY
isothermal
200m01
TATrs 300k isothermal
VB4.00L
VAVB2.00L
a Calculate
DSAandDSB
FAI
YZ
PAffftffe.ph
3Tn i 300Ktnz 90010
f
I
f
B
a
If
DSAnCvenf
12m01
Dsps
b
nRln
20 myenf930o0o t2mo1 8.3l44k.ymoDdh ZI 50.69
2m01 8.3144kmoi
DAB DU DS TDS
3457.7J
Bby
c
tnRln
DH DS TAS
34577J
DSsys DSatASB
A VETS
11.534k
ln
130014111.53
4
1300kV11.534k
39.24K
DA du Tds SAT
gtS SdTdA
pdv.TN
tots
PDU Sdt
lt
dU dgPdV
Tds Pdv
PITI P 13 s
Hat 13 F
298 248214m01t 205.15 44 25677
290.4
DSzgg
a
2337
01 K
313.4
25677Enotk 248.21
0110
2 205154mmolK
94.024mg.k
DS so
b
k 233.74miK
65.54mgK
313.44mg
1290.4 4K
mo1
DHzag 395
774 41206RO
98
97k4m
DG
DH
4
Tbg 70.9Dbza
98.97
298
0.094024
4m01
DHsso28to ol 21B5tmTy
7.44kmT
744 Kool DGyso
0.065
56.57Mmol
ftp
JP
AtAzPtAzpp2tAyP3tAsP4
q
AzPi Azp2t AyP2 ty
asp f.pe
yp2tteAgp4
AzPtIA3P7tzn
wgtpilp.lt
satmp3 4lzxio
3atmp_s
6atm
ip2ti.gxo
8xio
oo7xio
p f4
yfugality
Pressure
Sloper
bsr
DSz
z4
cpenftff cpenf.IE
fCpt6.28k4mna k
en 29a8j
DGTDS5fcptb2814motkdenfzfff.fi
tCpenf3z4g8g
zY
0.983141mg
DSz
D
K
509141m01
Whamoe
348.27 146
DStof0477141kmob
WITH
FA
revet
DH 5512 18 Hmd gal
w 1.45 10
5
FT
w
DAH FI
1.45 10514
j9
1 t.ggsk
7.6x104k
n
Fg
7
YoYYg
g
jszj 779
103 m
Cp Rf0.082.17 15415 6210415546.9104157 0 232
Rhombiu
DSzag5
3
0.082.17107 6.215512 69 1081
Sps
0.232
J
t
R
at
368.54
8T3
2T 6.2f sT2 6gxo
0.008ehtiz.mxo
s
39 0
Monoclinic
cpk.pelv9bxw2Lis 48xiosl1sJ2tY2xio8GsJ3
ps
2
351
R i a6 102T
4.8ft 2
2.3559
4.7
3158 7
4 37274mn10
The difference betweenthetwowas t73.4mol.ioThis answerdoes
support that they have zero entropies
ok
not