MSE 201
Problem Set 3
Q1. The isotropic composition of lead in atomic percent is given below. Calculate the
molar configurational entropy of lead.
Atomic Weight
204
206
207
208
Atomic percent
1.5
23.6
22.6
52.3
Q2. A rigid container is divided into two compartments of equal volume by a
partition. One compartment contains one mole of ideal gas A and the other contains 1
mole of ideal gas B, both at 1 atm pressure. Calculate the increase in entropy, which
occurs when the partition between the two compartments is removed. What should be
the increase in entropy if there were two moles of A present in the first compartment
at 1 atm pressure and 1 mole of B in the second compartment at 1 atm presure?
Q3. Gaskell Problem 4.3
Q4.
(a) An ideal gas moves irreversibly from State 1 to State 2 causing an Entropy
change (S) of the ideal gas from S1 to S2. Making use of Maxwells equation(s),
derive an equation for S as a function of temperature and pressure.
(b) Based on the above equation, prove that isentropic (equal entropy)
processes are adiabatic.
Q5. Prove that the difference between molar heat capacities at constant pressure and
at constant volume for a substance is given by
VT 2
CP CV =
where and are thermal expansion coefficient and isothermal compressibility
respectively and are given by
1 " V %
$ '
V # T &P
and
1 " V %
= $ '
V # P &T
Q6. Prove that:
$ U '
if dU = TdS PdV + i dni where, i = &
)
% ni (S,V ,n ji
i
$ A '
$ G '
$ S '
then & )
=& )
= i and & )
= i
T
% ni (T ,V ,n ji % ni (T ,P,n ji
% ni (U,S,n ji
Q7. Which of the following two reactions is more exothermic?
1. C(graphite) + O2 = CO(g) at 1000K
2. C(Diamond) + O2 = CO(g) at 1000K
The heat of formation of diamond from graphite at 298K is 1900J/mol and the molar
heat capacities (in J/mol-K) at constant pressure for graphite and diamond are:
Graphite: 0.11 + 38.94x10-3T 1.48x105 T-2 17.38x10-6T2 {298-1100K}
Diamond: 9.12 + 13.22x10-3T 6.19x105 T-2 {298-1200K}
Q8. Calculate the change in enthalpy and the change in entropy at 1000K for the
reaction CaO(s) + TiO2(s) = CaTiO3(s) given that no reactant or product undergo any
phase change between 298K to 1000K. Use the following data for your calculations:
H298K
S at 298K
(J/mol)
(J/K-mol)
CaO
-634900
38.1
49.62 + 4.51x10-3T 6.95x105 T-2
TiO2
-944000
50.6
75.19 + 1.17x10-3T 18.20x105 T-2
CaTiO3
-1660600
93.7
127.49 + 5.69x10-3T 27.99x105 T-2
CP (J/mol-K)
Q9. An adiabatic vessel contains 1000g of liquid aluminum at 700C. Calculate the
mass of Cr2O3 at room temperature that, when added to the liquid Al (with which it
reacts to form Cr and Al2O3) raises the temperature of resulting mixture of Al2O3,
Cr2O3 and Cr to 1600K. Use the following data:
H298K (J/mol)
CP (J/mol-K)
Al (s)
--
20.67 + 12.38x10-3T {298-934K}
Al (l)
--
31.76 {934-1600K}
Cr (s)
--
24.43 + 9.87x10-3T 3.67x105 T-2 {298-2130K}
Al2O3
-1675700
106.6 + 17.78x10-3T 28.53x105 T-2 {298-2325K}
Cr2O3
-1134700
119.37 + 9.3x10-3T 15.65x105 T-2 {298-1800K}
Atomic weights of Al, Cr and O are 26.98, 52 and 16 respectively.
Q10. The adiabatic flame temperature is that temperature reaches if all of the heat of
an oxidation reaction is used to increase the temperature of the products of the
reaction. Calculate the adiabatic flame temperature attained when methane, at 298K,
is combusted with oxygen in the molar ration O2/CH4 of 2. Assume that CO2 and H2O
are the products of the combustion.
H298K (J/mol)
CP (J/mol-K)
H2O (g)
-241800
30 + 10.71x10-3T + 0.33x105 T-2 {298-2500K}
CO2 (g)
-393500
44.14 + 9.04x10-3T 8.54x105 T-2 {298-2500K}
CH4 (g)
-74800
O2
--3
29.96 + 4.18x10 T 1.67x105 T-2 {298-3000K}
--
Q11. Calculate the Gibbs free energy (G) of the following reaction:
Si3N4 + 3O2 = 3 SiO2 (-quartz) + 2N2
H at 298K
S at 298K
(J/mol)
(J/K-mol)
Si3N4
-744800
113
70.54+98.74x10-3T {298-900K}
SiO2()
-910900
41.5
43.89+1x10-3T6.02x105 T-2 {298-847K}
N2
--
191.5
27.87+4.27x10-3T {298-2500K}
O2
--
205.1
29.96 + 4.18x10-3T 1.67x105 T-2 {298-3000K}
CP (J/mol-K)
