南昌大学2014上半年材料热力学重点题
————————————————————————————————作者：————————————————————————————————日期：
solution:)
/(81.6810ln 314.877.45277.6282.4)
/(152940)()/(67.4977.45277.6282.4)()
/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlN
N Al a -=+-?-?=?=?-=-?-?=?=?=+
Solution ：
)
/(173104190059210222mol cal H O
H Cu H O Cu =-=?+=+,exothermic reaction
For the reaction :
)()(2
1
2g H g H → 035
.3312
1
314.823212,)(,-=?-?=-=?H P g H P p C C C
J
C H dT C H H P o
P o
o 2128241702035.3217990)2982000(2982000
298
2982000=?-=-??+?=?+?=??
J
C H dT C H H P o
P o
o 2128241702035.3217990)
2982000(2982000
298
2982000=?-=-??+?=?+?=??
J C S dT C S S P o
P o
o 57.43298
2000
ln
035.335.492982000
ln
2982000298
2982000=?-=??+?=?+?=??
J
S T H G 12568457.432000212824020000200002000=?-=?-?=?
56
.72000
314.8125684
ln
125684ln
ln ln
ln )
(2
/1)
(2
/1)
(2
/12/1)(2000
2
2
2
2=?=
→==-=-=?g H H g H H g H H H g H o P P P P RT P P RT P P RT K RT G
atm
P P P P P H g H H g H g H 0005.0562.71ln
1,1)
(2)(2)(2=→=≈=+
Solution: (a) When the equilibrium is reached,
0P ln 2
1
ln 2=?=+?=?O o o RT G J RT G G －
RT T P O 2
1)06.1539850(18.4ln 2+-?=
T = 500?C = 773K
atm
P P O O 262
21014.169
.36773314.82
1
)
77306.1539850(18.4ln -?=-=???+-?=
(a) at T=300?C=573K,
Although the equilibrium P O2 is very low, kinetically the reaction is
not favoured and reaction speed is very slow. So 300?C is not suitable at
At T=800?C=1073K, lnP O2 =-22.2, P O2 =2.28?10-10 atm. At 800?C, if the equilibrium is reached, nitrogen can be of high purity level. However, at this high temperature , particles of Cu will weld together to reduce effective work surface. So it is not suitable to use this high temperature in purification either.
(c ) The equilibrium oxygen pressure remains the same when the total pressure increases, which means a higher purity level of N 2 .
Solution: N 2 =2N, H 2 = 2H
[]21
221
,N N a P K N = ， [],21
221
,H H a P K H =
For N2 dissolving :
For H2 dissolving
：2/1'2'
2/12
)(][][H H
P H P H =
（a ）For dissolving N2, P N2 = 1 atm, [N]=35cm 3/100g melt,
melt g cm P N P N N N 100/75.24)5.0(35][)(][32/12
/12
2/1'2=?==‘
similarly: [H]’ =24.75cm 3
/100g melt
total gas : [H]'＋[N]' = 49.5 cm 3/100g melt (b) [H]' =24.75 cm 3/100g melt (c ) [H]'＋[N]' =
2/1'2'
2/12
)(][][N N P N P N = [N](0.33)1/2 /1+[H](0.33)1/2
/1=20.10+20.10 = 40.2cm 3/100g melt
Solution ： （
1
）
0.2
H
? （2）
G
?=max W （3）
56KJ
.696}{0.21P P P P RTln
G KTlnk G G 2
3
O OH CH CO 20H 2322=+?=+?=?）（）
（）（）
（θ
θ
Solution:(a)mol
J
F Z -6252E
G =-=?θθ
(b)0.370
a ln RT G Cd
==? (c))(3.42P P
P
Cd
mmHg ==
θ
Solution:(b)
Pa
P P g Cl g Cl 21
)()(10
*86.8-RTln G 125.4T -605000G 22-==?=?θθ(c)-2.485V
ZF
-G
E =?=
答案更正为：
Temperat ure Phase Composit
ion
Fraction
1300 Liquid 60 61.5
α8 38.5
β99 0 1000+ Liquid 70 50.8
α9 49.2
β98 0 1000- Liquid _ 0
α7 63.7
β98 36.3
Solution:
)
/
(
363
10
2.
207
2
1
]
10
8.4
)
25
327
(
3.
29
[
2
1
2
1
)
(
2
3
3
2
2
s
m
V
v
n
n
W
Q
nMv
mv
W
H
T
C
n
Q
Q
Q
absorb
melting
p
melt
increase
absorb
=
?
=
?
+
-
?
=
=
=
?
+
?
=
+
=
-
Solution
)
/
(
245
60
20
8.9
75
)
/
(
121
60
60
24
10467000
/
/
)
(
10467000
1868
.4
10
2500
sin
3
S
J
t
h
mg
P
S
J
t
Q
t
W
P
J Q
g
increa
Burning
Burning
=
?
?
=
?
=
=
?
?
=
=
=
=
?
?
=
Solution：
)
(6.
436
)
10
6
10
3(
10
75
.
72
)
(
10
6
)
10
5.0(
4
)
10
5.0(
3
4
)
10
1(
2
3
2
5
2
3
2
6
3
6
3
1
J
S
W
m nS
S
Single
total
=
?
-
?
?
?
=
?
=
?
=
?
?
?
?
?
?
?
?
=
=
-
+
-
-
-
-
σ
π
π
Solution
)
(
25
.
6
)
(
746
60
10
4
273
90
)
25
90
(
2
4
m
S
W
t
W
P
St
Q
T
T
T
W
H
H
L
H
=
=
=
?
?
+
-
=
-
=
Solution:
)
(
643
746
25
.0
20
35
20
273
4
3
75
.0
W
P
P
T
T
T
P
Q
T
T
T
W
L
L
L
L
H
H
H
L
H
=
?
?
+
-
?
=
-
=
-
=
solution:
1
,2,2
,1,2
12
,2
,2
,2,21
,1,1
,1,198.82527352527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=
-=
.
,)(6286.0)
(1,2,not is b ok is a c P P b H H =
Solution:
mol
J
P P RT G water
O H ice O H /9.1089523.0ln 268314.8163
.3012
.3ln
)5273(314.8ln
,,22-=??=-?==?
mol J H /1085.53?=?
)
/(23.22268
)
9.108(5850K mol J T G H S S
T H G ?=--=?-?=?∴?-?=?
Solution: (a) At 0?C, ?G =0, ∴ T m ?S = ?H
)./(09.22273
6030
K mol J T H S m ==?=
? (b) At 0?C, ?G =0 ?
)
./(62.37)./(1818.45.0)./(5.0,K mol J K mol J K g cal C ice P =??==
./(1818.40.1)./(0.1,K
mol J K g cal C water P ??==a reversible process can be designed as follows to do the calculation:
mol
J H
dT C C dT
C H dT C H H H H
water P ice p water p ice p fu
/9.584160305)24.7562.37()(273
268
,,268
273
,273
268
,)3()2()1(=+?-=?+-=
+?+=?+?+?=??
??
（d ）
)
./(39.2109.22268
273
ln )24.7562.37()
(3273268
,,268
273
,273
268
,)
3()2()1()4(K mol J S
dT T
C C dT
T
C S dT T
C S S S water P ice p water
p ice
p =+?-=?+-=+?+=?+?+?=??
?
?
38.10939.212689.5841)4()4()4(=?-=?-?=?S T H G
Ice
wate
water
ice, （1
（2
（3
（4
Solution: (a) diamond graphite C C =
mol
kJ H H H o
graphite f o diamond f /1897,,=?-?=?
)
./(36.338.274.5,,K mol J S S S o graphite f o diamond f -=+-=?-?=?
mol
J S T H G /28.2898)36.3(2981897=-?-=?-?=?
(b) No, diamond is not thermodynamically stable relative to graphite at 298K. (c )
mol
J P V G diamand /29.34101309951
.310126
=???=?=?-
(c ) Assuming N atm , ?G = 0, reversible processes as following can be designed to realize this,
（4
graphite,
diamond,
)
(14939028.2898194.028.28981013051
.3101225.2101228.289810130)1)(V V ()(V 28.2898V 6
6)
3()2()1()4(atm N N N N P P G G G G diamond graphite diamond graphite ==+-=+????
? ?
??-?-=+?--=?-++?????--＝＋＋＝
,0,0'
'
=?=?=??
?dT T
C dT C C T T
p T T p p
mol J H H /1897298900=?=?
K mol J S S ./36.3298900-=?=?
mol J S T H G /492136.39001897=?+=?-?=?