1
?/p>
1
?/p>
化学热力学基本定?/p>
1
?/p>
1mol
双原子理想气体在
300 K
?/p>
101 kPa
下,
经恒外压恒温压缩至平衡态,并从此状态下
恒容升温?/p>
370 K
、压强为
1 010 kPa
。求整个过程?/p>
U
?/p>
?/p>
H
?/p>
?/p>
W
?/p>
Q
?/p>
?/p>
(答案:?/p>
U = 1455 J
,△
H = 2037 J
?/p>
W=17727 J
?/p>
Q = -16272 J
?/p>
解:
第一步:恒外压恒温压缩至平衡态,
U
?/p>
=0
?/p>
H
?/p>
=0
V
1
=8.314
×
300/101=24.695dm
3
,
此平衡态的体积就是末态的体积
V
2
?/p>
V
2
=8.314
×
370/1010= 3.046dm
3
此平衡态的压强
P?8.314
×
300/(3.046
×
10
-3
)=818.84kPa
W=-
P?V
2
-V
1
)=-818.92
×
10
3
×
(3.046-24.695)
×
10
-3
=17727 J=17.727 kJ
-Q=W=17.727 kJ
Q=-17.727 kJ
第一步:
因恒?/p>
W=0
U
?/p>
=Q
v
=C
v,m
(T
2
-T
1
)
=20.79
×
(370-300)=1455.3 J=1.455 kJ
H
?/p>
=(20.79+R)
×
70=2037.3 J=2.037 kJ
整个过程?/p>
W=17.727 kJ
?/p>
Q= -17.727+1.455= -16.27 kJ
?/p>
U
?/p>
=1.455 kJ
?/p>
H
?/p>
=2.037 kJ
?/p>
2
?/p>
设有
0.1 kg N
2
?/p>
温度?/p>
273.15 K
?/p>
压强?/p>
101325 Pa
?/p>
分别进行下列过程?/p>
?/p>
U
?/p>
?/p>
H
?/p>
?
Q
?/p>
W
?/p>
?/p>
(1)
恒容加热至压强为
151987.5 Pa
?/p>
(2)
恒压膨胀至原体积?/p>
2
倍;
(3)
恒温可逆膨胀至原体积?/p>
2
倍;
(4)
绝热可逆膨胀至原体积?/p>
2
倍?/p>
?/p>
(答案:
①△
U = Q
V
= 1.01
×
10
4
J
,△
H = 1.42
×
10
4
J
?/p>
W = 0
?/p>
②△
H = Q
P
= 28.4 kJ
,△
U = 20.20 kJ
?/p>
W= -8.11 kJ
?/p>
?/p>
Q = 5622 J
?/p>
W = -5622 J
,△
H =
?/p>
U = 0 J
?/p>
?/p>
Q = 0
?/p>
W =
?/p>
U = -4911 J
,△
H = - 6875 J
?/p>
解:
?/p>
N
2
气视为双原子理想气体,则
C
p,m
=29.10 J
·
mol
-1
·
K
-1
;
C
v,m
=20.79 J
·
mol
-1
·
K
-1
(1)
W=0,
末态温?/p>
T
2
=1.5T
1
=1.5
×
273.15 K
?/p>
U
?/p>
=Q
v
=n C
v
(T
2
-T
1
) =(100/28)
×
20.79
×
(1.5
×
273.15-273.15)=1.01
×
10
4
J