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Water: Enthalpy versus Temperature(℃)
Water: CP versus Temperature
6000 J/mol
40,000 J/mol
40 kJ mol-1
微分値定圧熱容量
圧一定なので
�W = �P�V = �P (Vg � Vl) = �RT + 1.013� 105 180.96
10�6
= �8.314� 373 + 1.013� 105 180.96
10�6 = �3099J
�H = �Q = 40 kJ
�S =�Q
T=
40000373
= 107 J K�1
�U = �Q + �W = 40000� 3099 = 36901 J�G = �H � T�S = �H ��Q = 0
水1molが蒸発するときの仕事? ↓水の密度から水の体積
µg(T, P ) = µl(T, P )µg(T + dT, P + dP ) = µl(T + dT, P + dP )
µg(T, P ) +�
�µg
�T
�
P
dT +�
�µg
�P
�
T
dP = µl(T, P ) +�
�µl
�T
�
P
dT +�
�µl
�P
�
T
dP
��G
�T
�
P
= �S,
��(G/n)
�T
�
P
=�
�µ
�T
�
P
= �S̄
��G
�P
�
P
= V,
��(G/n)
�P
�
T
=�
�µ
�T
�
P
= V̄
�S̄gdT + V̄gdP = �S̄ldT + V̄ldP
(V̄g � V̄l)dP = (S̄g � S̄l)dT
dP
dT=
�vapS̄
�vapV̄=
�vapH̄
T�vapV̄
Clapeyron equation クラペイロンの式
Clapeyron equation
�vapV̄ = V̄g � V̄l � V̄g =RT
PdP
dT=
P�vapH̄
RT 2
dP
P=
�vapH̄
R
dT
T 2� P2
P1
P�1dP =�vapH̄
R
� T2
T1
T�2dT
lnP2 � lnR1 =�vapH̄
R
�� 1
T2+
1T1
�
lnP2
P1=
�vapH̄
R
�1T1� 1
T2
�
Clausius-Clapeyron equationクラウジウス・クラペイロンの式
-1 hPa / 10 m0 m 1013 hPa 富士山頂 3776 m 1013 - 378=635 hPa
℃
ln635 ⇥ 1001013 ⇥ 100
= �40 ⇥ 10008.314
�1
373� 1
T1
⇥
0.467 ⇥ 8.31440000
+1
373=
1T1
T1 = 360K = 87
富士山の山頂でお湯は何度で沸く?
ΔHv = 40 kJ mol-1
Vg � Vl ⇥ Vg =nRT
PdP
dT=
(�Htrans/n)PT 2R
dP
P=
dT
T 2
�Hv
R⇤ P2
P1
P�1dP =�Hv
R
⇤ T2
T1
T�2dT
lnP2
P1=
�Hv
R
�� 1
T2+
1T1
⇥
lnP2
P1= ��Hv
R
�1T2� 1
T1
⇥
P1:
P2:
T2 373
T2
T2
--
T1=373 K
エベレストの頂上(8848 m)では?
Vg � Vl ⇥ Vg =nRT
PdP
dT=
(�Htrans/n)PT 2R
dP
P=
dT
T 2
�Hv
R⇤ P2
P1
P�1dP =�Hv
R
⇤ T2
T1
T�2dT
lnP2
P1=
�Hv
R
�� 1
T2+
1T1
⇥
lnP2
P1= ��Hv
R
�1T2� 1
T1
⇥
0 m 1013 hPa
300 hPa (実測値)
P1:
P2: ΔHv = 40 kJ mol-1
T1=373 K
℃
ln3001013
= �400008.314
�1T2� 1
373
�
T2 = 341 K = 68
1 atm ベンゼンの沸点 80℃
ΔHv = 31 kJ mol-1
Vg � Vl ⇥ Vg =nRT
PdP
dT=
(�Htrans/n)PT 2R
dP
P=
dT
T 2
�Hv
R⇤ P2
P1
P�1dP =�Hv
R
⇤ T2
T1
T�2dT
lnP2
P1=
�Hv
R
�� 1
T2+
1T1
⇥
lnP2
P1= ��Hv
R
�1T2� 1
T1
⇥
80+273= 353 K
P2 atm ベンゼンの沸点 25℃25+273= 298 K
(P1, T1)
(P2, T2)
lnP2
1= �31000
8.31
�1
298� 1
353
�= �1.950
exp(ln P2) = P2 = exp(�1.950) = 0.14(2) atom