### 第５回

```課題 １

dvB /dx = (5.146－14.29x ) dx
(5.146x－14.29x2 ) dx
＝ 18.079 －(1.802×10-2)× (5.1246/2 x2 －14.29/3 x3)
＝ 18.079 －0.0462x2 + 0.858x3

p [atm]
n [mol]
T [K]
V [m3]
H2
N2
H2 + N2
2.0
2.0
298
3.0
4.0
298
PH2 + PN2 = P
6.0
298
V1
V2
V1 + V2

V1 = (2.0 RT) / (2A) = RT/A [m3]
V2 = (4.0 RT) / (3A) = 4RT/3A [m3]
よって、混合後の体積は V = V1 + V2 = 7RT/3A [m3]

H2, N2の分圧はそれぞれPH2 = (2.0/6.0) P = (6/7)A [Pa], PN2 = (12/7)A [Pa] となる
これをそれぞれ (6/7)B [bar], (12/7)B [bar]と表すと、混合のギブズエネルギーは
ΔmixG = (2.0 RT){ ln (6/7)B －ln 2B} + (4.0 RT){ ln (12/7)B －ln 3B}
= (2.0 RT)×ln (3/7) + (4.0 RT)× ln (4/7)
= 2.0×8.31×298×(－0.847) + 4.0×8.31×298×(－0.560)
= －9.74×103 [J] = －9.7 [kJ]
p [atm]
n [mol]
T [K]
V [m3]
H2
N2
H2 + N2
2.0
2.0
298
2.0
4.0
298
PH2 + PN2 = P
6.0
298
V1
V2
V1 + V2

V1 = (2.0 RT) / (2A) = RT/A [m3]
V2 = (4.0 RT) / (2A) = 2RT/A [m3]
よって、混合後の体積は V = V1 + V2 = 3RT/A [m3]

H2, N2の分圧はそれぞれPH2 = (2.0/6.0) P = (2/3)A [Pa], PN2 = (4/3)A [Pa] となる
これをそれぞれ (2/3)B [bar], (4/3)B [bar]と表すと、混合のギブズエネルギーは
ΔmixG = (2.0 RT){ ln (2/3)B －ln 2B} + (4.0 RT){ ln (4/3)B －ln 2B}
= (2.0 RT)×ln (1/3) + (4.0 RT)× ln (2/3)
= 2.0×8.31×298×(－1.10) + 4.0×8.31×298×(－0.405)
= －9.45×103 [J] = －9.5 [kJ]
```