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課題　１

課題　１. 例題 5 ・ 1 と同様に、　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 塩 B では、 V B = v B = 6.128 + 5.146 x － 7.147 x 2 より、 d v B / dx = (5.146 － 14.29 x ) dx ( 5.146 x － 14.29 x 2 ) dx ＝　 18.079 － (1.802×10 -2 )× (5.1246/2 x 2 － 14.29/3 x 3 ) ＝ 18.079 － 0.0462 x 2 + 0.858 x 3. 課題　２. H 2.

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課題　１

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  1. 課題　１

  2. 例題5・1と同様に、　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　例題5・1と同様に、　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　 塩Bでは、VB= vB=6.128 + 5.146 x－7.147 x2 より、 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

  3. 課題　２

  4. H2 N2 H2 + N2 p [atm] 2.0 3.0 PH2 + PN2 = P n [mol]2.0 4.0 6.0 T[K] 298 298 298 V [m3] V1V2V1 + V2 混合前のH2, N2の圧力を 2A, 3A [Pa] とおくと、pV = nRTより、 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] 混合後の全圧は P = (6.0 RT) / (7RT/3A) = (18/7)A [Pa] となり、 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]

  5. H2 N2 H2 + N2 p [atm] 2.0 2.0 PH2 + PN2 = P n [mol]2.0 4.0 6.0 T[K] 298 298 298 V [m3] V1V2V1 + V2 混合前のH2, N2の圧力を 2A, 2A [Pa] とおくと、pV = nRTより、 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] 混合後の全圧は P = (6.0 RT) / (3RT/A) = 2A [Pa] となり、 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]

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