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課題 1.

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Presentation on theme: "課題 1."— Presentation transcript:

1 課題 1

2 例題5・1と同様に、                                 塩Bでは、VB = vB=  x-7.147 x2 より、 dvB /dx = (5.146-14.29x ) dx (5.146x-14.29x2 ) dx =  -(1.802×10-2)× (5.1246/2 x2 -14.29/3 x3) = -0.0462x x3

3 課題 2

4 H2 N2 H2 + N2 p [atm] PH2 + PN2 = P n [mol] T [K] V [m3] V V V1 + 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] PH2 + PN2 = P n [mol] T [K] V [m3] V V V1 + 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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