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課題　１

課題　１. 【 解答 】 例題 4 ･ 1 と同様に、　 Δ μ 　 ＝ V m Δ p = MΔp / ρ を用いて計算する M = 44.0 [g mol -1 ] = 44.0×10 -3 [kg mol -1 ] Δ p = 1.00 [bar] = 1.00×10 5 [Pa] 　である ρ (l) = 2.35 [g cm -3 ] = 2.35×10 3 [kg m -3 ] より、 Δ μ (l) = (44.0×10 -3 )×( 1.00×10 5 ) / ( 2.35×10 3 )

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課題　１

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  1. 課題　１

  2. 【解答】 例題4･1と同様に、　 Δμ　 ＝ VmΔp=MΔp/ ρ を用いて計算する M= 44.0 [g mol-1] = 44.0×10-3[kg mol-1] Δp= 1.00 [bar] = 1.00×105 [Pa]　である ρ(l)= 2.35 [g cm-3] = 2.35×103[kg m-3]より、 Δμ(l)= (44.0×10-3)×(1.00×105) / (2.35×103) = 1.872 = 1.87 [J mol-1] ρ(s)= 2.50 [g cm-3] = 2.50×103[kg m-3]　より、 Δμ(s)= (44.0×10-3)×(1.00×105) / (2.50×103) = 1.76 [J mol-1] 従って、Δμ(l) > Δμ(s) となり、固体の方が化学ポテンシャルが低いので、 液体が固体に転移（凝固）する。

  3. 課題　２

  4. 【解答】 数値例4･1と同様に、　 　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　を用いて計算する Vm = M / ρ(l) ベンゼン(C6H6):M = (12.01×6 + 1.01)×6 = 78.12 [g mol-1] = 78.12×10-3 [kg mol-1] ρ(l) = 0.879 [g cm-3] = 879 [kg m-3] ΔP = 100 [bar] = 1.00×107 [Pa] R = 8.314 [J mol-1 K-1] T = 25 + 273.2 = 298.2 [K] より、 M ×ΔP = ρ(l)×R ×T (78.12×10-3) ×(1.00×107) == 0.3584 [-] 879×8.314 × 298.2 よって p = p*× e0.3584 = 1.431 p*となり、蒸気圧は 43% 増加する。

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