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(解答) 式 (6.12) Δp = ( ΔH / ΔV )×ln ( T f / T i )

(解答) 式 (6.12) Δp = ( ΔH / ΔV )×ln ( T f / T i ) ここで、 Δ trans H = 0.368×10 3 [J mol -1 ] 、 ΔV = 6.957×10 -6 [m 3 mol -1 ] T i = 95.5 + 273.2 = 368.7 [K], T f = 100 + 273.2 = 373.2 [K] より、   2 従って、 Δp = (0.368×10 3 ) / (6.957×10 -6 ) ×ln(373.2/368.7) = 6.416×10 5 [Pa]

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(解答) 式 (6.12) Δp = ( ΔH / ΔV )×ln ( T f / T i )

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  1. (解答) 式(6.12) Δp = (ΔH / ΔV )×ln (Tf / Ti) ここで、 ΔtransH = 0.368×103 [J mol-1]、 ΔV = 6.957×10-6 [m3 mol-1] Ti = 95.5 + 273.2 = 368.7 [K], Tf= 100 + 273.2 = 373.2 [K] より、   2 従って、Δp = (0.368×103) / (6.957×10-6) ×ln(373.2/368.7) = 6.416×105 [Pa] 必要な圧力は (1.01 + 6.42)×105 [Pa] = 7.43×105 [Pa] = 7.4 [atm] 問題6.19の結果とよく一致している

  2. 解答 クラウジウス・クラペイロン式が適用できるのは、気相と他の相の変化である 従って、上で適用できるのは、(a), (b), (h) である

  3. 解答 クライジウス・クラペイロン式 1-ブタノールの25℃における蒸気圧 p, 標準沸点ではpo (=1 atm) ln (p / po) = -(45.90×103) / 8.314 ×{(1/ (273.2+25) -1/ (273.2+117.2)} = -4.372    よって p / po = e-4.372 = 1.262 ×10-2 [-] p = 1.26×10-2 [atm]  (= 1.27 [kPa] = 9.59 [mmHg]) 他の物質についても同様に計算し、結果をまとめると表のようになる。 よって蒸気圧の順序は t-BuOH > 2-BuOH > i-BuOH > 1-BuOH となり、 蒸発エンタルピー、沸点の低い順と一致する。

  4. 解答 (a) 2つ (斜方晶、単斜晶)   (b) 斜方晶 (298 K, 1atm) (c) 斜方晶→単斜晶→液体→気体    の順に相変化する

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