50、水的常壓沸點為100 oC,水的蒸發熱為40.67 kJ mol-1,假設..-阿摩線上測驗
1F 高昇賢 小三上 (2017/02/27)
用 Clausius–Clapeyron equation 解 dP/dT = PL/(T2R) (1/P)dP = (L/R)(1/T2)dT ∫P2P1 (1/P) dP = (L/R)∫T2T1 (1/T2) dT ln(P2/P1) = (L/R)〔(1/T1) -(1/T2)〕,L為潛熱(蒸發熱)J/mol、R為理想氣體常數8.314 J/(mol·K)、T為溫度K 100℃水蒸氣壓 = 1atm = 760torr = 76cmHg = 10332mmH2O ln(P150℃/1) = (40.67×1000/8.314)〔(1/373) -(1/423)〕(蒸氣壓單位:atm) ln(P150℃/760) = (40.67×1000/8.314)〔(1/373) -(1/423)〕(蒸氣壓單位:torr) ln(P150℃/76) = (40.67×1000/8.314)〔(1/373) -(1/423)〕(蒸氣壓單位:cmHg) ln(P150℃/10332) = (40.67×1000/8.314)〔(1/373) -(1/423)〕(蒸氣壓單位:mmH2O) P150℃ = 4.712atm = 3581torr = ... 查看完整內容 |