【用戶】高昇賢
【年級】小三上
【評論內容】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) dTln(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K100℃水蒸氣壓 = 1atm = 760torr = 76cmHg = 10332mmH2Oln(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 = ...