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甲醇在337.65K的饱和蒸气压p*=101.325kPa,在此温度压力下,甲醇的摩尔蒸发焓△vapHm=35.32kJmol-1。求在337.65K、101.325kPa下,2mol气态甲醇全部凝结成液态甲醇时的Q、W、△U、△H、△S、△A和△G。(12分)

发布时间：2025-10-29 11:37:02

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答案：解:甲醇在337.65K的饱和蒸气压p*=101.325kPa,在此温度压力下,气态甲醇凝结成液态甲醇为可逆相变。n=2mol ,摩尔凝结焓为△Hm =-△vapHm=-35.32kJmol-1。△H=n*△Hm=-2*35.32kJ=-70.64kJ (2分)可逆相变热为恒压热,即Q=△H=-2*35.32kJ=-70.64 kJ (1分)W = -P*(Vl-Vg)≈-P*(-Vg)=P*(nRT/p)=nRT=2*8.315*337.65J = 5.618kJ (2分)△U = Q W = (-70.64 5.618) kJ = -65.02 kJ (2分)可逆相变△S = △H/T = {-70.64/337.65} kJ·K-1 = -0.2091 kJ·K-1 (2分)△A = △U- T*△S = △U - △H = -65.02 kJ-(-70.64kJ) = 5.62KJ (2分)可逆相变,由吉布斯判据△G = 0 (1分)