2mol的水蒸气在100°C，101325Pa下变为水，求Q,W,ΔH。已知水的汽化热为2258J

过程为恒温恒压可逆过程,则：
Q = △H = 2 × (-2258) × 18 = -81288 J
W = -p外?[V(l) - V(g)] ≈ p?V(g) = nRT
= 2 × 8.314 × 373 = 6202.2 J