H2O(l)⟶100∘CH2O(g) Δvap Hs=Δvap Es+ΔngRT For the above reaction, Δng=np−nr=1−0=1∴40.66 kJ mol−1=ΔvapEs+1×8.314×10−3×373ΔvapEs=40.66 kJ mol−1−3.1 kJ mol−1=+37.56 kJ mol−1
Standard enthalpy of vaporisation ΔvapHs for water at 100∘C is 40.66 kJ mol−1. The internal energy of vaporisation of water at 100∘C (in kJmol−1) is (Assume water vapour to behave like an ideal gas).
Held on 30 Apr 2012 · Verified 9 Jul 2026.
+37.56
-43.76
+43.76
+40.66
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