What is the change in the internal energy when 2 moles of liquid water vaporises at 100°C? The heat of vaporisation, ΔH vap. of water at 100°C is 40.66KJmol-1.
106ΔU = ΔH - PΔV
= 40660*2 - PV ... (Vinitial ≈ 0)
= 81320 - 1.01*105*(373*44.8/273)*10-3 ... (find volume by using ideal gas equation)
= 75138 J (approx)
11Asish work is done by the gas for expansion therefore pdv is negative So you have to subtract it.
1Hey guys,
As there is no change in temperature((delta)T=0), so change in internal energy is zero as internal energy si a function of temperature.
All the heat is used up for change of state frm liquid to gas.
106@ ADITYA
what u r talking abt is valid for gaseous processes only.. i.e. internal energy is directly proportional to temperature only when it is gaseous... here there is a transition from liquid to gaseous so it is not valid... u hav to find as done...
