What is `Delta U` when `2.0` mole of liquid water vaporises at `100^(@)C`? The heat of vaporisation `(Delta H_("vap".))` of water at `100^(@)C` is `40.66 KJmol^(-1)`.

What is DeltaU when 2.0 mole of liquid water vaporise at 100^(@)C ? The heat of vaporistaion , DeltaH vap. of water at 100^(@)C is 40.66 kJ mol^(-1) .

Calculate w and DeltaU for the conversion of 1 mol of water at 100^(@)C to steam at 1 atm pressure.Heat of vaporisation of water at 100^(@)C is 40.670 kJ mol^(-1) .Assume ideal gas behaviour.

Work done for converson of 0.5 mole of water of 100^(@)C to steam at 1 atm pressure is (heat of vaporisation of water at 100^(@)C is 4070J"mol"^(-1) )

If Delta H_("vap") of pure water at 100^(@)C is 40.627 kJ mol^(-1) . The value of Delta S_("vap") is

Assuming that, water vapour is an ideal gas the internal energy (Delta U) when 1 mol of water is vapourised at 1 bar pressure and 100^(@)C (given : molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol^(-1) and R = 8.3 J K^(-1) mol^(-1) )) will be

Delta H (vap) for water is 40.7 kJ mol^(-1) . The entropy of vaporisation of water is :