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)`.

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 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) .

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) )

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