The molar enthalpy change for `H_(2)O(l)hArrH_(2)O(g)` at 373 K and 1 atm ius 41 kJ/mol. Assume ideal behaviour, the internal energy change for vaporization of 1 mol of water at 373 K and 1 atm in KJ `"mol"^(-1)` is :

The DeltaH for vaporization of a liquid is 20kJ/mol. Assuming ideal behaviour, the change in internal energy for the vaporization of 1 mol of the liquid at 60^(@)C and 1 bar is close to -

Standard enthalpy of vaporization of water at 1 atm pressure and 100°C is 40.66 kJ mol^(-1) . The internal energy change of vaporization of 2 moles of water at 100°C (in kJ) is

18g of water is takento prepare the tea. Find out the internal energy of vaporisation at 100^(@) C. (Delta_(vap)H for water at 373 K 373 K is 40.66kJ mol^(-1))

If water vapour is assumed to be a perfect gas, molar enthalpy change for vapouration of 1 mole of water at 1 bar and 100^(@)C is 41 kJ mol^(-1) . Calculate the internal energy change when a) 1 mol of water is vapourised at 1 bar and 100^(@)C b) 1 mol of water liquid is converted into ice.

Calculatte the molal elevation constant of water if molar enthalpy of vapourisation of water at 373K is 40.585 kJ mol^(-1) .

If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of 1 mol of water at 1 bar and 100^(@)C is 41 kJ mol^(-1) .Calculate the internal energy change, when (a) 1 mol of water is vaporised at 1 bar pressure and 100^(@) C. (b) 1 mol of water is converted into ice.