For water moI`triangle _(vap) H = 41 kj mol^(-1)` at 373 Kand 1 bar pressure. Assuming that water vapour is an ideal gas fhat occupies a much larger volume than liquid water, the internal energy change during evaporation of water is __________kJ `mol^(-1)`.

Assuming that water vapour is an ideal gas, the internal energy change (DeltaU) when 1 mol of water is vapourised at vapourisation of water at 1 bar and 373K=41 kJmol^-1 and R=8.3 Jmol^-1K^-1 will be

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

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

DeltaH (vap) for water is 40.7 kJ mol^(-1) . The entropy of vapourisation of water is

Water of mass m=1kg and M (mol mass) =18 turns completely into saturated vapour at standard atmospheric pressure. Assuming the saturated vapour to be an ideal gas find increment of internal energy of the system. Specific latent heat of steam is L=2550kJ//kg

90 g of water spilled out from a vessel in the room on the floor. Assuming that water vapour behaving as an ideal gas, calculate the internal energy change when the spilled water undergoes complete evaporation at 100^(@)C . (Given the molar enthalpy of vaporisation of water at 1 bar and 373 K = 41 kJ mol^(-1) ).

The molar enthalpy change for H_(2)O(l)hArrH_(2)O(g) at 373 K and 1 atm is 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 :

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))