The normal boiling point of water is 373 k. vapour of waterr at temperature T is 19 mm hg. If enthalpy of vapoursed is 40.67 kJ/mol, them temperature Twould be
(use : log 2 = 0.3, R : 8.3` Jk^(-1)mol^(-1)`):

The normal boiling point of water is 373 k. Vapour pressure of water at temperature T is 19 mm hg. If enthalpy of vapourization is 40.67 kJ/mol, then temperature T would be (use : log 2 = 0.3, R : 8.3 Jk^(-1)mol^(-1) ):

The normal boiling point of water is 373K . Vapour pressure of water at temperature T is 19 mmHg . If enthalpy of vaporization is 40.67 kJ//"mol" , then temperature T would be

Normal boiling point of water is 373 K . Vapour pressure of water at 298 K is 23 mm enthalpy of vaporisation is 40.656 kJ mol^(-1) if atmopheric pressure becomes 23 mm , the water will boil at:

The normal boiling point of watr is 373 K (at 760 mm), vapour pressure of water at 298 K is 23 mm. If entyalpy of vaporization is 40.656 kj//mol.the boiling point of water at 23 mm pressure will be :

When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be (DeltaH_("vap") (for water) =40.8kJ"mol"^(-1) )