For the reversible isothermal evaporation of 90.0 g of water at 100∘C. The value of ΔU for the reaction is:
Determine the value of ∆H and ∆U for the reversible isothermal evaporation of 90.0g of water at 100°C. Assume that water vapour behave as am ideal gas and heat of eveporation of water is 540cal.
MY PROBLEM
Only tell me How ∆H is calculated here, because if I know∆H then I can use the equation
∆H = ∆U + nRT
In my book it is given that
∆H = 90.0 x 540
= 48600 cal
But I want to know that why they have multipled the mass to heat of eveporation to find out ∆H