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

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Solution

∆H = heat of evaporation×amount evaporated

= 540×90

= 48600 cal (C)
Hope you know the rest of problem.
Good luck.

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∆H = heat of evaporation×amount evaporated = 540×90 = 48600 cal (C) Hope you know the rest of problem. Good luck.

∆H = heat of evaporation×amount evaporated

= 540×90

= 48600 cal (C)
Hope you know the rest of problem.
Good luck.