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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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Answer» 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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