What is value of △U (heat change at constant volume) for reversible isothermal evaporation of 90 g water at 100oC. Assuming water vapour behaves as an ideal gas and △Hvap)water=540 cal g−1
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.
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
Heat of reaction for C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O(v) at constant pressure is −651 K cal at 17∘ C. What is the heat of reaction at constant volume at 17∘ C?