[cseil]

Hi,
I'd like to have a confirm about how I solved an exercise.

It asks me to find the work, the variation of H and U during the vaporization of 1 mol of water at 1atm and 100°C. I know that the adsorbed heat during the process (p is constant) is 9706 cal/mol and the book gives me the density of water at 100°C (0.958g/mL).

I firstly calculated the work. The pressure is constant so it's equal to:

[tex]W=-p \int_{V1}^{V2} dV[/tex]

where V1 is the volume of the liquid water (18.78x10^-3 L) and V2 is the volume of the gas water (30.59L). The work is -740cal.

Talking about the enthalpy, ΔH = q at constant pressure so I've already got the value.
ΔU = ΔH - ΔnRT => ΔU = ΔH.

Is it right?
Thank you

[mjc123]

ΔU = ΔH - ΔnRT => ΔU = ΔH.

Why does that make ΔU = ΔH? What is Δn in this context?

[cseil]

Isn't it 1? Because there were 0 moles of gas water, now there's one.

Edit: ops, it's ΔH - RT then!