Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: cseil on August 14, 2015, 12:08:49 PM
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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
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ΔU = ΔH - ΔnRT => ΔU = ΔH.
Why does that make ΔU = ΔH? What is Δn in this context?
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Isn't it 1? Because there were 0 moles of gas water, now there's one.
Edit: ops, it's ΔH - RT then!