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### Topic: Work, dU and dH for vaporization of water  (Read 3929 times)

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#### cseil

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##### Work, dU and dH for vaporization of water
« on: August 14, 2015, 12:08:49 PM »
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:

$$W=-p \int_{V1}^{V2} dV$$

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

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##### Re: Work, dU and dH for vaporization of water
« Reply #1 on: August 14, 2015, 12:32:47 PM »
Quote
ΔU = ΔH - ΔnRT => ΔU = ΔH.
Why does that make ΔU = ΔH? What is Δn in this context?

#### cseil

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##### Re: Work, dU and dH for vaporization of water
« Reply #2 on: August 14, 2015, 12:34:41 PM »
Isn't it 1? Because there were 0 moles of gas water, now there's one.

Edit: ops, it's ΔH - RT then!