# Homework Help: Change in energy for vaporization

1. Oct 16, 2005

### Pengwuino

Ok I got this problem...

Now I figured Vi = 33.7L and to find Vf, i used the density to determine what volume that 1.10 moles of water would need...

Vf = (1.1mol * (2(1.0079) + 15.999)g/mol) * (1 cm^3/0.996g) * (1L/1x10^6 cm^3) = 1.878 x 10^-5 L

Using w = -P delta V or w = -P(Vf-Vi)

I got...

w= -(1.0atm)(1.878 x 10^-5) - 33.7) * 101.3J/(L*atm) = 3413.81J

delta E = 3413.81 J + 44730 J = 48.143kJ

But supposedly I'm wrong... Anything wrong in my reasoning?

2. Oct 16, 2005

### Gokul43201

Staff Emeritus
Haven't checked the math but the enthalpy is defined by dH = dE + PdV.

Is that what you used ?

3. Oct 16, 2005

### Pengwuino

Yes. It gave me the q value already, was just looken for the work.

I'm not ruling out the book being wrong because it's been wrong on a lot of questions (and this is after talking to various other grad students and professors about some of the questions).

4. Oct 16, 2005

### Gokul43201

Staff Emeritus
q or dH should be a negative number, since heat is released.

5. Oct 16, 2005

### Pengwuino

Well i THINK i started plugging in every combonation of -q or +q or -w or +w and still got nothing... but i'll check again afterwards.

Come to think of it that may very well be my problem, i'll check it once i get back home

6. Oct 16, 2005

### Gokul43201

Staff Emeritus
In fact w = + PdV, but in the final expression, you want E = q - w

7. Oct 16, 2005

### Pengwuino

Oops, i guess I did overlook that little nugget. Got it right after re-checking :D

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