Change in energy for vaporization

[Pengwuino]

Ok I got this problem...

1.10 moles of H2O(g) at 1.00 atm and 100. ºC occupies a volume of 33.7 L. When 1.10 moles of H2O(g) is condensed to 1.10 moles of H2O(l) at 1.00 atm and 100. ºC, 44.73 kJ of heat is released. If the density of H2O(l) at this temperature and pressure is 0.996 g/cm3, calculate ?E for the condensation of 1.10 moles of water at 1.00 atm and 100. ºC.

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?

 

[Gokul43201]

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

Is that what you used ?

 

[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).

 

[Gokul43201]

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

 

[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

 

[Gokul43201]

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

 

[Pengwuino]

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