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Change in energy for vaporization

  1. Oct 16, 2005 #1

    Pengwuino

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    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. jcsd
  3. Oct 16, 2005 #2

    Gokul43201

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

    Is that what you used ?
     
  4. Oct 16, 2005 #3

    Pengwuino

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    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).
     
  5. Oct 16, 2005 #4

    Gokul43201

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    q or dH should be a negative number, since heat is released.
     
  6. Oct 16, 2005 #5

    Pengwuino

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    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
     
  7. Oct 16, 2005 #6

    Gokul43201

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    In fact w = + PdV, but in the final expression, you want E = q - w
     
  8. Oct 16, 2005 #7

    Pengwuino

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