Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Ice-cream on May 30, 2005, 06:36:52 AM
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can any1 give me some help on this question:
What pressure would have to be applied to steam at 300 degrees C to condense it into liquid water (delta H(vap) = 40.7 kJ/mol)?
Ans. 97.6 atm
I really have no idea wat to do, i can't even start it!! i don't even know which formulae would be used...any1 got any ideas?
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You need to take the clasius clapyron equation, and take it from there.
use the fact that at boiling point, the pressure of the vapor is 1 atmosphere.
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Grafter
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If i remember correctly, i answered one of your earlier questions using the clausius clapeyron eqn. Very similar question.
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Just looked back - yes, very similar question.
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ahhhh ic ic, thanks guys
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the clausius-clayperon equation contains the perfect gas assumption, which is normally valid below 5bars. Personally, I prefer a more real answer, so I would look up the steam table (ie. Thermodynamic and Transport Properrties of Fluids by GFC Rogers & YR Mayhew)
P1 = 85bar, Ts1 = 299.2C
P2 = 90bar, Ts2 = 303.3C
(Ts: Saturated Temperature)
a more real value for the required pressure can be extracted by interpolation of the above data:
P = P1 + (300-299.2)/(303.3-299.2) x (P2 - P1) = 85.97 ~ 86bar
I wouldn't treat steam as a perfect gas because water is polar molecule that exhibits hydrogen bonding. moreover, the required pressure to condense steam at 300K is clearly above 5bars.
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pretty good. i thought of the table, but couldnt lay my hands on it. this will, of course, give a more accurate answer.