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Topic: Using the Clausius-Clapeyron equation  (Read 28444 times)

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Offline Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #15 on: May 26, 2009, 12:10:10 PM »
I sorta thought that the point of this experiment is to make a curve of pressure as a function of temperature

only to

to determine its heat of vaporization

your own words :)

Is :delta: Hvap the slope value, -4044? Ie, can I substitute :delta: Hvap with -4044?

Twice wrong, but you are very close. Just look at the equation again.
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Offline noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #16 on: May 26, 2009, 12:18:28 PM »
Haha... umm, crap. Ok, we'll what about -4044 / 8.314 as the slope... ie, -486.4?

Offline Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #17 on: May 26, 2009, 12:46:59 PM »
That would be the correct approach - that is, assuming you fitted data correctly.
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Offline noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #18 on: May 26, 2009, 01:12:15 PM »
So could you verify if I'm on the right track still? Using the following equation:

ln P = -  ( delta Hvap / R) T + C

Then I can fill in what I know...

ln P = - (-4044 / 8.314) T + 16.13
ln P = 486.4 T + 16.13

Would this basically be the equation used for solving for pressure or temperature?

Offline Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #19 on: May 26, 2009, 02:08:17 PM »
Not T!
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Offline noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #20 on: May 26, 2009, 03:34:53 PM »
Ok, so let me ask this then. If the boiling point for fluids is 760 mmHg, then I could plug in 760 mmHg for ln P in this equation and get the boiling point? (I'd check on my calculator but I haven't one since I'm sneaking on at work)... that sound right?

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Re: Using the Clausius-Clapeyron equation
« Reply #21 on: May 26, 2009, 03:50:28 PM »
That's how it is supposed to work (assuming you do it right).
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Offline noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #22 on: May 26, 2009, 04:26:00 PM »
Gotcha. Thanks. You're right, I was making that hard. I just really didn't understand what value my slope value was supposed to replace, if that makes sense. Thanks for your help and patience!

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