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Author Topic: Using the Clausius-Clapeyron equation  (Read 26022 times)

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noiseordinance

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Using the Clausius-Clapeyron equation
« on: May 23, 2009, 08:01:32 AM »

Good morning (or afternoon depending on where you are). We did an experiment a couple days ago that involved heating ethanol to determine its heat of vaporization. We collected data for both temperature (in Celsius) as well as pressure (in kPa). We monitored between 85 degrees C to 35 degrees C. We ended up with a couple values that I'm having a hard time stringing together. From the CRC, I know that the Heat of Vaporization for Ethanol is:

38.56 kJ/mol @ 78.3 degrees C
42.32 kJ/mol @ 25.0 degrees C

As for our data from our first trial, we go:

m (slope) : -4044
b (y-intercept) : 16.13
Correlation : -0.9986
RMSE : 0.02940

I realize that Correlation and RMSE are both values for accuracy that just simply need to be reported. However, I'm having a hard time fitting m (slope) and b (y-intercept) into the Clausius-Clapeyron equation:

ln P = - (:delta:Hvap/RT) + C

I don't know which values convert to which. It looks like "C" in the Clausius-Clapeyron equation is related to b, the y-intercept. I'd also imagine that R is the gas constant, 8.314 J mol−1K−1. What I'm not sure is where I'm supposed to fit in the slope. I am also guessing that since our data was originally collected in C, we'll need to convert to K since we're using the gas constant, but I don't know which data to convert to K. Argh.

Anyways, from this data, I need to know how to report relationships between vapor pressure and temp, the significance of the slope, and I'm also supposed to be able to calculate the heat of vaporization for ethanol and the boiling point.

Could anyone lend some pointers?
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #1 on: May 23, 2009, 12:45:26 PM »

I guess to ask more specific a question, do I have enough data to determine the boiling point and heat of vaporization or am I mission something?

From what it sounds like, my slope and intercept alone are supposed to be enough to determine both values. Afterwards, I'm supposed to compare my boiling point and heat of vaporization to the CRC values.

So I'm attempting to start with the empty equation:

ln P = -  :delta: Hvap / RT + C

Then I fill in what I know how to fill in...

ln P = -  :delta: Hvap / (8.314 J mol−1K−1) T + 16.13

This leaves me with ln P,  :delta: Hvap and temperature. Problem is, I only have my slope value, which I haven't filled in cause I don't know where it's supposed to go, or even what it means. Perhaps it's my  :delta: Hvap value in joules? If this is the case, I could plug in T in Kelvin and end up with pressure, and plug in pressure and get T... but how does that tell me the heat of vaporization? Or am I all messed up?
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #2 on: May 24, 2009, 06:00:46 AM »

I'm going to guess that this thread totally doesn't make sense... I could use any pointers anyone may have to offer.
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Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #3 on: May 24, 2009, 06:12:16 AM »

ln P = -  :delta: Hvap / RT + C

ln P = -  ( :delta: Hvap / R) T-1 + C = m T-1 + b

However, you don't know what you are doing, yet you have fitted something to something. What to what?
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #4 on: May 24, 2009, 06:23:03 AM »

ln P = - ( :delta: Hvap / RT ) + C is the equation that my teacher gave us to use. Are you saying that this is not a valid equation?
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Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #5 on: May 24, 2009, 07:00:56 AM »

This is the same equation, just written in different form.
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #6 on: May 24, 2009, 07:15:22 AM »

I realize that. I wanted to make sure that the original equation that I provided was written in the exact way it was specified by the instructor. Yes, this is the same equation as I originally wrote. That said, is it a valid equation?
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Borek

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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #8 on: May 24, 2009, 07:41:59 AM »

Thanks. Unfortunately I've looked at all the Clausius-Clapeyron equations I could find on the internet, as well as the material in my text book. Unfortunately, none that I could find involved integrating the experimental slopes and intercepts to determine heats of vaporization and boiling points...
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Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #9 on: May 24, 2009, 08:37:57 AM »

Wiki page lists the C-C equation in the form identical to the one you were given.

Find out what m&b stand for just by comparing coefficients side by side in the equation I have listed earlier.
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #10 on: May 24, 2009, 05:32:48 PM »

Gosh, I still really don't know what you mean. I'd like to think I'm not 100% dense though. Is there a more plain english way that this can be explained?
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Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #11 on: May 24, 2009, 08:51:45 PM »

ln P = -  ( :delta: Hvap / R) T-1 + C = m T-1 + b

Can you see what m & b must be equal to for the equation to be correct?

I suppose you are making it harder than it is in reality. It is pretty obvious.
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #12 on: May 25, 2009, 01:03:58 PM »

I don't know what that means... Considering I was top of the class for the majority of my math classes, I'm not feeling too dumb about this. It really isn't that obvious to me. So ln P = m T-1 + b, then ln P = -4044 T-1 + 16.13... that still leaves two unknown variables, which in algebra, can be difficult to figure out. I really don't know. You're right, I'm making this too hard, and with the holiday weekend, there hasn't been any tutor center, and the instructor has not emailed me back. After fighting this problem for 3 days, I'm really starting to get discouraged.
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Borek

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Re: Using the Clausius-Clapeyron equation
« Reply #13 on: May 25, 2009, 09:15:54 PM »

As far as I can guess P & T are not unknowns.

From the equations you wrote so far C=16.13... can you tell what  :delta: Hvap/R is equal to?

Trick is, you use m&b values from SOME fitting - but it is not clear what you have fitted to what. Seems to me you did it not having a slightest idea about the reason behind the operation.
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noiseordinance

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Re: Using the Clausius-Clapeyron equation
« Reply #14 on: May 26, 2009, 05:30:21 AM »

I sorta thought that the point of this experiment is to make a curve of pressure as a function of temperature. So the temperature would be the x-axis and the kPa would be the y-axis, and since the y-intercept is 16.13, that would mean maybe that at 0 degrees C, the pressure is 16.13 kPa? I dunno, I could have this all wrong.

So maybe I'm totally wrong, and I'll probably be totally wrong asking the next question... Is :delta: Hvap the slope value, -4044? Ie, can I substitute :delta: Hvap with -4044?

I appreciate your help.
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