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Topic: Solving for Hvap when using the Clausius Clapeyron equation  (Read 12138 times)

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Offline geode

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Solving for Hvap when using the Clausius Clapeyron equation
« on: September 19, 2009, 11:03:29 PM »
Hi Everyone,

This is my first time posting.  I did a quick search before I posted this question, so thanks in advance for your help.

In reviewing a problem in my textbook, I'm having trouble understanding how they were able to isolate the "Hvap" in order for solve for it.  I know (in most instances)how to algebraically solve equations, but this one is very confusing and I just can't follow how they did it.  Also compiled with that is I'm having trouble preventing my calculator from automatically rounding up my answers - which doesn't help when trying to get the answer that the textbook has.  Here's the problem:

Calculate the heat of vaporization of diethyl ether, C4H10O, form the following vapor pressures: 400 mmHg at 18 degrees celsius and 760 mmHg at 35 degrees celsius.

The solution is: 

ln 760 mmHg/400 mmHg = (change)Hvap/8.31 J/k*mol) x (1/291K-1/308K)
                        0.642 = (2.28 x 10e-5) x (change)Hvap/(J/mol)
             (change)Hvap = 2.82 x 10e4 J/mol

When I attempted this on paper, I multiplied by the reciprocal : 8.31 J/k*mol/(change)Hvap on both sides.  This left me with 8.31 j/k*mol/(change)Hvap x 760 mmHg/400 mmHg  = 1/291K - 1/308K.  I left off the "ln" because in searching for the answer in this forum, I read a post where someone said that you only use the "ln" when solving for the pressure and since I had both pressures, I omitted the "ln".  Once I had the above, it still didn't look correct because I still hadn't isolated the Hvap.  If I went any further, I would end up cancelling out the Hvap and that wouldn't be correct.

I'm not sure what to do at this point.  Any help would be appreciated.  And if anyone knows how to stop my calculator from rounding up, I'd appreciate that as well.  I have a Casio fx-300ms.


Offline raezur

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Re: Solving for Hvap when using the Clausius Clapeyron equation
« Reply #1 on: September 21, 2009, 04:22:11 PM »
You would not eliminate the "ln". 
Solving for Hvap involves rearranging (or deriving) the equation to
the following:

Hvap = ((ln p2/p1) X r)/(1/T1 - 1/T2)

Then plug in the values and solve:

Hvap = (ln 760/400 x 8.314 J/k*mol)/(1/291.15K - 1/308.15K)
Hvap = (5.336 J/mol /1.895 x 10^-4)
Hvap = 2.816 x 10^4 J/mol or 2.812 x 10^-7 kJ/mol

Hope this helps!

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