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Topic: Converting Clausius-Clapeyron Equation  (Read 5361 times)

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Offline Coastie17

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Converting Clausius-Clapeyron Equation
« on: January 26, 2011, 09:11:42 PM »
Not sure if this is a Chemistry or Calculus question but here goes....

The standard form of this equation in natural logarithm form is: ln P = (-â–²Hvap./R)(1/T) + C

Using the conversion factor 2.30 log x = ln x, convert the equation to log base 10 format.

 *I am taking chemistry and calculus concurrently but unfortunately, I have not started the calculus class yet and thus have no idea how to do this. Any advice would be greatly appreciated. Thank you!

Offline opti384

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Re: Converting Clausius-Clapeyron Equation
« Reply #1 on: January 26, 2011, 09:37:45 PM »
Since you have 2.30 log x = ln x, wouldn't ln P= 2.30 log P?

Offline Coastie17

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Re: Converting Clausius-Clapeyron Equation
« Reply #2 on: January 26, 2011, 09:49:20 PM »
Would anything have to be done to the other side of the equation?

*If it is that simple, I give up!  ;D

Offline rabolisk

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Re: Converting Clausius-Clapeyron Equation
« Reply #3 on: January 26, 2011, 10:09:36 PM »
Don't let the logs fool you. This is simple algebra of the form
a = b
a = 2.30c
c = ??? (in terms of b)

Offline Coastie17

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Re: Converting Clausius-Clapeyron Equation
« Reply #4 on: January 26, 2011, 10:43:06 PM »
I plugged in some values from other questions and the answer was the same regardless of whether I used ln P or 2.30 log P.

Thanks for all your assistance!

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