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Topic: Change in enthalpy of vaporization and normal boiling point  (Read 2501 times)

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Offline shafaifer

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If you get a bunch of data consisting of temperatures with associated pressures then it should be possible to calculate ΔvapH if you plot ln(p) vs temperature-1 because the obtained equation for the slope would be ΔvapH/R, so you should multiply both sides of the equation by the gas constant to get that value. If you would then like to know the normal boiling, couldn't you then set p/p0 = 0 and isolate the temperature?

Offline Plontaj

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Re: Change in enthalpy of vaporization and normal boiling point
« Reply #1 on: June 15, 2015, 02:05:51 PM »
I consider it as analogical to Van't Hoff equation:

dln(P/P0)/d(1/T) = - dHvap/R = slope
dSvap/R = interception

Examples of calculations has been attached in appendix (ODS file).

Offline shafaifer

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Re: Change in enthalpy of vaporization and normal boiling point
« Reply #2 on: June 17, 2015, 11:40:47 AM »
Thank you very much Plantaj, I look forward to open it. I think I suggested something that is wrong, my correction is: you should plot ln(p/p0), where p0 is the reference pressure (often atmospheric pressure), true?

Best regards,

MVS

Offline Plontaj

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Re: Change in enthalpy of vaporization and normal boiling point
« Reply #3 on: June 19, 2015, 10:09:13 AM »
Generally yes. But it must be kept in mind that the equation like this is only approximation (usually good in narrow range of temperature). It has been shown on the picture below on example of calculation pressure constant in function of temperature by similar equation and by comprehensive approach (using Gibbs free energy change calculated from enthalpy and entropy change of reaction variable in respect of temperature).

I'm sorry for polish language.



 

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