(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Calculate the vapor pressure of iso-propyle alcohol over iso-propyle alcohol in pure liquid state at 298,15K.

2. Relevant equations

A possible equation to solve this problem, I think, would be to use this formula put together by two definitions of free energy. [tex]\Delta G = \Delta H - T \cdot \Delta S[/tex] and

[tex]\Delta G = RT ln \frac{p}{p*}[/tex]

[tex]p = exp(\frac{\Delta H_{vap}-T \cdot \Delta S_{vap}}{R\cdot T})[/tex]

And for this we know the following data.

[tex]\Delta H_{vap} = 44,2 \frac{kJ}{mol}

[/tex]

[tex]\Delta S_{vap} = 127 \frac{J}{mol \cdot K}

[/tex]

3. The attempt at a solution

If we put in the data we get the following solution.

[tex]p = exp(\frac{44200 \frac{J}{mol}-298,15K \cdot 127 \frac{J}{mol \cdot K}}{8,315\frac{J}{mol \cdot K}\cdot 298,15K}=0,078 atm[/tex]

The unit of the vapor pressure is given in atm because you always compare a result to its predefined standard. In this case 1 atm. [tex]\frac{p}{p_{ref}}[/tex]

The result is therefore 0,078 atm, however in my solutions manual for the coursework question, the solution should be 0,18 atm. And this is my problem, I simply can't get to that solution no matter how many times I twist and turn it.

Sorry if the language is a bit off. If you don't understand my question or some of the midway solutions I come to, please do ask and I will respond as soon as possible.

Nikolaj Rahmberg

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Vapor pressure question in chemical thermodynamics

**Physics Forums | Science Articles, Homework Help, Discussion**