# Vapor pressure question in chemical thermodynamics

1. Jan 2, 2009

### Vunde

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. $$\Delta G = \Delta H - T \cdot \Delta S$$ and
$$\Delta G = RT ln \frac{p}{p*}$$

$$p = exp(\frac{\Delta H_{vap}-T \cdot \Delta S_{vap}}{R\cdot T})$$

And for this we know the following data.
$$\Delta H_{vap} = 44,2 \frac{kJ}{mol}$$
$$\Delta S_{vap} = 127 \frac{J}{mol \cdot K}$$

3. The attempt at a solution

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

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

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. $$\frac{p}{p_{ref}}$$

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

2. Jan 2, 2009

### Vunde

My error has been found by myself.. The value I had for vapor entropy was wrong. It was really 134,04 J/mol*K..

But thanks anyway..

3. Jan 2, 2009

### Redbelly98

Staff Emeritus
Glad you worked it out. Welcome to PF!