# Gibbs of vaporization of a substance at a non-standard temperature

I'm trying to solve for the gibbs energy of vaporization of ethanol at a temperature lower than its boiling point. I am given the heat capacities of liquid and gaseous ethanol, the standard enthalpy of vaporization, and the standard boiling point of ethanol.

$\ce{dH=(C_{p,g}-C_{p,l})dt}$

Following integration, I get the form:

$\ce{\Delta H=\Delta H_{vap} +\Delta T(C_{p,g}-C_{p,l})}$

This form gives me the new enthalpy of vaporization at a non-standard temperature. However, When I try to enter this into the form:

$\ce{\Delta G=\Delta H - T\Delta S}$, with $\ce{\Delta S=\frac{\Delta H}{T}}$

This solution always results in an answer of zero (equilibrium). Intuitively, I understand that phase changes are typically reversible, however using another representation:

$\ce{\Delta G=-RTln(K)}$, with $\ce{K=\frac{P}{P_{0}}}$, the ratio of vapor pressure to standard pressure of the liquid (liquid activity = 1).

My question is, why do these formulas apparently conflate?

Alright, this is kind of fast but after thinking about it I think the issue with the problem is the definition of $\ce{\Delta S}$. My second try to compute this value involves starting with the form:

$\ce{dS = \frac{C_{p,g}-C_{p,l}}{T}*dT}$

Following integration, I arrive at the form:

$\ce{\Delta S = (C_{p,g}-C_{p,l})ln\frac{T}{T_b}+\frac{\Delta H}{T}}$ where $\ce{\Delta H}$ is determined in the way as in the question above.

Placing everything into the form $\ce{\Delta G = \Delta H-T\Delta S}$:

$\ce{\Delta G = \Delta H_{vap} +(T-T_b)(C_{p,g}-C_{p,l}) - T((C_{p,g}-C_{p,l})*ln\frac{T}{T_b}+\frac{\Delta H}{T})}$ where $\ce{\Delta H}$ is the non-standard enthalpy of vaporization.

• Well, don't thank in posts or ask if they are ok. You can do it in comments, but as one can up/downvote answer or comment it then it could be said that all of them are waiting for review. As for your Q&A, for starters heat capacities don't change linearly. – Mithoron Mar 1 '17 at 22:51