For the process, $\ce{CH3OH(l) -> CH3OH(g)}$ $\Delta G^\circ = 4.30\ \mathrm{kJ/mol}$ at 25 °C. What is the vapor pressure of $\ce{CH3OH(l)}$ at 25 °C in mmHg?
(A) 0.176 mmHg
(B) 14.0 mmHg
(C) 134 mmHg
(D) 759 mmHg

This question is from the 2012 National Chemistry Olympiad (US).

Attempt at solution:

$\Delta G^\circ = -RT\ln K$
$4.3\times10^3\ \mathrm{J/mol}=-8.314\ \mathrm{J/(mol\cdot K)}\cdot 298\ \mathrm{K} \cdot\ln K$
$0.1763\ \text{(unitless)} = K$

Now, I’m supposed to know that $K$ is in units of atm and convert to mmHg accordingly. How would it know that this unitless quantity is atm and not pascal or mmHg?


1 Answer 1


This topic came up the other day but in regards to the evaporation of water. You might find my answer to that question helpful.

Here's how I would solve the problem.

  1. Explicitly note standard state used and assumptions. I will assume:

    • Methanol vapor is an ideal gas
    • Standard states for all species are 1 bar of partial pressure 25 °C. (This assumption matters a lot. The meaning of the $^\circ$ symbol in $\Delta G^\circ$ can't be understood without it.
    • The free energy of methanol liquid does not vary with pressure, i.e. $G_{liq} = G_{liq}^\circ$ at all pressures. I'm fairly confident that this assumption is equivalent to assuming methanol liquid is an incompressible fluid.
  2. Note the equilibrium condition. At equilibrium,

$$\Delta G = 0 = G_{vap} - G_{liq}$$

  1. Substitute the pressure-dependent free energy of an ideal gas

$$ 0 = G_{vap} - G_{liq} = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}$$

  1. Use the assumption of incompressibility of liquid methanol that we made above

$$0 = \left [G_{vap}^\circ + RT \ln{\frac{p}{p^\circ}}\right ] - G_{liq}^\circ$$

  1. Rearrange and substitute

$$0 = \left [G_{vap}^\circ - G_{liq}^\circ \right ] + RT \ln{\frac{p}{p^\circ}}$$

$$0 = \Delta G^\circ + RT \ln{\frac{p}{p^\circ}}$$

$$0 = 4.3 \frac{\textrm{kJ}}{\textrm{mol}} + RT \ln \frac{p}{p^\circ}$$

  1. Solve for $p$

$$\ln{\frac{p}{p^\circ}} = \frac{-4.3 \frac{\textrm{kJ}}{\textrm{mol}}}{RT} $$

$$p = p^\circ e^{\frac{-4300 \textrm{ J/mol}}{RT}}$$

$$p = p^\circ e^{\frac{-4300}{8.314\cdot 298}}= p^\circ e^{-1.73} = 0.176~p^\circ $$

  1. Figure out what $p^\circ$ is. Just like it does for $\Delta G$, adding a $^\circ$ indicates that we are referring to a standard state. Thus $p^\circ$ refers to the partial pressure of methanol vapor at its standard state. This is almost always defined as one atmosphere (or in more modern times, one bar, which is nearly but not exactly the same thing) of pressure. Assuming that the standard state is the usual one, then $p^\circ$ is one bar, which is equal to 750 mmHg. This is what we have assumed in step 1. If that is the case, then $p = 0.176~ p^\circ=0.176\cdot750~\textrm{mmHg}=134~\textrm{mmHg}$.

How would it know that this unitless quantity is atm and not pascal or mmHg?

Hopefully my answer above helps you answer your question. It has to do with the common use of $\circ$ to indicate a standard state of one bar of partial pressure for all gases.

  • $\begingroup$ Thanks for your very comprehensive answer! I adopted a somewhat less rigorous but easy to understand explanation, in that it is p over p standard, or 1 atm (or bar) but you help show how that is so $\endgroup$
    – Andy
    Apr 27, 2015 at 14:18
  • $\begingroup$ Seems familiar from my USNCO days. If you assume methanol is ideal, you can just use fugacity = 1. $\endgroup$
    – Zhe
    Nov 26, 2016 at 19:40
  • $\begingroup$ I'm not sure why you went through all the trouble of #2-5 instead of just using the equation ∆G° = -RT ln K. $\endgroup$ Apr 25, 2017 at 1:10

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