Calculating Vapor Pressure from Delta G/ Kp
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.


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 &deg;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.

Note the equilibrium condition.  At equilibrium, 


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


Substitute the pressure-dependent free energy of an ideal gas


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


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


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


Rearrange and substitute


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

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

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


Solve for $p$


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

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

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


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.