[madisonwi]

Hello all -

Using the Clausius-Clapeyron equation for determining enthalpy, I'm getting confused on the mathematical labeling for enthalpy.  To review:

ln P_vapor = -(change)H_vapor / R*T


In order to determine the units for -(change)H_vapor, I solved for it:

ln P * R * T = -(change)H_vapor

Inputing units for the left side to determine -(change)H_vapor using R = 0.082 (atm L / mol K):

ln atm * (atm L / mol K) * K

Now, since -(change)H_vapor is a specific heat, I know I can end with atm*L/mol.  The problem is that I end with atm²*L/mol since ln P is still atm in units.  

So, I'm thinking that ln P doesn't keep it's unit.  In that case, the units work out.  My question, then, is why does a natural log not keep its units.

[Demotivator]

The log is simply an exponent of some base. An exponent is dimensionless. The original number's units, however, are implicitly retained. So when taking antilog of the dimensionless exponent, the unit is reintroduced explicitly.
Like the pH for example.