Dimensional Analysis : Thermodynamics I was coming across some notes online for phase transitions. In one of the places, the author has written the Claussius-Clayperon equation in this form,
$$ \frac{d(ln P)}{d(ln T)} = \frac{T\Delta S}{P\Delta V} $$
I am wondering if this equation is valid, since both P and T have dimensions and quantities inside log or exp cannot be dimensional quantities (since they are a power series). 
 A: Unlike the exponential, the logarithm function can actually be defined on physical quantities with a dimension, it was discussed here. Not by power series, but that's not "the function" but just one often-useful way of calculating it! Mathematically, the preferred definition is via differential equations or integral identities.
Even if you avoid the "logarithm of kilometre", this differential makes sense independently. Let's write it explicitly:
$$
  \frac{\mathrm{d} \ln P}{\mathrm{d} \ln T}
    \equiv \lim_{\Delta l\to 0}
         \frac{ \ln (P|_{\ln T = l_0}) - \ln(P|_{\ln T = l_0 + \Delta l}) }
              { \Delta l }
$$
Now introduce some arbitrary reference pressure $P_0$.
$$\begin{aligned}
 \ldots =& \lim_{\Delta l\to 0}
         \frac{ \ln (P|_{l_0} \cdot \tfrac{P_0}{P_0}) - \ln(P|_{l_0 + \Delta l} \cdot \tfrac{P_0}{P_0}) }
              { \Delta l }
  \\=& \lim_{\Delta l\to 0}
         \frac{ \ln ( \tfrac{P|_{l_0}}{P_0}) + \ln(P_0) - \ln( \cdot \tfrac{P|_{l_0 + \Delta l}}{P_0}) - \ln(P_0) }
              { \Delta l }
  \\=& \lim_{\Delta l\to 0}
         \frac{ \ln ( \tfrac{P|_{l_0}}{P_0}) - \ln( \tfrac{P|_{l_0 + \Delta l}}{P_0}) }
              { \Delta l }
  \\=& \frac{\mathrm{d} (\ln \tfrac{P}{P_0})}{\mathrm{d} T}
         \cdot (\tfrac{\mathrm{d} \ln T}{\mathrm{d}T} )^{-1}.
\end{aligned}$$
Basically, as long as you only talk about differences / differentials of logs, the dimensions reduce to irrelevant, because cancelling, prefactors.
