Even seasoned professionals disagree on this one. Trialogue on the number of fundamental constants by M. J. Duff, L. B. Okun, G. Veneziano, 2002:

This paper consists of three separate articles on the number of
fundamental dimensionful constants in physics. We started our debate
in summer 1992 on the terrace of the famous CERN cafeteria. In the
summer of 2001 we returned to the subject to find that our views still
diverged and decided to explain our current positions. LBO develops
the traditional approach with three constants, GV argues in favor of
at most two (within superstring theory), while MJD advocates zero.

Okun's thesis is that 3 units (e.g. $c$, $\hbar$, $G$) are necessary for measurements to be meaningful. This is in part a semantic argument.

Veneziano says that 2 units are necessary: action $\hbar$ and some mass $m_{fund}$ in QFT+GR; or a length $\lambda_s$ and time $c$ in string theory; and no more than 2 in M-theory although he's not sure.

Finally, Duff says there is no need for units at all, all quantities are fundamentally subject to some symmetry, and units are merely conventions for measurement.

This is a very fun paper and answers your question thoroughly.

I don't think that you can assume that $I_c=\ln (p_0)$

A lot of equations have logs of dimensioned quantities in them. It's usually not hard to get rid of the log:

$$
\mathrm{ln}\left(\frac{p_1}{p_2}\right) = I_{c,1}-I_{c,2} - \left(\frac{\Delta H^{\circ}_{v,1}}{RT_1}-\frac{\Delta H^{\circ}_{v,2}}{RT_2}\right) + \mathrm{ln}\left(\frac{T_1^{\frac{\left(C^{g}_{p,1} - C_{p,1}^{l}\right)}{R}}}{T_2^{\frac{\left(C^{g}_{p,2} - C_{p,2}^{l}\right)}{R}}}\right)
$$

It may also be that $I_c$ is just a gas-specific constant, of the form $$I_c=\ln p_0-\frac{\Delta H^{\circ}_{v,0}}{R} -\frac{\left(C^{g}_{p,0} - C_{p}^{l,0}\right)}{RT_0}\mathrm{ln}\left(T_0\right) $$

for some $p_0,T_0$.

You can do the same for the other equations as well. When there's a log of a dimensioned quantity, one of two things is possible:

- The equation has the units implicitly assumed (this is generally not done in physics, but you see it a lot in chemistry)
- The equation is a change equation, and you need to subtract one copy of the equation at a different point from it.

The latter form arises when one takes an indefinite integral in the last step while solving a differential equation (Which is why subtracting is the right way to fix this, since that corresponds to tacking limits onto the integral)

## Best Answer

There is no discrepancy in the value of the

differenceof the logs $$ \log(V_1)-\log(V_2) $$ and the value of the log of theratio$$ \log{V_1/V_2}\;. $$ And neither of these depends on the choice of units. This is because, for both the difference and the ratio, the units cancel. For example, suppose I represent the "units" as a number "U" such that $$ V=\alpha U\;, $$ where $\alpha$ is unitless.Then $$ \log(V_1)-\log(V_2)=\log(\alpha_1 U)-\log(\alpha_2 U)=\log(\alpha_1)+\log(U)-\log(\alpha_2)-\log(U) $$ $$ =\log(\alpha_1)-\log(\alpha_2)\;. $$

And, similarly, $$ \log(V_1/V_2)=\log(\frac{\alpha_1 U}{\alpha_2 U})=\log(\frac{\alpha_1}{\alpha_2})\;. $$

Not much else to say but that... Clearly, trying to interpret $\log(V_1)$ on its own doesn't make sense since changing the unit changes the numerical value... but the interpretation and sensibleness of the

differenceof the logs certainly makes sense and is independent of the choice of units.