Why can't we use Boltzmann's ideal gas law using temperature in Celsius (intuition)? [closed]

Mathematically, i can see why we won't get the same results when inserting a temperature in Celsius rather than Kelvin(because K=C+273.15), but i can't understand how can a law not "work" for any man made unit of measure of a physical quantity. I mean, both say 20C and 293.15K say the same thing about the state of the system, so why can't we use whatever units we like in the law? Nature does not care what units we made up in order to measure her physical quantities, so what is going on here?

EDIT: To clarify, i am asking if there is a way, a kind of transformation to the law in order for us to use it in Celsius

closed as off-topic by Norbert Schuch, TheQuantumMan, user36790, Martin, GertJan 29 '16 at 23:26

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• While 20C and 293.15K correspond to the same temperature, that does not mean that you can plug "20" blindly into equations rather than "293.15". The absolute temperature (K) is directly related to kinetic energy. Some other scale with an offset in it is clearly NOT directly related to much of anything. – Jon Custer Jan 28 '16 at 22:32
• @JonCuster i did not say to just plug in "20", but surely there must be a way to "transform" the law in order for us to use Celsius also – TheQuantumMan Jan 28 '16 at 22:34
• Yes, use (273.15 + degrees Celsius) rather than (K). That is how they relate, after all... – Jon Custer Jan 28 '16 at 22:35

It should be obvious that the ideal gas law $$pV=nRT$$ is not invariant under the transformation $T\mapsto T+T_0$. But that is exactly what the relation between Kelvin and Celsius is - they are not, like most other choices of units, a different scaling, but they choose a different zero point of temperature, so you may not use them interchangably in formulae that take their specific form only for one choice of zero point (i.e. you may only use them interchangably in equations invariant under $T\mapsto T+T_0$, such as those only involving temperature differences).
Well, you could do it in any arbitrary units if you choose. Let's look at the perfect gas law: $P = \rho R T$. I can pick any consistent set of units that I want there, but it may involve offsets. For example, let's keep $P$ in Pascals, $\rho$ in kilograms per unit volume, $R$ in joules per kilogram per kelvin and let's use $T$ in Celsius. We know that $T_k = T_c + 273.15$ so we can just plug that in and get $P = \rho R (T_c + 273.15)$.