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If notions of thermodynamics emerge from statistical mechanics, is there a way to derive the ideal gas law?

And this derivation would be more "fundamental"?

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Yes, there are several. Check the last section of this Wikipedia page, where it is derived from the equipartition theorem. Another way would be through the partition function and computing quantities associated with it, such as internal energy, pressure, etc.

The derivation from statistical mechanics is more fundamental inasmuch as you start with fundamental laws, not by phenomenological ones as it's done in thermodynamics.

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A derivation I enjoy for how fundamental it is but didn't find anywhere else is based on combinatorics:

Say there are N particles which can be placed in a volume-lattice. The volume-lattice has cubic cells of side $a$. There are $M= \frac{V}{a^3}$ cells. Since we may assume particles are point-like and the cells are not, we can place particles into cells which are already occupied. Therefore, there are $W=M^N$ ways to spread $N$ distinguishable particles over $M$ cells.

Free energy is given by: $$F = const - T S$$ Pressure is given by: $$p = - \frac{\partial F}{\partial V}$$ Entropy is given by: $$S = k_B \ln(W) = k_B \ln(M)*N = k_B N (\ln V-3 \ln a)$$ Putting it all together: $$p = - \frac{\partial F}{\partial V} = - \frac{\partial}{\partial V} const + T k_B N \frac{\partial}{\partial V}(\ln V) - T k_B N \frac{\partial}{\partial V}(3 \ln a)$$ Only terms containing $V$ survive, resulting in: $$p = T k_B N \frac{\partial}{\partial V}(\ln V) = T k_B \frac{N}{V}$$

I believe this to be the most fundamental derivation, since it makes no assumption on particle properties, momentum, grid layout or even dimension. One could just as easily derive "one-dimensional pressure" this way.


Technicality: I assumed particles may be distinguished. Usually, they are assumed to be indistinguishable. This would result in $W=\frac{M^N}{N!}$. However, the derivation wouldn't change, since this would just add another $V$-independent term "$-\ln(N!)$". This factor is called Gibbs factor, resolving Gibbs' mixing paradox.

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