Derivation of ideal gas law

I looked up on the ideal gas law which our high school textbook derives with the empirical Combined Gas Law. However, the textbook did give a good explanation for this equation $$pV = \frac{N}{3}m\bar{v^2}$$ with which I only need to verify that $$K.E. = \frac{3}{2}k_BT$$ is true. I further looked up this link Average Molecular Kinetic Energy which deduces the result from the Boltzmann distribution $$f(E)=Ae^{-\beta \epsilon}$$ but I could not read any literature deriving $$\beta = \frac{1}{kT}$$ I was wondering if I am in a correct direction and how to derive the thermodynamic beta $\beta$.

• its a mere definition. $\beta$ is just a concise way of writing $1/kT$. – AccidentalFourierTransform Oct 31 '15 at 12:54
• but then why is the exponent $\frac{1}{kT}$? – MarcoXerox Nov 1 '15 at 2:13
• At some point in statical physics we must define what we mean by temperature. The modern approach is just to take $f_{MB}(E)\equiv Z^{-1} \mathrm e^{-E/kT}$ (where $Z=1/A$ in your notation). This means: we define temperature as the number $T$ that appears in the Boltzmann distribution (also, note that you have the formula wrong: the exponent is $\beta E$ instead of $\beta T$). The former approach (in the beginings of Thermodynamics) is to define temperature through the Ideal Gas Law, that is, we take $pV=nRT$ as an axiomatic rule that has (1/2) – AccidentalFourierTransform Nov 1 '15 at 11:03
• (2/2) to be obeyed by all gases (at sufficiently dim pressures). From this definition, it is not difficult to prove that $\beta=1/kT$. In any case, one way or another, we have to explicitly say what $T$ is. All definitions are equivalent, but the easier (and more theoretically meaningful) is to take the exponent in the Boltzmann distribution to be $-E/kT$ by definition. – AccidentalFourierTransform Nov 1 '15 at 11:08
• @qftishard thanks for reminding me of the wrong formula. So to sum up, in classical thermodynamics temperature is defined through average kinetic energy as $U = \frac{3}{2}RT$ but in statistical approach we define $T$ as a part of the exponent in Boltzmann distribution, and by using this newer definition we can derive back the classical definition? – MarcoXerox Nov 2 '15 at 14:35

The ideal gas law is a combination of many other laws about gases. Some assume the pressure to be constant, others assume the quantity stays constant and others. Now those laws have been set up mostly after experiment and it people working on it noticed that the pressure $P$ according to the small laws seemed to be proportional to the quantity $n$ (in moles), to the temperature $T$ (in kelvins) and inversely proportional to the volume. Know when we find such proportionality, we always need to multiply by a proportionality constant (here they called it R), that could be different than one and that would obviously nkt change the proportions. That proportionality constant, they have found its value by experiment and putting all this together gives the familiar $$P=nRT/V$$.
In classical thermodynamics, temperature $T$ is defined through ideal gas equation $$pV = nRT$$ from which we conclude that $$K.E. = \frac{3}{2}k_BT$$ is true for any ideal monatomic gas which cannot exist in real life anyways. Statistical mechanics provides postulates that is broader in context. It redefines the temperature through the second law $$dE = TdS$$ Now from $$p_i = \frac{e^{-\beta \epsilon_i}}{Z}$$ obviously we could obtain $$\beta = \frac{d \ln \Omega}{dE}$$ and by Boltzmann's assumption $$S = k_b \ln \Omega$$ we could have $$\beta = \frac{1}{k_BT}$$ so everything boils down to the definition of absolute temperature.