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 $\beta$.
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2its a mere definition. $\beta$ is just a concise way of writing $1/kT$. – AccidentalFourierTransform Oct 31 '15 at 12:54
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but then why is the exponent $\frac{1}{kT}$? – MarcoXerox Nov 01 '15 at 02:13
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2At 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 01 '15 at 11:03
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1(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 01 '15 at 11:08
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@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 02 '15 at 14:35
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Exactly. You could even try to prove the equivalence yourself! Anyway, remember that $U=\frac{3}{2}RT$ Is only valid for an Ideal Gas, and not true in general (I believe you already know this, but I wanted to say it just in case) – AccidentalFourierTransform Nov 02 '15 at 15:56
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@MarcoXerox, absolute temperature $T$ is defined by Kelvin's construction that relies on 2nd law of thermodynamics. The ideal gas equation is just special model of thermodynamic system, it does not define absolute temperature. It is however common method to measure temperature by measuring volume of a rarified gas. – Ján Lalinský Nov 02 '15 at 15:59
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@qftishard $U = \frac{3}{2}RT$ only applies to monatomic ideal gases right? Thanks for your help! – MarcoXerox Nov 03 '15 at 14:21
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@JánLalinský Let me evaluate and see if I'm wrong. The ideal gas equation only applies to ideal gases. The absolute temperature is defined by the relationship $dE = TdS$ which is a corollary of second law. We define $\beta = \frac{d \ln \Omega}{dE}$ and $\beta = \frac{1}{k_BT}$ is obtained through Boltzmann's assumption. – MarcoXerox Nov 03 '15 at 14:27
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@qftishard, my point was that absolute temperature $T$ is not,in general, defined through the special equation $U=3/2nRT$, which corresponds to ideal gas. This equation and the equation of state $pV=nRT$ is a model of gas behaviour. Although it is true both can be derived from other ideas such as those upon which statistical physics is based, this does not in any way change the fact it is a model. – Ján Lalinský Nov 04 '15 at 16:57
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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.
MarcoXerox
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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 $$P=nRT/V$$.
Hope you understand !
L.K.
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Case is ideal gas equation unfortunately is a result of too many constraints from classical thermodynamics and I'm curious how it can be derived from a modern standpoint. The combined gas law is purely empirical, built up from observations of different scientists centuries ago, and that constant is determined in the same way. I just wonder if there's a modern approach to the problem. – MarcoXerox Nov 04 '15 at 14:18
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Well there would maybe be, the problem is thermodynamics is really statistical and the ideal gas is as it says for ideal gases only which do not really exist. So that this is really an approximation that only work for low pressure and energy. But then you could take a non ideal gas law and arrive to the ideal one by assumption that the energy is low and the pressure is also low. – Nov 04 '15 at 14:29
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Thanks for confirming my assumption. I'd be really interested to see the actual derivation from a non-ideal gas law to the ideal one though. – MarcoXerox Nov 06 '15 at 14:43