# Thermodynamic Derivation of Ideal Gas Law

The ideal gas law is perhaps the best-known equation of state, and admits both a derivation via the kinetic theory of gases and via statistical mechanics. But these are both microscopic theories, and the ideal gas law is a macroscopic equation.

• Can the ideal gas law be derived from macroscopic thermodynamics alone, without recourse to microscopic theories? If so, how? If not, why not?

Assume now that $$U = \frac{3}{2}Nk_\text{B}T,$$ as is the case for a monatomic ideal gas.

• Can we now derive the ideal gas law?

Perhaps something to consider regarding the second question. From thermodynamics, $$P = -\left(\frac{\partial U}{\partial V}\right)_{S,N} = -\frac{3}{2}Nk_\text{B}\left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} \frac{Nk_\text{B}T}{V} \quad \Longleftrightarrow \quad \left(\frac{\partial T}{\partial V}\right)_{S,N} \stackrel{?}{=} -\frac{2}{3}\frac{T}{V},$$ where $\stackrel{?}{=}$ is read "is to be shown equal to." This is an interesting identity, I suppose, but I'm not sure if or where I can proceed from here. As usual, any help is appreciated.

• for the second part, you must interchange differentials $\frac{dU}{dT}=\frac{3}{2}Nk\frac{dV}{dV}=3nR$ , then $dU=\frac{3}{2}nRdT$--> $$U=\frac{3}{2}nRT$$ (within a constant, i suppose). If N=Navogadro, U=$\frac{3}{2}RT$
– user43021
May 9 '17 at 4:37

Classical thermodynamics does not admit that molecules exist thus if the gas law is derived it must be without recourse to any molecular ideas regarding the nature of matter.

The chemical potential of a perfect pure gas at temperature T and pressures p and $p^{\mathrm{o}}$ is defined as $$\mu = \mu^{\mathrm{o}} + RT\ln(p/p^{\mathrm{o}})$$ We can choose $p^{\mathrm{o}}$ to be unity then $$\mu = \mu^{\mathrm{o}} + RT\ln(p)$$ This is still dimensionally correct as $p^{\mathrm{o}} = 1 \pu{~bar}$ and as we have chosen the temperature, $\mu^{\mathrm{o}}$ is the standard chemical potential which is a function of temperature only.

Differentiating the chemical potential at constant T gives $$\left(\frac{\partial \mu}{\partial p} \right)_T = \frac{RT}{p}$$

To find another expression for the derivative we can use $$dG=-SdT+ Vdp +\Sigma\mu_idn_i$$ for $dn_i$ moles of species i gives. For a pure species we can drop the subscript i and replace the summation with $\mu dn$ then $$\left(\frac{\partial \mu}{\partial p} \right)_T=\frac{\partial}{\partial n}\left(\frac{\partial G}{\partial p} \right)_T= V_m$$ where $V_m$ is the volume per mole.

Comparing these two derivatives gives $V_m=RT/p$ or $$pV=nRT$$

Thus the definition of the chemical potential implicitly contains the perfect gas law.

• thank you for the response. In this framework, is it possible to motivate the definition of the chemical potential? May 9 '17 at 14:36
• Actually I find this derivation unsatisfactory as it is somewhat circular in nature, but at present I can't see a way round it. May 9 '17 at 19:59

The answer is almost: starting from the $1^{st}$ law $dU=TdS-pdV$ partial differentiation gives you $$\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial S}{\partial V} \right)_T - p \tag{1}\label{1}$$ and now apply the Maxwell relation to the right side and get $$\left(\frac{\partial U}{\partial V} \right)_T = T\left(\frac{\partial p}{\partial T} \right)_T - p \tag{2}\label{2}$$ Note that for an ideal gas (Joule's law) $U=f(T)$ and thus $$\left(\frac{\partial U}{\partial V} \right)_T = 0 \tag{3}\label{3}$$ so that $T\left(\frac{\partial p}{\partial T} \right)_T - p = 0$ and also $\left(\frac{\partial p}{\partial T} \right)_T = \frac{p}{T}$. This we can integrate to be $$\text{ln} p = \text{ln}T + g(V)$$ or $$ph(V) = T \tag{4}\label{4}$$ for some function $h(V)$ of the volume$V$.

Now to finish we need Boyle's law (this is why the answer is "almost") according to which for the ideal gas the product $pV$ depends only on the temperature. When applied to $(4)$ this gives us $V \propto h(V)$, and that is just the ideal gas equation of state $$pV \propto T \tag{5}\label{5}$$