# A peculiar thermodynamic relation for perfect gas

I was preparing lecture notes for an undergrad thermodynamics course, and (due to fiddling with some equations) I stumbled upon a peculiar thermodynamic relation that I couldn't find anywhere else. I have the feeling I'm rediscovering hot water.

The relation is as follows, and holds for a generic reversible process of a perfect gas: $$\frac{\Omega_1}{V_1\cdot T_1^{1/(\gamma-1)}}= \frac{\Omega_2}{V_2\cdot T_2^{1/(\gamma-1)}}=\mathrm{constant}$$ where $$\Omega$$ is the number of microstates, and $$V$$, $$T$$ the thermodynamic volume and temperature respectively, and $$\gamma=C_P/C_V$$ is the heat capacity ratio. I've found this equation to be very useful for computing the relative change (ratio $$\Omega_2/\Omega_1$$) of microstates for a generic process.

This relation is interesting, for instance if the microstates do not change, $$\Omega_1=\Omega_2$$, then the relation becomes: $$V_1\cdot T_1^{1/(\gamma-1)}= V_2\cdot T_2^{1/(\gamma-1)}$$ which describes an adiabatic process! This is consistent, because $$\delta Q=T\cdot dS=0$$, then if $$dS=0$$ the number of microstates remains constant, due to the Boltzmann relation: $$dS=k_B\cdot d\Omega/\Omega=0$$.

## Question(s)

Is this relation correct? Is it known (are there any sources), or useful? What is the constant quantity? Does it have any physical meaning?

## Derivation

We consider the Boltzmann definition of the Entropy between two generic states $$\Delta S = S_2-S_1 = R\cdot\ln\frac{\Omega_2}{\Omega_1}$$ where here $$R=k_BN_A$$ is the universal gas constant, such that $$\Delta S$$ has dimensions J/mol/K (molar terms) rather than J/K (atomic terms). Using the value for the entropy change for a generic process on a perfect gas $$\Delta S = \frac{R}{\gamma-1}\ln\frac{T_2}{T_1}+R\ln\frac{V_2}{V_1}$$ and the previous equation combined we have: $$\ln\frac{\Omega_2}{\Omega_1}= \frac{1}{\gamma-1}\ln\frac{T_2}{T_1}+\ln\frac{V_2}{V_1}$$ which translates directly to the aforementioned relation by removing the logarithms and bringing all the terms referring to each state on the two sides of the equality.

• Hi. It's called an ideal gas, not a perfect gas.
– Gert
Commented Nov 2, 2020 at 17:52
• @Gert Hi. I'm using the "engineering" definition of what you consider an ideal gas. In our sense an ideal gas (or ideal fluid) is a gas (fluid) which has equal and non-negligible interaction forces between all its constituents, regardless of the atom type. In that case the EoS is $PV=ZnRT$. In the "engineering" definition, a perfect gas is a gas which has negligible interaction energy between all its constituents with respect to the total kinetic energy; which in turn translates to the EoS $PV=nRT$. Commented Nov 2, 2020 at 17:55
• In an ideal gas, the constituent particles have ABSOLUTELY NO interactions and zero volume (among other things). The EoS $pV = ZnRT$ is for a compressible gas, not an ideal gas. Distorting the well-established conventions for the definitions and introducing yet another term ("perfect" gas) only causes confusion. Commented Nov 3, 2020 at 15:13
• @JeffreyJWeimer I'm sorry but I flunked. All the "ideal" part refers to ideal mixtures, not pure fluids (in that case the compressibility factor $Z$ of the mixture). While it is true that perfect gas and ideal gas can be used somewhat interchangeably, I don't see that in the "established" IUPAC gold book for instance; which nowhere mentions about interactions (goldbook.iupac.org/terms/view/I02935). In fact, what you say is perfectly fine for deriving the Ideal (or Perfect) Gas Law, even from statmech assumptions ($V(q)=0$ in the hamiltonian of the system). Commented Nov 3, 2020 at 15:48
• @JeffreyJWeimer However, since an ideal gas does not exist (interactions are always present between particles), our engineering definition of a perfect gas, states more "realistically" that the $V(q)$ term is negligible to $T(p)$, meaning that the hamiltonian of the system can be approximated to $T(p)$, as if $V(q)$ was zero. But again, this is an approximation since in real life $V(q)$ literally becomes negligible with respect to $T(p)$, and we can apply the model of the Ideal (or Perfect) Gas Law, or EoS. Commented Nov 3, 2020 at 15:52