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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.

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    $\begingroup$ Hi. It's called an ideal gas, not a perfect gas. $\endgroup$
    – Gert
    Commented Nov 2, 2020 at 17:52
  • $\begingroup$ @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 $PV=nRT$. $\endgroup$
    – TheVal
    Commented Nov 2, 2020 at 17:55
  • $\begingroup$ 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. $\endgroup$ Commented Nov 3, 2020 at 15:13
  • $\begingroup$ @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). $\endgroup$
    – TheVal
    Commented Nov 3, 2020 at 15:48
  • $\begingroup$ @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. $\endgroup$
    – TheVal
    Commented Nov 3, 2020 at 15:52

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