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In a reversible adiabatic process, $PV^\gamma=\text{const.}$ but reversible adiabat $\iff$ isentrope so does that mean $PV^\gamma$ can be treated as a function for entropy?

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No. In an isothermal process, $T$ is constant, but (for an ideal gas) isothermal processes don't change the internal energy $U$, but of course $U\neq T$, right? You should be able to tell right away that $PV^\gamma \neq S$ - dimensional analysis would be a good start.

The entropy of an ideal gas can be explicitly calculated. It turns out to be

$$S = N k_B \left[\log\left(\frac{VT^{\frac{3}{2}}}{N\Phi}\right)+\frac{5}{2}\right]$$

where $\Phi$ is a constant. You can see that the change in entropy between two states $(P_1,V_1,T_1)\rightarrow (P_2,V_2,T_2)$ is given by

$$\Delta S = S_2-S_1 = Nk_B\log\left[\frac{V_2}{V_1}\left(\frac{T_2}{T_1}\right)^\frac{3}{2}\right]$$ and since $T = \frac{PV}{Nk_B}$,

$$\Delta S = Nk_B \log\left[\frac{V_2}{V_1} \left(\frac{P_2 V_2}{P_1 V_1}\right)^\frac{3}{2}\right] = Nk_B \log\left[\left(\frac{P_2}{P_1}\right)^\frac{3}{2}\left(\frac{V_2}{V_1}\right)^\frac{5}{2}\right]$$

$$ = \frac{3}{2}Nk_B \log \left[\frac{P_2 V_2^\gamma}{P_1 V_1^\gamma}\right]$$

where $\gamma = \frac{5}{3}$. Therefore, for an ideal gas with constant $N$, isentropic processes are precisely those for which the combination $PV^\gamma$ doesn't change, as expected.


Upon further reflection, I don't want to say that entropy is non-ambiguous. Though this is reasonably accurate from the more fundamental POV of statistical mechanics, in classical thermodynamics we define entropy based on the differential

$$dE = \frac{\delta Q}{T}$$

which means that the zero point of the entropy is not fixed.

That being said, there are a number of reasons why entropy cannot be proportional to $PV^\gamma$. Most immediately in my mind, the entropy of a system should be extensive, which means that if you put two identical copies of the system together, the result should have double the entropy. You can see right away that something proportional to $PV^\gamma$ does not satisfy this requirement.

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  • $\begingroup$ Right, so although it could act as an entropy function, it is not the one we use i.e. any monotonic scaling of the entropy would still preserve its order and such. Am I right? In this case, the monotonic scaling is the exponential $\endgroup$ – Aakash Lakshmanan Dec 19 '18 at 7:16
  • $\begingroup$ I am motivating this by the fact that there are multiple acceptable empirical temperature functions but only thermodynamic temperature is nice so PVgamma would be an "empirical entropy" in some sense $\endgroup$ – Aakash Lakshmanan Dec 19 '18 at 7:18
  • $\begingroup$ I don't quite know what you mean by that. Neither the entropy nor the temperature of a thermodynamic system are ambiguous quantities. $\endgroup$ – J. Murray Dec 19 '18 at 7:21
  • $\begingroup$ @J.Murray your answer seems to prove exactly what you deny to be true: your last equation is saying that the difference of entropy can be obtained as difference of $\frac{3}{2}N k_B log(P V^{\gamma})$ at two states. Even more direct, if you feed the expression for $T$ coming from the equation of state in the first expression for $S$ you arrive to the same conclusion. The real reason $PV^{\gamma}$ cannot be used as a proxy for entropy is different. $\endgroup$ – GiorgioP Dec 19 '18 at 7:37
  • $\begingroup$ @AakashLakshmanan Please see my edit for additional clarification. $\endgroup$ – J. Murray Dec 19 '18 at 7:40
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$PV^{\gamma}$ cannot be used in place on entropy as it is, for reasons of physical dimensions as well as the lack of extensiveness, but no one of these reasons is really compelling. Units could be adjusted by introducing a suitable constant with dimensions and one could look for a function of $PV^{\gamma}$ containing, for example, a dependence on $N$ able to restore extensiveness.

The real problem is elsewhere. In thermodynamics, entropy is usually defined starting from an analysis of thermodynamic cycles, ending with the possibility of defining a function of the state. Now, a thermodynamic state can be specified using many different state variables. In principle any quantity depending on the state (and not on a process) could be used. However, it turns out that, if one would like to encode in a single function the maximum information about a thermodynamic system, the choice of the independent variable is not arbitrary and should be done carefully.

It turns out that, although one could consider entropy of a fluid system as a function of any triple of independent state variables, only when expressed in term of energy $U$, volume $V$ and number of particles $N$ the whole information about the system is encoded in the single entropy function.

Starting from $S(U,V,N)$, temperature, pressure and chemical potentials are obtained through $$ \frac{1}{T} = \left. \frac{\partial{S}}{\partial{U}} \right|_{V,N} ~~~~~~[1] $$ $$ \frac{P}{T} = \left. -\frac{\partial{S}}{\partial{V}} \right|_{U,N} ~~~~~~[2] $$ $$ \frac{\mu}{T} = \left. -\frac{\partial{S}}{\partial{N}} \right|_{U,V}. ~~~~~~[3] $$ It is always possible to use such relations to transform the function $S(U,V,N)$ into a function of other variables. For instance, one could invert [$1$] to obtain $U$ as a function of $T,V,N$. Thus, S could be written as $S(U(T,V,N),V,N)$, i.e. as a function of $T,V,N$. However it turns out that such a function does not contain the same information about the system. For example it is impossible to get back the full form of the internal energy since from $S(T,V,N)$ one can only get $ \left. \frac{\partial{U}}{\partial{T}} \right|_{V,N}$ and the integration over $T$ provides $U$ within an arbitrary function of V and N.

Therefore it is now understandable in which sense writing entropy as $PV^{\gamma}$ or as a function of this quantity cannot work, besides any problem of dimension or extensiveness. Starting as in the previous paragraph from $S(U,V,N)$ it is possible, as it has been shown in the J.Murray's answer, to write the entropy on an adiabat as a function of $N$ and $PV^{\gamma}$. However, such a formula cannot provide more than the value of the entropy on each adiabat. No hope to extract only from that equation the specific heat or isothermal compressibility or temperature.

In conclusion, the constance on adiabatic transformations of $PV^{\gamma}$ could be used as a parametrization of the entropy, but is not equivalent to the knowledge of entropy as a function of its own natural variables $U,V,N$.

It could be interesting to notice that, at variance with the case of $PV^{\gamma}$, the quantity $PV$ can be used as a kind of thermodynamic potential (i.e. contains the full information about the system), if expressed as a function of the set of variables $(T,V,\mu)$, and indeed it is the thermodynamic quantity associated by statistical mechanics to the grand canonical partition function.

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