5
$\begingroup$

In thermodynamics, quantities like pressure, temperature and entropy are associated with overall states of a macroscopic system. In that case, we do not talk about "the quantity $Q$ at the point $p$ of the system" thinking of it as a continuum, for instance, we rather say "the quantity $Q$ of the system" as if it were the same for all of it.

In that sense, does thermodynamics only deal with homogenous systems? That is, do we always suppose that pressure, temperature and all those quantities are the same on the entire system under study? If so, what is the need and motivation for this?

$\endgroup$

7 Answers 7

2
$\begingroup$

I can't think of any source that would claim you need homogeneity to do thermodynamics. Textbooks might usually assume systems have a single constant pressure and temperature because it's easier, but it's not a requirement.

Intensive variables can be functions of position. Any discussion of the buoyant force needs position-dependent pressure. Any discussion of the heat equation needs position-dependent temperature, etc.

You can't find the entropy at a certain point, but that's just because it's extensive; you need to find the entropy of the entire system. It's the same as how you can't find the energy at a certain point, or the mass at a certain point. What you can do is find densities of these things. Mass density, energy density, specific entropy, etc. You can then integrate these densities over the entire system to find the mass, energy, and entropy.

$\endgroup$
15
  • 1
    $\begingroup$ Problems with position dependent thermodynamic variables of the kind that you are describing are NOT part of thermodynamics. Indeed, without ad-hoc assumptions thermodynamics has nothing to say about them, at all. Buoyancy is not a thermodynamics, but a fluid mechanics problem. One can find engineering textbooks which do not make a clear distinction of where thermodynamics ends and where linearized-non-equilibrium thermodynamics models begin. This makes some students believe that these models have the same physical meaning as thermodynamics proper, but that's not the case. $\endgroup$
    – CuriousOne
    Sep 19, 2014 at 6:27
  • $\begingroup$ Would you like to give reasons to support your assertions? $\endgroup$ Sep 19, 2014 at 6:44
  • $\begingroup$ @CuriousOne, gradient of pressure does not necessarily mean the situation is non-equilibrium. As long as the state of the fluid in gravitational field changes quasi-statically and is close to thermodynamic equilibrium, thermodynamics (1.+2.law) is applicable. See for example Landau & Lifshitz, first chapters of Fluid mechanics. $\endgroup$ Sep 19, 2014 at 6:54
  • $\begingroup$ See particularly the sections 3,4 of the first chapter. $\endgroup$ Sep 19, 2014 at 7:04
  • 1
    $\begingroup$ @JánLalinský: I didn't say that the situation in buoyancy is non-equilibrium but that it is part of mechanics and not part of thermodynamics. There is no temperature and no pressure dependence in a basic buoyancy setup. It is simply Newtonian forces acting on Newtonian masses. $\endgroup$
    – CuriousOne
    Sep 19, 2014 at 18:29
2
$\begingroup$

Thermodynamics does deal with inhomogeneities and non-equilibrium conditions...

...but indeed it requires some "macroscopic smearing" of quantities. Problems like diffusion of particles, or heat are dealt with and very well explained by thermodynamics, specifically with Maxwell's Thermodynamic Potentials formalism. But the thing is, the thermodynamic variables of a system are defined over macroscopic portions of it, because Thermodynamics is a macroscopic theory. Variables like density, or chemical concentration are averages over certain macroscopic volume; and temperature has no meaning without equilibrium! (although it is used other fields of physics where analogies allow for convenient modified definitions)

I have emphasized macroscopic because I wanted to specify that it need not be a ~$10^{23}$ components system. A macroscopic system or portion of the system can be that whose size is large enough that the mean value Energy or Mass or any extensive variable (additive variable) has small enough fluctuations.

Summarizing: Thermodynamics deals with inhomogeneities and explains the evolution of a system towards equilibrium, but as long as these inhomogeneities are of macroscopic order.

$\endgroup$
0
1
$\begingroup$

Thermodynamics only deals with homogenous systems? That is, we suppose always that pressure, temperature and all those quantities are the same on the entire system under study?

Of course not. Thermodynamics would be a rather useless field of study if it only addressed homogeneous systems. That thermodynamics does far more than that is what makes it instead so incredibly useful.

Even at the most basic level, all but the zeroth law of thermodynamics addresses systems that are not homogeneous. The first law addresses heat flow, the second heat engines. How can you have heat flow or a heat engine if everything is of the same composition and is at the same pressure, temperature, and density?

Intrinsic variables such as pressure, temperature, and density are intrinsically local. Even extrinsic variables such as volume, mass, entropy, and energy can be made local either by looking at their thermodynamic conjugate or by looking at a ration of two extrinsic variables. It's only in elementary treatments that students are taught to look at a system as having one pressure, one temperature, one density, throughout some volume. This is done is because introductory level students don't yet have the mathematical prowess to understand a more thorough description.

$\endgroup$
0
$\begingroup$

You can break the system down into smaller subsystems. If we have a room with an inlet of air at one corner and an outlet at the opposite corner, we can make a grid of cells of whatever spacing we want. We can then assume the pressure and temperature are uniform within the cell and compute the flows between the cells. This is what the weather people do. The grid spacing is limited by our calculation power.

$\endgroup$
0
$\begingroup$

As thermodynamic deals with macroscopic properties like temperature , preesure, which doesn't change with time for system in equillibrium state it only occur if system is homogeneous b'coz heterogeneous system can't have constant P ,T

$\endgroup$
0
$\begingroup$

This topic makes some sense, “Thermodynamics quantities like pressure, temperature and entropy are associated with overall states of a macroscopic system” or associated with overall states of a local, this is a fact. In the equation \begin{align}dU=TdS-pdV+Ydx+\sum_j\mu_jdN_j\end{align} $T, S, p, V $ “are associated with overall states of a macroscopic system” or a local, $Y, x, \mu_j, N_j$ can be, but not a requirement. So thermodynamics now is a “Grey box theory”. In some new theoretical models[1], the Intensive variables $T, p$ can be instead, by the distribution of the extensive variables associated with them. For example, consider an idea gas or photon gas, using $q$ denotes the energy of thermal motion within the system, name as the internal heat energy, then we have \begin{align}dU=dq.\end{align} The entropy of the system \begin{align}dS=\frac{dU}{T}+\frac{pdV}{T}=\frac{dq}{T}+\frac{pdV}{T}.\end{align} For an idea gas, $T=q/iNk$ and $pV=NkT$, such that \begin{align}dS=\frac{iNk}{q}dq+\frac{Nk}{V}dV.\end{align} For a photon gas, $q=3pV$ and $T=q/3NR_{p}$, where $R_{\{p\}}$=$[\zeta (4)/\zeta (3)]k$, $\zeta (3)$ and $\zeta (4)$ are the Riemann zeta functions. Such that, we get \begin{align}dS=\frac{3NR_{\{p\}}}{q}dq+\frac{NR_{\{p\}}}{V}dV.\end{align} It implies, for an idea gas or photon gas \begin{align}dS=\frac{iNR_{\{k\}}}{q}dq+\frac{NR_{\{k\}}}{V}dV. \end{align} Where $i$ is number of the degrees of freedom of the particles, and $R_{\{k\}}$ denotes the system constant, for an idea gas $R_{\{k\}}$=$k$, and for a photon gas $R_{\{k\}}$=$R_{\{p\}}$. All of the variables in the equation are extensive variables, and an inference for the first term can be made to dynamics \begin{align} \frac{iNR_{\{k\}}}{q}dq\rightarrow R_{\{k\}}\sum_{j=1}^{N}\sum_{s=1}^s d\ln\epsilon_{j\{s\}}.\end{align} Where $\epsilon_{j}$ is the kinetic energy of the particle $j$, and $s$ is the degrees of freedom of the particles. So, thermodynamic quantities are associated with overall states of a macroscopic system, but this “associated with” is not a requirement.

[1] https://arxiv.org/abs/1201.4284v5

$\endgroup$
0
$\begingroup$

TL; DR: the probability density is homogeneously spread over all the states with the same energy

Thermodynamics vs. Statistical physics
Let me first point out that thermodynamics and statistical physics are not the same thing: these are two descriptions of the same phenomena, one of which is phenomenological (thermodynamics) and the other is microscopic (statistical physics).

Non-equilibrium thermodynamics
Further, the question seems to refer to equilibrium thermodynamics/statistical-physics. Both of these have extensions (in fact multiple extensions) for treating non-equilibrium (and thus non-homogeneous) systems.

So, homogeneous or not?

In that sense, does thermodynamics only deal with homogenous systems?

Directly extending phenomenological thermodynamic description to inhomogeneous systems might be hard. From the point of view of statistical physics, we do make homogeneity assumption, but somewhat indirect: we assume that all the phase space configurations with the same energy (and same value of some other parameters) are equally probable, and that system will visit all of them, so that we can replace the time averaging by ensemble averaging. Tu put it into more "homogeneity" language: we do assume that the probability density is homogeneously spread over all the states with the same energy. This is known as microcanonical ensemble, and serves to build canonical and grand canonical cases.

Homogenuity of quantities

That is, do we always suppose that pressure, temperature and all those quantities are the same on the entire system under study?

Pressure and temperature are intrinsic variables that characterize the system as a whole. As such, they say nothing about whetehr the system is homogeneous or not. What make them seem like specific to homogeneous system is the interpretation of these quantities as in terms of ideal gas, where pressure comes from the molecules colliding against the container walls, while the temperature is the average kinetic energy of these molecules. These quantities (and many other, such as, e.g., magnetic moment) can be however defined in very general thermodynamic ways, without resorting to their microscopic interpretation (note how thermodynamics here is more convenient than statistical physics):

  • Temperature as the derivative of the internal energy in respect to entropy $$T=\left(\frac{\partial U }{\partial S}\right)_{V,N}\text{ or } \frac{1}{T}=\left(\frac{\partial S }{\partial U}\right)_{V,N}$$
  • Pressure as the derivative of the internal energy in respect to volume $$P=\left(\frac{\partial U}{\partial V}\right)_{S,N}$$

Remark: equating these with the mechanistic notions of pressure and temperature as defined for the ideal gas may sometimes be tricky however - see, e.g., Significance of Stokes hypothesis.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.