Why is choosing a suitable thermodynamic potential important? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-07-19T01:47:43Z https://physics.stackexchange.com/feeds/question/480488 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/480488 4 Why is choosing a suitable thermodynamic potential important? sangstar https://physics.stackexchange.com/users/153135 2019-05-16T19:12:48Z 2019-05-16T22:14:21Z <p>Say we are undergoing an isothermal process that eventually settles at equilibrium. <span class="math-container">$F$</span> will minimize at equilibrium, and if the entropy of the system is not held constant <span class="math-container">$U$</span> will not. Why do we care though? What do we afford to lose if we use <span class="math-container">$U$</span> anyway? What use do we have for <span class="math-container">$F$</span> if it has this property for isothermal processes?</p> <p>Gibbs free energy minimizes <em>only</em> at constant pressure <strong>and</strong> temperature, so to me it sounds like a less general version of Helmholtz free energy. So why is it used?</p> https://physics.stackexchange.com/questions/480488/-/480489#480489 3 Answer by eranreches for Why is choosing a suitable thermodynamic potential important? eranreches https://physics.stackexchange.com/users/71341 2019-05-16T19:29:03Z 2019-05-16T19:56:28Z <p><span class="math-container">$U$</span> (or actually <span class="math-container">$S$</span>) is only relevant for an isolated system. This is not the case, for example, when you keep the temperature constant, since then heat must flow from the environment to your system (or vice versa) for <span class="math-container">$T$</span> to remain fixed.</p> <p>But let's say the system you are interested in is small compared to the environment. Keeping track after all the thermodynamics of the surroundings is difficult, so we want to somehow treat the system alone without having to think all the time on what is happening outside. The way to do it, in the case of thermal contact only, is to define the Helmholtz free energy</p> <p><span class="math-container">$$F=U-TS$$</span></p> <p>Here <span class="math-container">$U$</span> is the energy of the system, <span class="math-container">$S$</span> is its entropy but <span class="math-container">$T$</span> is the temperature of the <strong>environment</strong> (and it is fixed since your system is so small so that practically it has no effect on it). When <span class="math-container">$F$</span> is minimized, it can be shown (see <a href="https://en.wikipedia.org/wiki/Helmholtz_free_energy#Formal_development" rel="nofollow noreferrer">here</a>) that the entropy of the system + the environment is maximized, which is the condition you would expect for a system in thermal equilibrium.</p> <p>Similarly, if your system can only exchange <strong>volume</strong> with the environment, the quantity that is minimized is known at the enthalpy</p> <p><span class="math-container">$$H=U+PV$$</span></p> <p>In the case of both <strong>heat</strong> and <strong>volume</strong> exchange, you would use the Gibbs free energy</p> <p><span class="math-container">$$G=U-TS+PV$$</span></p> <hr> <p>There is also a mathematical-inclined argument for this. You know that</p> <p><span class="math-container">$${\rm d}F=-S{\rm d}T-P{\rm d}V$$</span></p> <p>so the natural variables of <span class="math-container">$F$</span> are <span class="math-container">$T$</span> and <span class="math-container">$V$</span>, <em>i.e.</em> <span class="math-container">$F=F\left(T,V\right)$</span>. It means that <span class="math-container">$F$</span> is the right function to use if you can control both <span class="math-container">$T$</span> and <span class="math-container">$V$</span>, which is the case for a system of constant volume in contact with a thermal bath. In the other cases</p> <p><span class="math-container">$${\rm d}H=T{\rm d}S+V{\rm d}P\Longrightarrow H=H\left(S,P\right)$$</span></p> <p>and</p> <p><span class="math-container">$${\rm d}G=-S{\rm d}T+V{\rm d}P\Longrightarrow G=G\left(T,P\right)$$</span></p> <p>You can clearly see which variables you can control and consequently which situations each thermodynamic potential fits. This mathematical trick of changing the variables of your function is known as the <a href="https://en.wikipedia.org/wiki/Legendre_transformation" rel="nofollow noreferrer">Legendre transformation</a>. This is exactly the same as the relation between Lagrangians and Hamiltonians in classical mechanics.</p> <hr> <p>To finish with an example, when treating liquid-gas phase transitions, it is customary to use the Gibbs free energy. This is the right thermodynamic potential since your system can exchange both its entropy <span class="math-container">$S$</span> and volume <span class="math-container">$V$</span> with the outside, such that the variables you control (the environment) are the corresponding conjugate variables - the temperature <span class="math-container">$T$</span> and the pressure <span class="math-container">$P$</span> respectively.</p> https://physics.stackexchange.com/questions/480488/-/480511#480511 8 Answer by Jeffrey J Weimer for Why is choosing a suitable thermodynamic potential important? Jeffrey J Weimer https://physics.stackexchange.com/users/228177 2019-05-16T20:43:39Z 2019-05-16T22:14:21Z <h1>Foundations</h1> <p>In a closed system, the four combined laws of thermodynamics are as follows:</p> <p><span class="math-container">$$dU = TdS - pdV$$</span> <span class="math-container">$$dH = TdS + Vdp$$</span> <span class="math-container">$$dA = -SdT - pdV$$</span> <span class="math-container">$$dG = -SdT + Vdp$$</span></p> <p>You can obtain these in one of two ways. You can start from the first law (the first equation above) and apply the definitions of enthalpy, Helmholz energy, and Gibbs energy. Alternatively, you can start from the postulates of <span class="math-container">$U(V,n)$</span> and <span class="math-container">$S(U,n)$</span> and work through Legendre transforms. The imperative appreciation here is that derivations of these equations is <strong>NOT</strong> based on statements about what kind of process is occurring. They are fundamental and absolute expressions.</p> <p>The other imperative statement is that we always need two and only two parameters to define the exact mechanical, thermal, chemical, and therefore thermodynamic state of a closed system with a pure substance. It is a statement of the Gibbs phase rule <span class="math-container">$F_{max} = (C - R) - \Pi_{min} + 2 = (1 - 0) - 1 + 2 = 2$</span>. This establishes why the expansions below consider only two parameters and no more as requirements to set the thermodynamic criteria for equilibrium or spontaneity.</p> <h1>Equilibrium</h1> <p>At equilibrium, a system must show no change in any state property. By review of the combined laws</p> <ul> <li>A process that occurs at constant entropy and volume will show no change in <span class="math-container">$U$</span></li> <li>A process that occurs at constant entropy and pressure will show no change in <span class="math-container">$H$</span></li> <li>A process that occurs at constant temperature and volume will show no change in <span class="math-container">$A$</span></li> <li>A process that occurs at constant temperature and pressure will show no change in <span class="math-container">$G$</span></li> </ul> <p>This establishes the four criteria for equilibrium of any process.</p> <ul> <li><span class="math-container">$\Delta U = 0$</span> at constant <span class="math-container">$(S, V)$</span></li> <li><span class="math-container">$\Delta H = 0$</span> at constant <span class="math-container">$(S, p)$</span></li> <li><span class="math-container">$\Delta A = 0$</span> at constant <span class="math-container">$(T, V)$</span></li> <li><span class="math-container">$\Delta G = 0$</span> at constant <span class="math-container">$(T, p)$</span></li> </ul> <p>We live (mostly) in a constant temperature + pressure world. Therefore, to determine whether a system is or is not at equilibrium, we work with (engineer processes using) changes in the Gibbs energy more than any of the other thermodynamic state functions.</p> <h1>Spontaneity</h1> <p>A process that is occurring in one of the four realms and is not at equilibrium must have a change in the respective state function that itself is not zero. We apply a systematic analysis starting from the Clausius form of the second law (heat flows spontaneously from hot to cold) to establish the convention for <span class="math-container">$dU$</span> in spontaneous processes. The rest of the expressions follow from it.</p> <ul> <li><span class="math-container">$\Delta U &lt; 0$</span> at constant <span class="math-container">$(S, V)$</span></li> <li><span class="math-container">$\Delta S &gt; 0$</span> at constant <span class="math-container">$(U, V)$</span></li> <li><span class="math-container">$\Delta H &lt; 0$</span> at constant <span class="math-container">$(S, p)$</span></li> <li><span class="math-container">$\Delta A &lt; 0$</span> at constant <span class="math-container">$(T, V)$</span></li> <li><span class="math-container">$\Delta G &lt; 0$</span> at constant <span class="math-container">$(T, p)$</span></li> </ul> <h1>Utility</h1> <p>One or the other of the expressions above is never more (or less) "useful". The framework is always first to define what parameters are held constant as the process occurs. That step defines which state function is to be used to establish equilibrium or spontaneity.</p> <p>The mechanical work of a reversible process is always <span class="math-container">$w = \pm \int p\ dV$</span> (I leave the sign convention here as a hot discussion point that IUPAC folks and engineers banter over). <em>Other</em> work may be obtained from a process when it is spontaneous in constant <span class="math-container">$T, V$</span> or constant <span class="math-container">$T, p$</span> space. This is how we relate changes in Helmholtz or Gibbs energy to "free" work of a process.</p>