Concerning the definitions of the enthalpy, the Helmholtz free energy and the Gibbs free energy of a system

The definition of enthalpy, $$H = U + PV,$$ assumes that the system is in a constant-pressure environment. Similarly, the definition of the Helmholtz free energy of the system, $$F = U - TS,$$ assumes a constant-temperature environment. The definition of the Gibbs free energy, $$G = U + PV - TS,$$ takes for granted both of the aformentioned assumptions. Does this mean, for example, that the enthalpy of a system is undefined for a system with a volume-dependent environmental pressure? I have similar questions about the Helmholtz free energy and the Gibbs free energy.

• I can understand where your puzzle comes from. Some textbook (such as Schroeder's book, section 5.1), for the convenience of presenting, uses phrases like constant pressure or constant temperature. But it doesn't exclude other conditions. – user115350 Apr 2 at 16:31

Those are the definitions of each. They don't assume anything about the system and can always be applied. You are getting mixed up with the scenarios in which they are usually applied since nice things happen. For example, for a system at constant pressure (and number of particles) $$\Delta H=Q$$, where $$Q$$ is the heat that enters or leaves the system.

To add some more detail, this can be seen by substituting in the thermodynamic identity $$\text dU=T\text dS-P\text dV+\mu\text d N$$ into the differential of one of your thermodynamic potentials. For example, as mentioned above we have $$\text dH=\text dU+P\text dV+V\text dP$$ so then $$\text dH=T\text dS+V\text dP+\mu\text dN$$ i.e. at constant pressure and number of particles $$\text dH=T\text dS=\text dQ$$

You also say that the Gibbs free energy takes both mentioned assumptions "for granted", but see what happens if you do this process with the Gibbs free energy at constant temperature and pressure. It is a very important relation.

These processes are more generally called Legendre transformations

• Sorry, I'm still confused. Are you claiming that if I don't assume anything about the environment those definitions still hold? – PiKindOfGuy Apr 3 at 0:15
• @PiKindOfGuy Yes that is exactly right. As long as the system has well defined Pressure, Volume, Entropy, etc. then those definitions are fine. Why wouldn't they be? They are just definitions in terms of state variables. They don't refer to any processes. It is just how we define work to be $W=\int\mathbf F\cdot\text d\mathbf x$. As long as we have a force acting on an object over a distance the definition applies without any further assumptions of the system. – Aaron Stevens Apr 3 at 3:06
• The pressure in those equations refers to the pressure of the environment, not that of the system, no? (See your comment.) – PiKindOfGuy Apr 3 at 3:13
• @PiKindOfGuy We usually assume quasi-static processes where at any point in time the system is at equilibrium, so the pressure the system exerts on its boundary is the same as the pressure exerted on the system boundary by the environment – Aaron Stevens Apr 3 at 9:24
• @PiKindOfGuy Is there anything else I can do to make things more clear here? – Aaron Stevens Apr 4 at 17:10

I agree with @Aaron Stevens. There are no built in assumptions of constant pressure or temperature in these definitions.

The Hemlholtz free energy, Gibbs free energy, Enthalpy and Internal Energy are sometimes referred to as thermodynamic potentials. For a discussion of these check out

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/helmholtz.html

Hope this helps.

It is important to keep separate definitions of thermodynamic potentials (TP) from the specific cases where one particular TP is more convenient than others.

Let me use the Gibbs free energy $$G$$ as an example (the cases of Helmholtz free energy and enthalpy can be obtained by a parallel discussion).

Let's start from the fact that the most convenient way of representing the internal energy $$U$$ of a fluid system as a function of the thermodynamic state is to use as independent variables the entropy $$S$$, the volume $$V$$ and the number of particles $$N$$. Such a choice does not imply any constraint on the system and in principle it is possible to use thermodynamic relations to express the same value of the internal energy as a function of different state variables (say $$T$$ and/or $$P$$). However, it turns out that only the description in terms of the set $$S,V,T$$ allows to encode the full thermodynamic behavior in one function of three variables. From this "privileged" choice of variables one can derive other descriptions using different state variables, but it is not possible to go the other way around. For instance, knowing $$U(T,V,N)$$ does not allow to reconstruct $$U(S,V,N)$$ (this is discussed in detail in Callen's textbook on Thermodynamics).

The right way to change the independent variables while keeping the same information about the system as contained in $$U(S,V,N)$$ is to perform a Legendre transform (better a Legendre-Fenchel transform, due to the possibility of discontinuity of first derivatives). Legendre(-Fenchel) transforms of the internal energy are the so-called thermodynamic potentials.

The Legendre transform to obtain the Gibbs free energy is briefly indicated as $$G = U -TS +PV$$ Actually, the precise meaning of the above expression is that one starts with $$U(S,V,N)$$ and then $$G$$ is defined as $$G(T,P,N) = \inf_{S,V} ( U(S,V,N) - TS + PV )$$ Therefore, at level of definition, $$T$$ and $$P$$ are fixed just in the sense that one has to fix the values of the independent variables to identify the point where the function is evaluated. $$T$$ and $$P$$ can be changed at will and correspondingly one finds a value for $$G$$. Once one is using $$T,P,N$$ as independent variables, this does not imply that the conjugate variables $$S,V,\mu$$ are not definite or do not have a precise value. Their values are fixed by the function $$G$$ and can be obtained by the relations $$S=-\left.\frac{\partial{G}}{\partial{T}}\right|_{P,N},~~~~ V=\left.\frac{\partial{G}}{\partial{P}}\right|_{T,N}, ~~~~ \mu=\left.\frac{\partial{G}}{\partial{N}}\right|_{T,P}.$$

After introducing thermodynamic potentials with their natural variables ($$T,P,N$$ for $$G$$), since the differential of each thermodynamic potentials is obviously a sum of therms multiplying the differential of the independent variables, in the case of $$G$$ $$dG = -SdT + VdP + \mu dN$$ where $$S=-\left.\frac{\partial{G}}{\partial{T}}\right|_{P,N}$$, $$V=\left.\frac{\partial{G}}{\partial{P}}\right|_{T,N}$$ and $$\mu=\left.\frac{\partial{G}}{\partial{N}}\right|_{T,P}$$.

Therefore, in the case of a transformation keeping constant some of the variables $$T,P,N$$ one gets a simplified formula with respect to the full differential.