Published on *PNG 520: Phase Behavior of Natural Gas and Condensate Fluids* (https://www.e-education.psu.edu/png520)

Thermodynamics cannot tell about the rate (kinetics) of a process, but it can tell whether or not it is possible for a process to occur. For this, we make use of the first and second laws of thermodynamics.

From our basic courses in thermodynamics, we recall that the first law of thermodynamics for a closed system is written as follows:

$$\partial W=PdV$$ (14.1)

where,

U = internal energy,

Q = heat added to or extracted from the system,

W = work done by or to the system.

If we want to improve the definition of the first law of thermodynamics stated in (14.1), we need to come up with an expression for the amount of work and heat. The amount of work need to accomplish a *reversible process* is given by the expression:

$$\partial Q=TdS$$ (14.2)

where:

P = pressure,

V = volume.

Additionally, we can compute the heat required to accomplish a reversible process by virtue of the second law of thermodynamics:

$$\partial Q=TdS$$ (14.3)

where:

T = temperature,

S = entropy.

We substitute (14.2) and (14.3) into (14.1) to get:

$$dU=TdS-PdV$$ (14.4)

which is the fundamental thermodynamic relationship used to compute changes in **Internal Energy (U)** for a closed system. This is only a restatement of the first law of thermodynamics.

Although equations (14.2) and (14.3) are applicable strictly to* reversible processes*, equation (14.4) is quite general and does not have such a constraint. Equation (14.4) applies to reversible and irreversible processes. Are you surprised by this? Okay, it is about time that we start to get a feeling for the implications of the fact that most thermodynamic properties are *state functions*. Internal energy (U) is a state function, and as such, its changes (dU) do not depend on the path that was taken (reversible/irreversible), but on the end points of process. Hence, if taking any path to compute “dU” is okay, why not take the reversible path? We do not have nice, explicit equations to describe work and heat for irreversible processes. Why should we bother? Generally speaking, “reversibility” is a thermodynamic trick that helps us to get away with many thermodynamic manipulations by taking advantage of the properties of state functions.

The formal, fundamental thermodynamic definition of **Enthalpy (H)** is the following:

H = U + P $\tilde{v}$ (14.5)

Enthalpy is a state function as well. If we are interested in computing changes in enthalpy, we write:

$$H=U+P\tilde{v}$$ (14.6)

Since we already have a way to compute dU (equation 14.4), we can now write the fundamental thermodynamic relationship used to compute changes of enthalpy in a closed system:

$$dH=dU=PdV=VdP$$

By the same token, the formal, fundamental definition of **Gibbs Free Energy (G)** is the following:

$$G=H=TS$$ (14.8)

from which we see that changes in this property may be calculated as:

$$dG=dH=TdS=SdT$$ (14.9a)

We now substitute (14.7) to get the fundamental thermodynamic relationship for changes in Gibbs free energy in a closed system:

$$dG=VdP=SdT$$ (14.9b)

We proceed with **Helmholtz Free Energy (A)** accordingly. Its definition:

$$A=U=TS$$ (14.10)

The expression of change,

$$dA=dU=TdS=SdT$$ (14.11)

We substitute (6.4) into it and get:

$$dA=-PdV-SdT$$ (14.12)

We have just derived the following fundamental thermodynamic relationships for fluids of constant composition:

$$dU=TdS-PdV$$ (14.4)

$$dH=TdS+VdP$$ (14.7)

$$dG=VdP-SdT$$ (14.9)

$$dA=-PdV-SdT$$ (14.12)

All these properties (U, H, G and A) are *state functions* and extensive properties of the system. Notice that the only assumption that we took throughout their development is that the system was closed (for the derivation of 14.4). Hence, these equations strictly apply to systems of constant composition.