Other state variables useful for characterizing different classes of processes can be defined from \(E_{\text {int }}, S, P\), and \(V\). These include the enthalpy, \(H=E_{\text {int }}+p V\), the Helmholtz free energy, \(A=E_{\text {int }}-T S,\) and the Gibbs free energy, \(G=E_{\text {int }}+p V-T S\). a) Write the differential equations for \(d H, d A,\) and \(d G\). b) All of these are also exact differentials. What relationships follow from this fact? Use the First Law to simplify.

To find the differential equations for H, A, and G, we will differentiate their definitions with respect to the respective variables. Enthalpy (H): \(H=E_{\text {int }}+p V\) Differentiate with respect to its variables: \(dH = dE_{\text{int}} + pdV + Vdp\) Helmholtz free energy (A): \(A=E_{\text{int}}-T S\) Differentiate with respect to its variables: \(dA = dE_{\text{int}} - TdS - SdT\) Gibbs free energy (G): \(G=E_{\text{int }}+p V-T S\) Differentiate with respect to its variables: \(dG = dE_{\text{int}} + pdV + Vdp - TdS - SdT\)

Exact differentials preserve the work done in a thermodynamic process. Being exact differentials, they don't depend on path but solely on the initial and final states. The given state functions are exact differentials, and we use it to find relationships between them. According to the First Law of Thermodynamics, \(dE_{\text{int}} = TdS - pdV\) Now, substitute the expression of \(dE_{\text {int }}\) into the expressions for \(dH, dA,\) and \(dG\): \(dH = (TdS - pdV) + pdV + Vdp = TdS + Vdp\) \(dA = (TdS - pdV) - TdS - SdT = -pdV - SdT\) \(dG = (TdS - pdV) + pdV + Vdp - TdS - SdT = Vdp - SdT\) These are the relationships we were looking for: 1. \(dH = TdS + Vdp\) 2. \(dA = -pdV - SdT\) 3. \(dG = Vdp - SdT\) These differential forms give us relationships between the changes in the state functions (enthalpy, Helmholtz free energy, and Gibbs free energy) and other thermodynamic variables (entropy, pressure, volume, and temperature).