Question

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 \(d G .\) Use the First Law to simplify. b) All of these are also exact differentials. What relationships follow from this fact?

Short answer

## Short Answer Given the definitions of enthalpy (H), Helmholtz free energy (A), and Gibbs free energy (G), we derived the differential equations for each as: \(dH = \delta Q + Vdp\) \(dA = \delta Q - pdV - SdT\) \(dG = \delta Q - SdT + Vdp\) We then showed that these differentials are exact by finding the conditions: \(\frac{\partial (\delta Q)}{\partial V} = \frac{\partial V}{\partial p}\) for enthalpy H \(\frac{\partial (\delta Q - p)}{\partial V} = \frac{\partial (-S)}{\partial S}\) for Helmholtz free energy A \(\frac{\partial (\delta Q - S)}{\partial p} = \frac{\partial V}{\partial T}\) for Gibbs free energy G These relationships reflect the conditions necessary for their exact differentials and show how these state variables are connected to one another and to thermodynamic properties.

Step-by-step solution

Define enthalpy H, Helmholtz free energy A, and Gibbs free energy G

We are given the following definitions: Enthalpy, \(H = E_{\text{int}} + pV\) Helmholtz free energy, \(A = E_{\text{int}} - TS\) Gibbs free energy, \(G = E_{\text{int}} + pV - TS\)

Write the differentials for H, A, and G

To find the differential equation for H, A, and G, we need to take the differential of each equation we got in step 1: \(dH = d(E_{\text{int}} + pV)\) \(dA = d(E_{\text{int}} - TS)\) \(dG = d(E_{\text{int}} + pV - TS)\)

Apply the First Law of Thermodynamics

The First Law of Thermodynamics can be expressed as: \(dE_{\text{int}} = \delta Q - \delta W\) Considering an isobaric process, we can write the work done as \(\delta W = pdV\). Then, we can rewrite the First Law as: \(dE_{\text{int}} = \delta Q - pdV\) Now we can substitute this expression into the differentials of H, A, and G: \(dH = (\delta Q - pdV) + pdV + Vdp\) \(dA = (\delta Q - pdV) - TdS - SdT\) \(dG = (\delta Q - pdV) + pdV + Vdp - TdS - SdT\)

Simplify the differential equations

Now, we can simplify the differential equations for H, A, and G: \(dH = \delta Q + Vdp\) \(dA = \delta Q - pdV - SdT\) \(dG = \delta Q - SdT + Vdp\)

Show that dH, dA, and dG are exact differentials

If a differential is exact, it can be written in the form: \(df = M dx + N dy\) Comparing this with the expressions obtained in step 4, we can identify the functions M, N, and the variables x and y as follows: For \(dH = \delta Q + V dp\), let \(H=f(p,V)\), then \(M=\delta Q\) and \(N=V\). Since \(dH\) is exact, we need: \(\frac{\partial M}{\partial V} = \frac{\partial N}{\partial p}\), which implies \(\frac{\partial (\delta Q)}{\partial V} = \frac{\partial V}{\partial p}\). For \(dA = \delta Q - pdV - SdT\), let \(A=f(S,V)\), then \(M=\delta Q - p\) and \(N=-S\). So we have: \(\frac{\partial M}{\partial V} = \frac{\partial N}{\partial S}\), which means \(\frac{\partial (\delta Q - p)}{\partial V} = \frac{\partial (-S)}{\partial S}\). For \(dG = \delta Q - SdT + Vdp\), let \(G=f(T,p)\), and we get \(M=\delta Q - S\) and \(N=V\). In this case: \(\frac{\partial M}{\partial p} = \frac{\partial N}{\partial T}\), that is \(\frac{\partial (\delta Q - S)}{\partial p} = \frac{\partial V}{\partial T}\). These conditions ensure that dH, dA, and dG are exact differentials.

Deduce the relationships from the exact differentials

As we have shown that the differentials are exact, we can now deduce the relationships that follow: For enthalpy H, we have the relationship: \(\frac{\partial (\delta Q)}{\partial V} = \frac{\partial V}{\partial p}\) For Helmholtz free energy A, we obtain: \(\frac{\partial (\delta Q - p)}{\partial V} = \frac{\partial (-S)}{\partial S}\) And for Gibbs free energy G, we find: \(\frac{\partial (\delta Q - S)}{\partial p} = \frac{\partial V}{\partial T}\) These relationships reflect the conditions that must be satisfied for each of these state variables to be exact differentials.