Q3(a)(i). Chemical potential from U, H, A and G

1. In a mixture whose composition can change, U depends on S, V and every amount nᵢ. Its total differential is dU=(∂ U/∂ S)_(V,n)dS+(∂ U/∂ V)_(S,n)dV+Σᵢ(∂ U/∂ nᵢ)_(S,V,nⱼ)dnᵢ For fixed composition dU=T dS-p dV, so the first two coefficients are T and -p. The third coefficient defines the chemical potential: dU=T dS-p dV+Σᵢμᵢ dnᵢ ⇒ μᵢ=(∂ U/∂ nᵢ)_(S,V,nⱼ)
2. Enthalpy. H=U+pV, so dH=dU+p dV+V dp. Putting in dU: dH=T dS+V dp+Σᵢμᵢ dnᵢ, and μᵢ=(∂ H/∂ nᵢ)_(S,p,nⱼ)
3. Helmholtz energy. A=U-TS, so dA=dU-T dS-S dT: dA=-S dT-p dV+Σᵢμᵢ dnᵢ, and μᵢ=(∂ A/∂ nᵢ)_(T,V,nⱼ)
4. Gibbs energy. G=H-TS, so dG=dH-T dS-S dT: dG=-S dT+V dp+Σᵢμᵢ dnᵢ, and μᵢ=(∂ G/∂ nᵢ)_(T,p,nⱼ) This is the fundamental equation of chemical thermodynamics.
5. Each transform swaps one natural variable (V→ p or S→ T) and leaves the term Σμᵢ dnᵢ unchanged. So the same μᵢ appears in all four: 5pt] μᵢ=(∂ U/∂ nᵢ)_(S,V,nⱼ)=(∂ H/∂ nᵢ)_(S,p,nⱼ)=(∂ A/∂ nᵢ)_(T,V,nⱼ)=(∂ G/∂ nᵢ)_(T,p,nⱼ)
6. A partial molar quantity is a derivative at constant T and p, the intensive variables that can be held fixed while material is added. Only (∂ G/∂ nᵢ)_(T,p,nⱼ) is one: μᵢ is the partial molar Gibbs energy. The other three hold S or V (extensive) constant, so they are not the partial molar internal energy, enthalpy or Helmholtz energy.
7. Because of this, G=Σᵢ nᵢμᵢ at constant T and p, and comparing dG with this sum gives the Gibbs–Duhem equation, Σᵢ nᵢ dμᵢ=0. No such simple sum holds for U, H or A.

Application. at constant T and p, dG=Σμᵢ dnᵢ, so a component moves from the phase where its μᵢ is higher to where it is lower. Phase equilibrium needs μᵢ(α)=μᵢ(β) for every component, the starting point of the phase rule. The same sum is the maximum non-expansion (electrical) work of a cell reaction.