Question

Derive the Maxwell equations for F (Helmholtz Free Energy) and G (Gibbs Free Energy) with “for all mathematical stages"? What is the condition of being a state function, and which of the parameters E (Internal energy), Q (Heat), W (Work), S (Entropy) are not the state function? Why is that?

Answer 1

We knows that Helmholtz free energy ( F) of a system is defined as F=U−TS

differential form of above equation is dF = dU−d(TS) = dU−TdS−SdT

Substituting dU = TdS−PdV

dF = TdS−PdV−TdS−SdT

dF = −PdV−SdT

Consider F = F(V,T)

⇒ dF = (∂F/∂V)_TdV + (∂F/∂T)_VdT

⇒(∂F/∂V)_TdV + (∂F∂T)_VdT = −PdV − SdT

comparing

⇒ (∂F/∂V)_T = −P,   (∂F/∂T)_V = −S

⇒ ∂ /∂T)_V(∂F/∂V)_T = −(∂P/∂T)_V,

⇒ ∂ /∂T)_V(∂F/∂V)_T = −(∂P/∂T)_V,   ∂/∂V)_T(∂F/∂T)_V = −(∂S/∂V)_T

⇒    (∂P/∂T)_V = (∂S/∂V)_T

Gibbs free energy is defined as the G = H−TS

differential form dG = dH−d(TS) = dH−TdS−SdT

Substituting dH=TdS+VdP

dG = TdS+VdP−TdS−SdT

⇒ dG = VdP−SdT

Consider G=G(P,T)

⇒ dG = (∂G/∂P)_TdP + (∂G∂T)_PdT

⇒ (∂G/∂P)_TdP + (∂G∂T)_PdT = VdP−SdT

⇒ (∂G∂P)_T = V, (∂G∂T)_P = −S

⇒ ∂ / ∂T)_P(∂G/∂P)_T = (∂V∂T)_P,

∂ / ∂P)_T(∂G/∂T)_P = −(∂S/∂P)_T

⇒(∂V/∂T)_P = −(∂S/∂P)_T

State functions do not depend on the path by which the system arrived at its present state.

Internal energy and Entropy are state functions because they only depend upon initial and final state of the system not on the path.