Thermodynamics Potentials

Enthalpy

In constant pressure, enthalpy, $H$, is defined as $$H = U + PV$$ At constant pressure, $PV$ is the work done on the system. Thus, enthalpy is the energy required for creating a system with internal energy $U$ and volume $V$ at constant pressure $P$.
By thermodynamic identity, $dU = TdS - PdV + \mu dN$, so $$dH = dU + PdV + VdP = TdS + VdP + \mu dN$$

Helmholtz Free Energy

In constant temperature, the Helmholtz free energy, $F$, is defined as $$F = U - TS$$ When there is no work done on or by the system, $dU = dQ$, so $Q = TS$. Thus, Helmholtz free energy is the energy required for creating a system with internal energy $U$ at constant temperature $T$.
By thermodynamic identity, $dU = TdS - PdV + \mu dN$, so $$dF = dU - TdS - SdT = -SdT - PdV + \mu dN$$

Gibbs Free Energy

In constant temperature and pressure, the Gibbs free energy, $G$, is defined as $$G = U - TS + PV$$ By thermodynamic first law and temperature definition, $SdT = dQ + PdV$, so $Q = TS - PV$. Thus, Gibbs free energy is the energy required for creating a system with internal energy $U$ at constant temperature $T$ and pressure $P$.
By thermodynamic identity, $dU = TdS - PdV + \mu dN$, so $$dG = -SdT + VdP + \mu dN$$

Maxwell Relations

Consider system with constant number of particles, i.e. $dN = 0$ $$dU = TdS - PdV$$ So, $$T = \left(\frac{\partial U}{\partial S}\right)_V$$ and $$P = -\left(\frac{\partial U}{\partial V}\right)_S$$ Since $$\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial S}\right) = \frac{\partial}{\partial S}\left(\frac{\partial U}{\partial V}\right)$$ we have $$\bbox[5px,border:2px solid #666] { \left(\frac{\partial T}{\partial V}\right)_S = - \left(\frac{\partial P}{\partial S}\right)_V }$$ Similarly, by $dF = -SdT -PdV$, $$\bbox[5px,border:2px solid #666] { \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V }$$ By $dH = TdS + VdP$, $$\bbox[5px,border:2px solid #666] { \left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P }$$ By $dG = -SdT + VdP$, $$\bbox[5px,border:2px solid #666] { \left(\frac{\partial S}{\partial P}\right)_T = - \left(\frac{\partial V}{\partial T}\right)_P }$$ These are called Maxwell relations.

Clausius-Clapeyron Relation

When the number of particles is a constant, i.e. $dN = 0$, entropy depends on $V$ and $U$ hence $T$ only, so $$dS = \left(\frac{\partial S}{\partial V}\right) _T dV + \left(\frac{\partial S}{\partial T}\right)_V dT$$ During phase change, temperature is constant $$dS = \left(\frac{\partial S}{\partial V}\right) _T dV$$ By Maxwell relations, $$dS = \left(\frac{\partial P}{\partial T}\right)_V dV$$ As pressure and temperature should be constant, the partial derivative should be the same as total derivative $$dS = \frac{dP}{dT} dV$$ So, $$\frac{dP}{dT} = \frac{S_{p1} - S_{p2}}{V_{p1} - V_{p2}}$$ Enthalpy at constant pressure is $$dH = dU + PdV$$ Substitute first law of thermodynamics, at constant pressure $$dH = Q - PdV + PdV = Q$$ So the latent heat $L$ is equal to $$L = \Delta H$$ As $dN = 0$, $$dH = TdS + VdP = TdS$$ So, $$\Delta S = \frac{\Delta H}{T} = \frac{L}{T}$$ Then, $$\bbox[5px,border:2px solid #666] { \frac{dP}{dT} = \frac{L}{T\Delta V} }$$ This is called Clausius-Clapeyron relation.