The Gibbs free energy is one of the 4 thermodynamic potentials whose partial derivatives produce all other thermodynamics state properties; its differential is d g = ∂ ∂ p g d p + ∂ ∂ T g d T = v d p − − s d T . Integrating this over an isotherm from p s , v f to p s , v g , noting that the pressure is the same at each endpoint, and setting the result to zero yields g g − − g f = ∑ ∑ j = 1 3 ∫ ∫ p j p j + 1 v j d p = − − ∫ ∫ v f v g p d v + p s ( v g − − v f ) = 0 Here because v is a multivalued function, the p integral must be divided into 3 parts corresponding to the 3 real roots of the vdW equation in the form, v ( p , T ) (this can be visualized most easily by imagining Fig. 1 rotated 90 ∘ ∘ ); the result is a special case of material equilibrium. The last equality, which follows from integrating v d p = d ( p v ) − − p d v , is the Maxwell equal area rule which requires that the upper area between the vdW curve and the horizontal through p s be equal to the lower one. This form means that the thermodynamic restriction that fixes p s is specified by the equation of state itself, p = p ( v , T ) . Using the equation for the Gibbs free energy obtained previously for the vdW equation applied to the saturated vapor state and subtracting the result applied to the saturated liquid state produces, R T s [ v g v g − − b − − v f v f − − b − − ln ⁡ ⁡ ( v g − − b v f − − b ) ] − − 2 a ( 1 v g − − 1 v f ) = 0 This is a third equation that along with the two vdW equations above can be solved numerically. This has been done given a value for either T s or p s , and tabular results presented; however, the equations also admit an analytic parametric solution obtained most simply and elegantly, by Lekner. Details of this solution may be found in the Maxwell Construction ; the results are T r s ( y ) = ( 27 8 ) 2 f ( y ) [ cosh ⁡ ⁡ y + f ( y ) ] g ( y ) 2 p r s = 27 f ( y ) 2 [ 1 − − f ( y ) 2 ] g ( y ) 2 v r f = ( 1 3 ) 1 + f ( y ) e y f ( y ) e y v r g = ( 1 3 ) 1 + f ( y ) e − − y f ( y ) e − − y where f ( y ) = y cosh ⁡ ⁡ y − − sinh ⁡ ⁡ y sinh ⁡ ⁡ y cosh ⁡ ⁡ y − − y g ( y ) = 1 + 2 f ( y ) cosh ⁡ ⁡ y + f ( y ) 2