From the van der Waals equation applied to the saturated liquid, v f < v m i n , and vapor, v g > v m a x , states p s = R T s v f − − b − − a v f 2 p s = R T s v g − − b − − a v g 2 These two equations specify 4 variables so they can be solved for p s , T s in terms of v f , v g . This results in p s = p ∗ ∗ v ∗ ∗ 2 [ v f v g − − v ∗ ∗ ( v f + v g ) ] v f 2 v g 2 T s = T ∗ ∗ v ∗ ∗ ( v f + v g ) ( v f − − v ∗ ∗ ) ( v g − − v ∗ ∗ ) v f 2 v g 2 where p ∗ ∗ = a / b 2 , v ∗ ∗ = b , and T ∗ ∗ = a / ( R b ) are a characteristic pressure, molar volume, and temperature defined by the constants (note that p ∗ ∗ v ∗ ∗ / T ∗ ∗ = R ). Applying the Maxwell construction to the van der Waals equation gives − − T s ln ⁡ ⁡ ( v g − − v ∗ ∗ v f − − v ∗ ∗ ) + T ∗ ∗ v ∗ ∗ ( v g − − v f ) v f v g + p s ( v g − − v f ) / R = 0 These three equations 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 that, according to Lenkner, was obtained by Gibbs. Lenkner himself devised a simple, elegant, method to obtain this solution, by eliminating p s and T s from the equations, and writing them in terms of a stretched dimensionless number density, ϱ ϱ = v ∗ ∗ / ( v − − v ∗ ∗ ) , that varies between ∞ ∞ and 0 as v varies from v ∗ ∗ to ∞ ∞ ; this produces ln ⁡ ⁡ ( ϱ ϱ f ϱ ϱ g ) = ( ϱ ϱ f − − ϱ ϱ g ) ( ϱ ϱ g + ϱ ϱ f + 2 ) ϱ ϱ f + ϱ ϱ g + 2 ϱ ϱ f ϱ ϱ g Although transcendental, this equation has a simple analytic, parametric solution obtained by writing the left side of the equation, which is just ( s g − − s f ) / R , as ln ⁡ ⁡ ( ϱ ϱ f ϱ ϱ g ) = s g − − s f R = δ δ = 2 y Then ϱ ϱ f = e δ δ ϱ ϱ g and when used to eliminate ϱ ϱ f from the right hand side a linear equation for ϱ ϱ g is obtained, whose solution is ϱ ϱ g ( y ) = f ( y ) e − − y and ϱ ϱ f ( y ) = f ( y ) e y where f ( y ) = y cosh ⁡ ⁡ y − − sinh ⁡ ⁡ y sinh ⁡ ⁡ y cosh ⁡ ⁡ y − − y Accordingly, the fundamental variable that specifies all the others in this phase transition process is ( s g − − s f ) / ( 2 R ) = y . This solution to the saturation problem is easily extended to encompass all its variables T s ( y ) = T ∗ ∗ 2 f ( y ) [ cosh ⁡ ⁡ y + f ( y ) ] g ( y ) 2 p s ( y ) = p ∗ ∗ f ( y ) 2 [ 1 − − f ( y ) 2 ] g ( y ) 2 v f ( y ) = v ∗ ∗ 1 + f ( y ) e y f ( y ) e y v g ( y ) = v ∗ ∗ 1 + f ( y ) e − − y f ( y ) e − − y where g ( y ) = [ 1 + ϱ ϱ g ( y ) ] [ 1 + ϱ ϱ f ( y ) ] = 1 + 2 f ( y ) cosh ⁡ ⁡ y + f ( y ) 2 .