The three-term virial equation or a cubic virial equation of state Z = 1+B\rho+C\rho^2 has the simplicity of the Van der Waals equation of state without its singularity at . Theoretically, the second virial coefficient represents bimolecular attraction forces, and the third virial term represents the repulsive forces among three molecules in close contact. With this cubic virial equation, the coefficients B and C can be solved in closed form. Imposing the critical conditions: \frac{\mathrm{d}P}{\mathrm{d}v}=0 \qquad \text{and} \qquad \frac{\mathrm{d}^2P}{\mathrm{d}v^2}=0 the cubic virial equation can be solved to yield: B = -v_c , C = \frac{v_c^2}{3}, and Z_c = \frac{P_c v_c}{RT_c} = \frac 1 3. Z_c is therefore 0.333, compared to 0.375 from the Van der Waals equation. Between the critical point and the
triple point is the saturation region of fluids. In this region, the gaseous phase coexists with the liquid phase under saturation pressure P_\text{sat}, and the saturation temperature T_\text{sat}. Under the saturation pressure, the liquid phase has a molar volume of v_\text{l}, and the gaseous phase has a molar volume of v_\text{g}. The corresponding molar densities are \rho_\text{l} and \rho_\text{g}. These are the saturation properties needed to compute second and third virial coefficients. A valid equation of state must produce an isotherm which crosses the horizontal line of P_\text{sat} at v_\text{l} and v_\text{g}, on T_\text{sat}. Under P_\text{sat} and T_\text{sat}, gas is in equilibrium with liquid. This means that the PρT isotherm has three roots at P_\text{sat}. The cubic virial equation of state at T_\text{sat} is: P_\text{sat} = RT_\text{sat} \left(1 + B\rho + C\rho^2\right) \rho It can be rearranged as: 1 - \frac{RT_\text{sat}}{P_\text{sat}} \left(1 + B\rho + C\rho^2\right) \rho = 0 The factor RT_\text{sat}/P_\text{sat} is the volume of saturated gas according to the
ideal gas law, and can be given a unique name v^\text{id}: v^\text{id} = \frac{RT_\text{sat}}{P_\text{sat}} In the saturation region, the cubic equation has three roots, and can be written alternatively as: \left(1 - v_\text{l} \rho \right) \left(1 - v_\text{m} \rho \right) \left(1 - v_\text{g} \rho \right) = 0 which can be expanded as: 1 - \left(v_\text{l} + v_\text{g} + v_m\right)\rho + \left(v_\text{l} v_\text{g} + v_\text{g} v_\text{m} + v_\text{m} v_\text{l}\right)\rho^2 - v_\text{l} v_\text{g} v_\text{m} \rho^3 = 0 v_\text{m} is a volume of an unstable state between v_\text{l} and v_\text{g}. The cubic equations are identical. Therefore, from the linear terms in these equations, v_m can be solved: v_\text{m} = v^\text{id} - v_\text{l} - v_\text{g} From the quadratic terms,
B can be solved: B = -\frac{\left(v_\text{l} v_\text{g} + v_\text{g} v_\text{m} + v_\text{m} v_\text{l}\right)}{v^\text{id}} And from the cubic terms,
C can be solved: C = \frac{v_\text{l} v_\text{g} v_\text{m}}{v^\text{id}} Since v_\text{l}, v_\text{g} and P_\text{sat} have been tabulated for many fluids with T_\text{sat} as a parameter,
B and
C can be computed in the saturation region of these fluids. The results are generally in agreement with those computed from Benedict-Webb-Rubin and Starling equations of state. == See also ==