Phases and Property Tables of Pure Substances

Saturated liquid-vapour mixture properties

A saturated liquid-vapour mixture (wet region) consists of saturated liquid and saturated vapour at the same pressure and temperature. Its state is described by the quality (dryness fraction) x = m_g/m, the vapour mass fraction, with 0 \le x \le 1. Any specific property y of the mixture follows the linear relation y = y_f + x\,y_{fg}, where y_{fg} = y_g - y_f. Thus v = v_f + x\,v_{fg}, u = u_f + x\,u_{fg}, h = h_f + x\,h_{fg} and s = s_f + x\,s_{fg}. The mixture is genuinely wet only when v_f \le v \le v_g, and the quality is recovered by inverting the volume relation, x = (v - v_f)/(v_g - v_f). In a closed container of volume V the specific volume is v = V/m, so the mass follows from m = V/v. Quality found this way then gives h and the total enthalpy H = mh.

When the two phases each occupy half of the total volume, the volume balance m_f v_f = m_g v_g gives x = v_f/(v_f + v_g) directly. When two saturated masses in separate compartments are combined, the total volume is V = \sum m_i v_i and the final specific volume is v = V/m, which then fixes the state. If liquid is drained at constant pressure from a rigid tank, the tank volume stays fixed, so the new quality is obtained from v = V/m_{\text{new}} using the same v_f and v_g. In a rigid closed tank no work is done, so Q = \Delta U with U = mu, and the final state is found from the specific volume and then u_2. In a sealed rigid vessel cooled at constant volume, v stays constant, the quality follows from v = v_f + x\,v_{fg}, and the liquid-to-vapour mass ratio is (1-x)/x.

At saturation the specific Gibbs energies of coexisting phases are equal, g_f = g_g, and this is an intensive property that does not depend on how much of each phase is present.

• quality (dryness fraction) — vapour mass fraction of a wet mixture. Example: a mixture with m_g = 0.3 kg in total m = 1 kg has x = 0.3.
• wet region — the saturated liquid-vapour zone under the dome, where v_f \le v \le v_g.
• intensive property — a property independent of the amount of substance, such as g, T or P.

Clausius-Clapeyron relation

The Clausius-Clapeyron relation links the slope of the saturation curve to the latent change of a phase transition. For two phases in equilibrium, \dfrac{dP_{sat}}{dT} = \dfrac{s_{fg}}{v_{fg}} = \dfrac{h_{fg}}{T\,v_{fg}}, where v_{fg} and s_{fg} are the differences in specific volume and specific entropy between the two phases. Rearranging gives h_{fg} = T\,v_{fg}\,(dP/dT)_{sat}, which is used to find a latent enthalpy from a measured slope and a known volume change.

If the saturation pressure is fitted as \ln P_{sat} = A - B/T, differentiation gives \dfrac{dP_{sat}}{dT} = \dfrac{B\,P_{sat}}{T^2}. Hence s_{fg} = v_{fg}\,B\,P_{sat}/T^2. Note that P_{sat} appears to the first power and the temperature enters as T^2 in the denominator.

For melting (solid-liquid), the same form reads \dfrac{dT}{dP} = \dfrac{T\,\Delta v}{L}, where \Delta v = v_{liq} - v_{sol} and L is the latent heat of fusion. If the liquid is denser than the solid, \Delta v < 0, so the melting point falls as pressure rises, as for ordinary water. With constant properties, \Delta T \approx T\,\Delta v\,\Delta P / L.

A substance that expands on freezing has \Delta v > 0 in the sense that its solid is less dense, so the solid-liquid line on a P-T diagram has negative slope. Liquid can then exist at temperatures below the triple-point temperature when the pressure is above the triple-point pressure. The triple-point temperature is the highest temperature at which melting can occur, and the critical pressure lies far above the triple-point pressure.

For a vapour with compressibility factor z, v_g = z\,R_u T/P, so \dfrac{d\ln P}{dT} = \dfrac{h_{fg}}{z\,R_u T^2}. Since d(1/T) = -dT/T^2, the slope of \ln P against 1/T is -h_{fg}/(z\,R_u). Its magnitude gives a quick estimate of latent heat from vapour-pressure data.

• triple point — the unique P-T state where solid, liquid and vapour coexist.
• latent heat of fusion — enthalpy change on melting at constant pressure, L = h_{liq} - h_{sol}.