Clapeyron and Clausius Clapeyron Equations To derive equations which describe lines of two phase equilibria in a 1-component phase diagram we begin with the differential expression for a change in the Gibbs’ free energy in an open system: dG = V dP - S dT + S ui dni Restricting our development to closed systems: dG = V dP - S dT Why did the 3rd term fall out for closed systems? Applying this equation to a general phase equilibria occuring at some fixed temperature and pressure: T, P phase 1 <=====> phase 2 we obtain: d (DG) = DV dP - DS dT Since at equilibrium, DG = 0, we get: 0 = DV dP - DS dT Solving for dP/dT yields the differential form of the Clapeyron equation, which describes any line of two phase equilibria in a closed 1-component phase diagram: dP / dT = DS / DV How is the derivative dP/dT related to lines of equilibria in a phase diagram which plots pressure versus temperature? 41.1

Using the fact that at equilibrium: 0 = DG = DH - T DS allows us to substitute for DS the expression: DS = DH / T to arrive at a 2nd version of the differential form of the Clapeyron equation: dP / dT = DH / (T DV) Separating variables and integrating from a reference pressure and temperature (P1, T1) to an arbitrary pressure and temperature (P, T): P1 P dP = T1  T (DH / T DV) dT gives the integrated form of the Clapeyron equation: P = P1 + (DH / DV) ln (T / T1) which could be used to plot how the pressure would vary with temperature along any line of two phase equilibria in a closed 1-component phase diagram. What assumption was made about DV and DH in evaluating the integral? 41.2

DVmelting = Vliquid - Vsolid = mass [1 / dliquid - 1 / dsolid] < 0 Let’s apply the Clapeyron equation to the melting of ice: H2O (s) -------> H2O (l) The enthalpy change for the melting will be endothermic: DHmelting > 0 The volume change on melting: DVmelting = Vliquid - Vsolid = mass [1 / dliquid - 1 / dsolid] < 0 will, however, be negative, since the density of ice is less than the density of liquid water. What are some of the consequences of the density of ice being less than the density of liquid water? These results imply that the slope of the line of equilibrium pressures and temperatures for the solid-liquid equilibrium for water will therefore be negative: dP / dT = (DH > 0) / [T (DV < 0)] < 0 liquid solid dP / dT < 0 pressure temperature Most substances expand on melting (water is nearly unique in that it contracts on melting). What would the slope of the solid-liquid equilibrium line be for a substance that expands on melting? 41.3

Will liquid He II float on or sink in liquid He I? He is the only pure element that exhibits two liquid phases one of which is the superfluid He II. The phase diagram for He is: 0.01 0.1 1 10 Pressure (bar) 100 liquid He I vapor Temperature (K) 2 3 4 5 liquid He II solid Will liquid He II float on or sink in liquid He I? A web site with a short description of the properties of superfluid He can be found at: http://www.eng.vt.edu/fluids/msc/super/super-f.htm 41.4

If one of the phases is the vapor phase: condensed phase -------> vapor phase the Clapeyron equation can be further developed by assuming that the vapor behaves ideally: DV = Vvapor - Vcondensed phase @ Vvapor = n R T / P Why is DV @ Vvapor? Substituting this value for DV into the differential form of the Clapeyron equation yields the differential form of the Clausius-Clapeyron equation: dP / dT = DH / (T DV) = P DH / (n R T2) Separating variables and integrating this equation from a reference pressure and temperature (P1, T1) to an arbitrary pressure and temperature (P, T): P1  P (1 / P) dP = T1  T (DH / n R T2) dT gives, assuming that DH is independent of temperature, the integrated form of the Clausius-Clapeyron equation: ln (P / P1) = - (DH / n R) [1 / T - 1 / T1] which could be used to plot how the pressure would vary with temperature along a line of two phase equilibria in a closed 1-component phase diagram in which one of the phases is a vapor phase. 41.5

log P (in torr) = a - b / T (in Kelvin) In the Chemical Rubber Company Handbook of Chemistry and Physics empirical equilibrium vapor pressure data for organic compounds are given in the form: log P (in torr) = a - b / T (in Kelvin) For the vaporization of benzene: C6H6 (l) ------> C6H6 (g) the normal boiling point is 353.2 K and the enthalpy of vaporization is + 8.146 kcal/mole. What are the constants “a” and “b” in the vapor pressure equation given above? Rearranging the integrated Clausius-Clapeyron equation: ln (P / P1) = - (DH / n R) [1 / T - 1 / T1] gives: log P = log P1 + (DH / 2.303 n R) (1 / T1) - (DH / 2.303 n R) (1 / T) Could you derive this equation from the one just above it? which identifies the constant “a” as: a = log P1 + (DH / 2.303 R) (1 / T1) = log (760) + [8.146 kcal/mole / (2.303)(0.001987 kcal/ mole K)(353.2 K)] = 7.921 and the constant “b” as: b = (DH / 2.303 n R) = [8.146 kcal/mole / (2.303)(0.001987 kcal/ mole K)] = 1,780 K 41.6

The values from the Handbook of Chemistry and Physics are: log P = 7.834 - 1780 / T and, once again, the agreement is pretty good. Use the following data for methane, CH4: DHovap, 111.7 K = + 8.17 kJ / mole Tovap = 111.7 K Tcritical = 190.6 K to calculate the critical point pressure for methane. 41.7