State the Clapeyron equation and explain the two fundamental approximations that transform it into the Clausius-Clapeyron equation.
The exact Clapeyron equation derived from the equality of chemical potentials (dμ_liq = dμ_vap) is: dP/dT = ΔS_vap / ΔV_vap = ΔH_vap / [ T · (V_m,gas − V_m,liq) ], where V_m,gas and V_m,liq are the molar volumes of the gas and liquid phases. Rudolf Clausius applied two key physical approximations: (1) Liquid Molar Volume Neglect: Far below the critical point, the molar volume of a gas is typically 1000 to 2000 times larger than that of the liquid (for ethanol at 78 °C, V_m,gas ≈ 28.8 L/mol while V_m,liq ≈ 0.058 L/mol, making V_m,liq < 0.2% of V_m,gas). Thus, ΔV_vap = V_m,gas − V_m,liq ≈ V_m,gas. (2) Ideal Gas Law for the Vapor: The vapor phase is assumed to obey the ideal gas equation of state: V_m,gas ≈ R T / P. Substituting these approximations yields: dP/dT = (ΔH_vap · P) / (R T²), which rearranges into the differential Clausius-Clapeyron equation: d(ln P) / dT = ΔH_vap / (R T²), or d(ln P) / d(1/T) = −ΔH_vap / R.