Clausius-Clapeyron Equation: Understanding Phase Transitions and Vapor Pressure, Study notes of Physical Chemistry

Download Clausius-Clapeyron Equation: Understanding Phase Transitions and Vapor Pressure and more Study notes Physical Chemistry in PDF only on Docsity! 1 Lecture 24 Chapt 14 Clausius-Clapeyron Equation Announcements: We’ll review for exam this afternoon – no quiz this week. Outline: Clapeyron Equation VT H V S dT dp m m φ φ ∆ ∆ = ∆ ∆= Clausius – Clapeyron Equation 2 1 nRT H dT dp p trans∆= ( ) constant ln +∆−= RT Hp trans Review Phase Diagrams Summarize Solid-Liquid-Gas Behavior of a Substance ∂ ∂ ∂ ∂ G p V and G T S T p       =     = − (equating coefficients in expressions for dG) Therefore, if we plot G vs. T, what is the slope? negative of the (molar) entropy G T TMelt TVap at fixed p solid liquid gas If we repeat for many pressures, then we can build up a full phase diagram – a P vs. T plot showing regions of thermodynamic stability. We have a generic one below: 2 solid liquid gas P T critical point Let’s look at phase transitions a little more quantitatively. Equilibrium: ∆G must be 0, so between two phases, α and β: µα(p,T) = µβ(p,T) Therefore, there is no thermodynamic driving force for change We can use this equality to get an equation for the phase boundary. If p and T are changed infinitesimally in a way that the two phases α and β remain in equilibrium. Then µα = µβ before the change µα + dµα= µβ + dµβ after the change and therefore dµα = dµβ (or equivalently for two phases of a pure substance) dGα,m = dGβ,m plug in (fundamental equation?) dGm = -SmdT + Vmdp for each phase: -Sα,mdT +Vα,mdp = -Sβ,mdT +Vβ,mdp and this rearranges (Vβ,m - Vα,m)dp = (Sβ,m - Sα,m)dT For measurable changes, this expression becomes the Clapeyron Equation: dp dT S V m m = ∆ ∆ Notice that this is a total derivative and applies for any phase change for any pure substance.