1 / 10

Lecture 10 Classical description of a single component system

Lecture 10 Classical description of a single component system. Gibbs free energy and equilibrium phases as a function of temperature Phase diagram Molar properties and Clapeyron equation Problem 11.4. Gibbs free energy.

mccormickl
Download Presentation

Lecture 10 Classical description of a single component system

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 10 Classical description of a single component system • Gibbs free energy and equilibrium phases as a function of temperature • Phase diagram • Molar properties and Clapeyron equation • Problem 11.4

  2. Gibbs free energy Typically we control pressure and temperature - therefore for a single component system to find equilibrium phase we need to consider Gibbs free energy, G (P, T). In general for each phase, : From chapter 4 we know that

  3. G as a function of T for a given pressure Entropy of vapor > entropy of liquid > entropy of solid vapor liquid solid At the melting point Gsolid = Gliquid At the boiling point Gliquid = Gvapor

  4. Enthalpy Since G = H-TS also vapor liquid H solid Latent heat - heat of fusion Heat of vaporization

  5. Phase diagram solid liquid vapor

  6. Molar properties - converting extensive into intensive variables Mole fraction Molar entropy etc.

  7. Gibbs-Duhem equation The original equation for each phase Dividing by Gives for two phases,  and 

  8. Clapeyron equation Along the coexistence line Also for a single component in  and  phases Therefore From which Since in equilibrium coexistence

  9. Coexistence with vapor slope

  10. Problem 11.4 Calculate the vapor pressure at 300 K of an element for which melting point is 1000 K, the normal boiling point is 2500 K, the heat of vaporization is 240, 000 J/mol and the heat of fusion is 12,000 J/mol.

More Related