Week 3 Friday October 12, 2012 page 1 ๐ ๐2 ๐ = ( 1)๐ถ๐,๐ ๐1 ๐2 ๐1๐1 ๐2๐2 = ๐๐๐ ๐๐๐๐๐๐๐ก ๐๐๐ ๐1 ๐2 ๐ ๐2 ๐2๐2 = ๐1 ๐1๐1 ๐ ๐ ๐ ๐ ๐ ๐ถ๐ ๐ ๐ ๐: 2 2 = ( 1)๐ถ๐,๐ = 1 ๐ , ๐1๐1 ๐2 ๐2๐ถ๐,๐ Assume CV,m is constant. ๐ ๐1๐1๐ถ๐,๐ ๐ ๐ถ๐ ๐ , +1 ๐ = ๐2๐2๐ถ๐,๐ +1 ๐ + ๐ถ๐ ๐ ๐ถ๐ ๐ , = , ๐ ๐๐๐๐ ๐ถ๐ ๐ − ๐ถ๐ ๐ = ๐ , , ๐ถ๐ ๐ ๐ถ๐ ๐ , , ๐ถ ๐พ = ๐, ๐ ๐พ ๐๐ ๐๐๐๐๐ (๐๐๐ก ๐๐๐๐๐๐ก๐๐๐) ๐ถ๐ ๐ , +1 = P1V1แต=P2V2แต Adiabatic process of increasing volume ends with lower pressure since the final temperature is lower. Lower T → Lower P q = 0 since adiabatic so โU=w โ๐ = ๐ถ๐ ∫ ๐๐ = ๐ค ๐2 โ๐ป = ∫ ๐ถ๐๐๐ = ๐ถ๐(๐2 − ๐1) ๐1 ๐๐ ๐ ๐ข๐๐๐๐ ๐๐๐๐ ๐ก๐๐๐ก ๐ถ๐ โ๐ป = โ๐ + โ(๐๐) = โ๐ = ๐๐ โ๐ ๐๐๐๐๐๐๐ก ๐๐๐ 1. Cyclic process a. โH=0 โT=0 โU=0 โP=0 โV=0 b. โฎUdU means cyclic integral c. โฎUdU=0 since U is a state function d. โฎdq isn’t necessarily 0 since q isn’t a state function 2. Reversible phase change at constant T and P a. Latent heat (of evaporation or melting…) b. q>0 if it’s melting c. q is latent heat (experiment, not from thermodynamics) d. q<0 if it’s freezing e. w=-PโV f. โ๐ ∝ ๐ ๐ g. โH=qP h. โU=qP+w 3. Constant pressure heating with no phase change 2 a. ๐ค = ๐ค๐๐๐ฃ = − ∫1 ๐๐๐ = −๐โ๐ ๐ ๐ ๐๐ ๐ ๐ b. โ๐ ∝ c. ๐ = ๐๐๐ ๐ ๐๐๐๐ ๐๐ ๐๐๐๐ข๐๐ ๐๐๐ ๐๐๐๐๐๐๐ก ๐๐๐ ๐ ๐ โ๐ = ๐2 − ๐1 ๐ d. โ๐ป = ๐๐ = ∫๐ 2 ๐ถ๐(๐)๐๐ ๐ถ๐(๐)๐๐ ๐ ๐๐๐๐๐๐๐ ๐ก๐ ๐ ๐ข๐๐ ๐ก๐๐๐๐ ๐๐๐ ๐โ๐๐ ๐ 1 e. โU=qP+w f. These formulas work as long as P1=P2 , not just constant P. 4. Constant volume heating with no phase change a. V=constant โV=0 w=0 โU=qV ๐ b. โ๐ = ∫๐ 2 ๐ถ๐๐๐ = ๐๐ 1 c. โH=โU+โ(PV)=โU+VโP=โU+nRโT only for perfect gas 5. Perfect gas change of state (T1V1 → T2V2) a. dU=CVdT perfect gas ๐ b. โ๐ = ∫๐ 2 ๐ถ๐๐๐ 1 c. ๐๐ป = ๐ถ๐๐๐ ๐ d. โ๐ป = ∫๐ 2 ๐ถ๐๐๐ 1 ๐ e. ๐ค = −๐๐๐ = −๐๐ ∫ ๐ ๐๐ ๐๐๐๐ ๐๐๐ก๐๐๐๐๐, ๐๐๐๐ ๐ = ๐(๐), ๐&๐ ๐๐๐ก ๐๐๐๐ ๐ก๐๐๐ก f. q=โU-w 6. Reversible isothermal process of a perfect gas a. T=constant b. โU=0 since U=U(T) for a perfect gas c. โH=0 2 ๐ ๐ d. ๐ค = − ∫1 ๐๐๐ = −๐๐ ๐๐๐ ๐2 = ๐๐ ๐๐๐ ๐1 1 e. q=-w 7. Reversible adiabatic process of a perfect gas a. q=0 โU=w ๐ b. โ๐ = ∫๐ 2 ๐ถ๐๐๐ ๐ ๐1 1 ๐ ๐ถ๐,๐ ๐ ๐2 c. ( 2) = ( 1) 2 ๐ โ๐ป = ∫๐ 2 ๐ถ๐๐๐ 1 ๐๐ ๐ ๐ข๐๐๐๐ ๐ถ๐ ๐ ๐๐ ๐๐๐๐ ๐ก๐๐๐ก , 8. Adiabatic expansion of a perfect gas into a vacuum (free expansion) a. w=0 q=0 โU=0 first law b. โH=โU+โ(PV)=โU+nRโT=0 c. also U=U(T) and โU=0 so โT=0 The molecular nature of internal energy U U=Utran+Urot+Uvib+Uelec+Uintermolecular+Urest mass Utran+Urot+Uvib are functions of T Uelec is constant at moderate T (<5000K) Urest mass is constant Motion in 3 dimensions: each axis is a degree of freedom thermodynamics is based on classical mechanics ๐ฅ: 1 ๐ ๐ 2 ๐ฆ: 1 ๐ ๐ 2 ๐ง: 1 ๐ ๐ 2 ๐ = 8.314 ๐ฝ ๐๐๐ ๐พ N is the # of atoms per molecule species Translation Rotation Vibration(1500K or higher) Atom 1.5 RT 0 0 Linear molecule 1.5 RT RT (3N-5)RT Nonlinear molecule 1.5 RT 1.5 RT (3N-6)RT For linear molecule, the rotation only has 2 degrees of freedom because rotation about the nuclear axis doesn’t contribute. N2,O2 and other diatomic linear molecules Um=1.5RT + RT + RT = 3.5RT (translation + rotation + vibration) ๐๐ ๐ (3.5๐ ๐) = 3.5๐ ๐ถ๐ ๐ = ( ๐ ) = , ๐๐ ๐ ๐๐ CO2 linear molecule but not diatomic Um=1.5RT+RT+4RT=6.5RT (translation + rotation + vibration) CV,m=6.5R ๐ถ๐ ๐ , = 6.5 ๐ ๐๐๐๐ ๐ก๐๐๐ก ๐๐ฆ ๐๐๐๐ ๐ ๐๐๐๐ ๐โ๐ฆ๐ ๐๐๐