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Physical equilibria : pure substances

Physical equilibria : pure substances. 자연과학대학 화학과 박영동 교수. Physical equilibria : pure substances. 5.1 The thermodynamics of transition 5.1.1 The condition of stability 5.1.2 The variation of Gibbs energy with pressure 5.1.3 The variation of Gibbs energy with temperature

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Physical equilibria : pure substances

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  1. Physical equilibria: pure substances 자연과학대학 화학과 박영동 교수

  2. Physical equilibria: pure substances 5.1 The thermodynamics of transition 5.1.1 The condition of stability 5.1.2 The variation of Gibbs energy with pressure 5.1.3 The variation of Gibbs energy with temperature 5.2 Phase diagrams 5.2.4 Phase boundaries 5.2.5 The location of phase boundaries 5.2.6 Characteristic points 5.2.7 The phase rule 5.2.8 Phase diagrams of typical materials 5.2.9 The molecular structure of liquids

  3. The condition of stability When an amount dnof the substance changes from phase 1 with Gm(1) to phase 2 with Gm(2), • Gi = n1Gm(1) +n2Gm(2) • ΔG< 0 to be spontaneous • if Gm(2) > Gm(1), dn < 0; 2→1 • if Gm(2) < Gm(1), dn> 0;1→2 • if Gm(2) = Gm(1), at equilibrium • n2 + dn n2 • ΔG= {Gm(2) − Gm(1)}dn phase 2 phase 2 n1 n1 - dn Gf = (n1 - dn)Gm(1) +(n2+dn)Gm(2) phase 1 phase 1

  4. Pressure dependence of G G = H – TS dG = dH – TdS – SdT = Vdp - SdT = V For liquid or solid, ΔG = VΔp For vapor, ΔG = ∫Vdp = nRT∫(1/p)dp =nRTln(pf/pi) ΔGm=RT ln(pf/pi) ch05f01

  5. Standard Gibbs Energy, ΔG⦵(p) = V ch05f02

  6. Calculate the Vapor pressure increase of water, when the pressure is increased by 10 bar (Δp = 1.0 × 10 6 Pa) at 25°C. water: density 0.997 g cm-3at 25°C , molar volume 18.1 cm3 mol-1. Gm,i(g) Gm,f(g) vapor, pi vapor, pf Gm,i(l) Gm,f(l) water, p1 = 1 bar water, p2= 11 bar

  7. Temperature dependence of G G = H – TS dG = dH – TdS – SdT = Vdp - SdT = -S For liquid or solid, ΔGm=-Sm ΔT 1. Sm> 0, so G will decrease as T increases. 2. Sm(s) < Sm(l) <<Sm(g) ch05f03

  8. Phase Diagram E D C B A H F G H G F ch05f05

  9. Vapor pressure ch05f06

  10. Vapor pressure and Temperature ch05f07

  11. Cooling and Thermal Analysis ch05f05

  12. The location of phase boundaries and Clapeyron equation dGm(1) = Vm(1)dp − Sm(1)dT dGm(2) = Vm(2)dp − Sm(2)dT dGm(1) = dGm(2), ch05f09

  13. pvap(T ) and Clausius–Clapeyronequation

  14. The significant points of a phase diagram ch05f12

  15. (a) Use the Clapeyron equation to estimate the slope of the solid–liquid phase boundary of water given the enthalpy of fusion is 6.008 kJ mol−1 and the densities of ice and water at 0°C are 0.916 71 and 0.999 84 g cm−3, respectively. Hint: Express the entropy of fusion in terms of the enthalpy of fusion and the melting point of ice. (b) Estimate the pressure required to lower the melting point of ice by 1°C.

  16. Usual and Unusual Substances ch05f13

  17. The phase rule, F=C-P+2 ch05f14

  18. Water - phase diagram ch05f15

  19. Water ch05f16

  20. carbon dioxide ch05f18

  21. helium-4 superfluid flows without viscosity ch05f19

  22. 열역학 제1법칙은 많은 사실에 적용된다. 전압이 1.2V 인 어떤 건전지가 있다. 이 전지가 1A의 전류로 1시간 동안 소형 모터를 작동하는데 사용되었다. • 이 건전지의 일을 계산해 보시오. • 이 건전지의 내부에너지 변화를 계산해 보시오.

  23. 다음과 같은 관계가Cp와 Cv사이에 성립한다. 이는 또한 다음과 같이 표현할 수도 있다. 이 사실을 이용하여 van der Waals 기체에 대하여 다음 사실을 밝히고, 이 값을 CO2기체에 대하여 25℃, 1기압에서 계산해 보시오.

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