Physical transformations of pure substances

Presentation on theme: "Physical transformations of pure substances"— Presentation transcript:

1 Physical transformations of pure substances
자연과학대학 화학과 박영동 교수

2 Physical transformations of 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 Phases

4 The condition of stability
phase 1 phase 2 When an amount dn of 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 ΔG = {Gm(2) − Gm(1)}dn n1 - dn n2 + dn phase 1 phase 2 Gf = (n1 - dn)Gm(1) +(n2+dn) Gm(2)

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

6 Standard Gibbs Energy, ΔG⦵(p)
ΔGm = RT ln(pf/pi)

7 water: density 0.997 g cm-3 at 25°C , molar volume 18.1 cm3 mol-1.
Calculate the Vapor pressure increase of water, when the pressure is increased by 10 bar (Δp = 1.0 × 106 Pa) at 25°C. water: density g cm-3 at 25°C , molar volume 18.1 cm3 mol-1. Gm,i(l) Gm,i(g) water, p1 = 1 bar vapor, pi Gm,f(l) Gm,f(g) water, p2= 11 bar vapor, pf

8 Temperature dependence of G
G = H – TS dG = dH – TdS – SdT = Vdp - SdT ( 𝜕𝐺 𝜕𝑇 ) 𝑝 = -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)

9 Phase Diagram D B E C A G H F H G F

10 Vapor pressure

11 Chemical potential and equilibrium

12 Vapor pressure and Temperature

13 Cooling and Thermal Analysis
ch05f05

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

15 pvap(T ) and Clausius–Clapeyron equation
∆( ln 𝑝)= ∆ 𝑣𝑎𝑝 𝐻 𝑅 𝑇 2 ×∆𝑇 ln 𝑝= ln 𝑝′+ ∆ 𝑣𝑎𝑝 𝐻 𝑅 𝑇 ′ − ∆ 𝑣𝑎𝑝 𝐻 𝑅𝑇 log 𝑝=𝐴 − 𝐵 𝑇

16 The significant points of a phase diagram

17 (a) Use the Clapeyron equation to estimate the slope of the solid–liquid phase boundary of water given the enthalpy of fusion is kJ mol−1 and the densities of ice and water at 0°C are and 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.

18 Usual and Unusual Substances

19 The phase rule, F=C-P+2

20 carbon dioxide

21 Water - phase diagram

22 Water

23 helium-4 superfluid flows without viscosity

24 열역학 제1법칙은 많은 사실에 적용된다. 전압이 1. 2V 인 어떤 건전지가 있다
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.

25 The Maxwell relations 𝑈 𝑆, 𝑉 ;𝑑𝑈=𝑇𝑑𝑆 −𝑝𝑑𝑉;𝑇= 𝜕𝑈 𝜕𝑆 𝑉 ;𝑝=− 𝜕𝑈 𝜕𝑉 𝑆 ; 𝜕𝑇 𝜕𝑉 𝑆 =− 𝜕𝑝 𝜕𝑆 𝑉 𝐻 𝑉, 𝑇 ;𝑑𝐻=𝑇𝑑𝑆+𝑉𝑑𝑝; 𝑇= 𝜕𝐻 𝜕𝑆 𝑝 ; 𝑉= 𝜕𝐻 𝜕𝑝 𝑆 ; 𝜕𝑇 𝜕𝑝 𝑆 = 𝜕𝑉 𝜕𝑆 𝑝 𝐺 𝑉, 𝑇 ;𝑑𝐺=𝑉𝑑𝑝 −𝑆𝑑𝑇; 𝑉= 𝜕𝐺 𝜕𝑝 𝑇 ; 𝑆=− 𝜕𝐺 𝜕𝑇 𝑝 ; 𝜕𝑉 𝜕𝑇 𝑝 = − 𝜕𝑆 𝜕𝑝 𝑇 𝐴 𝑉, 𝑇 ;𝑑𝐴=−𝑝𝑑𝑉 −𝑆𝑑𝑇;𝑝=− 𝜕𝐴 𝜕𝑉 𝑇 ;𝑆=− 𝜕𝐴 𝜕𝑇 𝑉 ; 𝜕𝑝 𝜕𝑇 𝑉 = 𝜕𝑆 𝜕𝑉 𝑇

26 ( 𝜕𝐺 𝜕𝑝 ) 𝑇 = V ( 𝜕𝐺 𝜕𝑇 ) 𝑝 = -S