Physical equilibria: pure substances 자연과학대학 화학과 박영동 교수.

Physical equilibria: pure substances

자연과학대학 화학과박영동 교수

Physical equilibria: pure sub-stances

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

The condition of stabilityWhen an amount dn of the substance changes from phase 1 with Gm(1) to phase 2 with Gm(2),

Gi = n1Gm(1) +n2Gm(2)

Gf = (n1 - dn)Gm(1) +(n2+dn) Gm(2)

ΔG < 0 to be sponta-neous

if Gm(2) > Gm(1), dn < 0; 2→1

if Gm(2) < Gm(1), dn > 0; 1→2

if Gm(2) = Gm(1), at equilib-rium

n1

n2

phase 1

phase 2

n1 - dn

n2 + dn

phase 1

phase 2

ΔG = {Gm(2) − Gm(1)}dn

Pressure dependence of G

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G = H – TSdG = dH – TdS – SdT = Vdp - SdT

= V

For liquid or solid, ΔG = VΔp

For vapor, ΔG = ∫Vdp = nRT ∫(1/p)dp=nRT ln(pf/pi)

ΔGm = RT ln(pf/pi)

Standard Gibbs Energy, ΔG⦵(p)

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= V

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-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

Temperature dependence of G

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G = H – TSdG = dH – TdS – SdT = Vdp - SdT

= -S

ΔGm = -Sm ΔT

For liquid or solid,

1. Sm > 0, so G will decrease as T increases.

2. Sm(s) < Sm(l) <<Sm(g)

Phase Diagram

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G FH

GH F

D BE C A

Vapor pressure

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Vapor pressure and Tempera-ture

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Cooling and Thermal Analy-sis

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The location of phase boundaries and Clapeyron equation

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dGm(1) = Vm(1)dp − Sm(1)dT

dGm(2) = Vm(2)dp − Sm(2)dT

dGm(1) = dGm(2),

pvap(T ) and Clausius–Clapeyron equation

log p=A −BT

∆ ¿ln p=ln p ′+¿

∆vap H

R T ′ −∆vap H

R T¿

The significant points of a phase diagram

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(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.

Usual and Unusual Sub-stances

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The phase rule, F=C-P+2

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Water - phase diagram

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Water

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carbon dioxide

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helium-4

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superfluidflows without viscosity

열역학 제 1 법칙은 많은 사실에 적용된다 . 전압이 1.2V 인 어떤 건전지가 있다 . 이 전지가 1A 의 전류로 1 시간 동안 소형 모터를 작동하는데 사용되었다 .

a. 이 건전지의 일을 계산해 보시오 .b. 이 건전지의 내부에너지 변화를 계산해 보시오 .

다음과 같은 관계가 Cp 와 Cv 사이에 성립한다 .

이는 또한 다음과 같이 표현할 수도 있다 .

이 사실을 이용하여 van der Waals 기체에 대하여 다음 사실을 밝히고 , 이 값을 CO2 기체에 대하여 25℃, 1 기압에서 계산해 보시오 .