thermodynamic properties of pure gas and ideal gas mixture

1. isothermal G-P relation of pure gas, dG=-SdT+VdP at fixed T

∫dG=∫VdP=∫(RT/P)dP=∫_ᴘ₀^ᴾ RTdlnP, ∆G=G-G⁰=RT(lnP-lnP₀) i.e. Choose P₀=1 atm, pure gasG⁰=0

dG=RTdlnP → G=G⁰+RTlnP

2. ideal gas mixture, at fixed T, P and V

partial pressure pₐ, p_ᵇ, p_ᶜ.... pᵢ=XᵢP

partial molar quantity: Qᵢ≡(∂Q∕∂nᵢ)_{ᴛ,ᴘ,nʲ...} e.g. V, S, U, H, A, G; Gᵢ≡(∂G∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}=μᵢ chemical potential

§5-6/5-7 G=G(T,P,nᵢ,n_ʲ...)

dG=-SdT+VdP+(∂G∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}dnᵢ+(∂G∕∂n_ʲ)_ᴛ,ᴘ,nᵢ_...dn_ʲ+....=-SdT+VdP+∑(∂G∕∂n_ʲ)_{ᴛ,ᴘ,ᶜₒₘₚ}dnᵢ

=-SdT+VdP+∑ μᵢdnᵢ 限制條件_{ᴛ,ᴘ,nʲ...}G與 Qᵢ相同

dU=TdS-PdV+∑(∂U∕∂n_ʲ)_{S,ᴠ,ᶜₒₘₚ}dnᵢ, (∂U∕∂n_ʲ)_{S,ᴠ,ᶜₒₘₚ≠} Uᵢ=(∂U∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}

dH=TdS+VdP+∑(∂H∕∂n_ʲ)_{S,ᴘ,ᶜₒₘₚ}dnᵢ, (∂H∕∂n_ʲ)_{S,ᴘ,ᶜₒₘₚ≠} Hᵢ=(∂H∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}

dA=-SdT-PdV+∑(∂A∕∂n_ʲ)_{ᴛ,ᴠ,ᶜₒₘₚ}dnᵢ, (∂A∕∂n_ʲ)_{ᴛ,ᴠ,ᶜₒₘₚ≠} Aᵢ=(∂A∕∂nᵢ)_{ᴛ,ᴘ,nʲ..}

3. pure gas (∂G∕∂P)_ᴛ=V

gas mixture: for i component, (∂Gᵢ∕∂P)_ᴛ,nᵢ_,nʲ...= Vᵢ

pf: (∂G∕∂P)_{ᴛ,ᶜₒₘₚ=V,} Vᵢ≡(∂V∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}=(∂[(∂G∕∂P)_{ᴛ,ᶜₒₘₚ}]/∂nᵢ)_{ᴛ,ᴘ,nʲ...}=(∂[(∂G∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}]/∂P)_{ᴛ,ᶜₒₘₚ}

₌(∂Gᵢ∕∂P)_{ᴛ,ᶜₒₘₚ...} ⸫(∂Gᵢ∕∂P)_ᴛ,nᵢ_,nʲ...= Vᵢ

因此pure gas的Gibbs free energy式子推導都可適用於gas mixture.

Pure gas: dG=RTdlnP → G=G⁰+RTlnP vs. Gas mixture: dGᵢ=RTdlnpᵢ → Gᵢ=Gᵢ⁰+RTlnpᵢ

Vᵢ₌(∂Gᵢ∕∂P)_{ᴛ,ᶜₒₘₚ...} → ∫_ᴳᵢᵒ^ᴳͥdGᵢ =∫_ᴘᵢ^pᵢVᵢdP=∫_ᴘᵢ^pᵢ(XᵢRT/pᵢ)∙(dpᵢ/Xᵢ)=∫_ᴘᵢ^pᵢ RTdlnpᵢ

→ Gᵢ-Gᵢ⁰=RT(lnpᵢ-lnPᵢ)

i.e. Vᵢ≡(∂V∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}=[∂(nᵢRT/P)∕∂nᵢ]_{ᴛ,ᴘ,nʲ...}=RT/P=RT/(pᵢ/Xᵢ)=XᵢRT/pᵢ

⸫∆Gᵢ=Gᵢ-Gᵢ⁰=RTlnpᵢ choose Pᵢ=1 atm, ∆Gₘᵢₓ=∑(nᵢ∆Gᵢ+n_ʲ∆G_ʲ+...)

i=a,b,c... 混合前Gₐ=Gₐ⁰+RTlnPₐ → Gᵢ=Gᵢ⁰+RTlnPᵢ(通式)

混合後Gₐ=Gₐ⁰+RTlnpₐ → Gᵢ=Gᵢ⁰+RTlnpᵢ(通式)

∆Gᵢ=Gᵢ-Gᵢ=RTln(pᵢ/Pᵢ)=RTln(pᵢ/P)= RTlnXᵢ

∆Gₘᵢₓ=nₐ∆Gₐ+n_ᵇ∆G_ᵇ+n_ᶜ∆G_ᶜ+...=∑nᵢ∆Gᵢ ⸫∆Gₘᵢₓ=RT∑nᵢlnXᵢ or ∆Gₘᵢₓ=RT∑XᵢlnXᵢ<0 ⸪Xᵢ<1

∆Hₘᵢₓ=∑nᵢ∆Hᵢ=∑nᵢ(Hᵢ-Hᵢ)

Gᵢ已知, 求Hᵢ=? 利用Gibbs-Helmholtz eq. ⸫[∂(Gᵢ/T)∕∂T]ₚ=-Hᵢ/T²

⸪ Gᵢ=Gᵢ⁰+RTlnpᵢ → Gᵢ/T=Gᵢ⁰/T+Rlnpᵢ → [∂(Gᵢ/T)∕∂T]ₚ=[∂(Gᵢ⁰/T)∕∂T]ₚ+[∂(Rlnpᵢ)∕∂T]ₚ i.e.pᵢ=XᵢP

→ -Hᵢ/T²=-Hᵢ⁰/T² → Hᵢ=Hᵢ⁰, 同理 Hᵢ=Hᵢ⁰ ⸫ Hᵢ-Hᵢ=0, ∆Hₘᵢₓ=0

∆Gₘᵢₓ=∆Hₘᵢₓ-T∆Sₘᵢₓ at fixed T, ∆Sₘᵢₓ=-∆Gₘᵢₓ/T=-R∑nᵢlnXᵢ or ∆Sₘᵢₓ=- R∑XᵢlnXᵢ>0

∆Vₘᵢₓ=∑nᵢ∆Vᵢ=∑nᵢ(Vᵢ-Vᵢ)=0

Vᵢ≡(∂V∕∂nᵢ)_{ᴛ,ᴘ,nʲ...}=[∂(nᵢRT/P)∕∂nᵢ]_{ᴛ,ᴘ,nʲ...}=RT/P, Vᵢ=nᵢRT/Pᵢ=nᵢRT/P → Vᵢ=RT/P

∆Uₘᵢₓ=? ⸪∆Hₘᵢₓ=∆Uₘᵢₓ+P∆Vₘᵢₓ ⸫∆Uₘᵢₓ=0

conclusion: one mole ∑Xᵢ=1

∆Gₘᵢₓ=RT∑XᵢlnXᵢ, ∆Sₘᵢₓ=- R∑XᵢlnXᵢ, ∆Hₘᵢₓ=0, ∆Vₘᵢₓ=0, ∆Uₘᵢₓ=0 ideal solution的結果相同

§real gas properties

pure(ideal) gas dG=RTdlnP  vs. Real gas dG≡RTdlnf f≠P

gas mixture      dGᵢ=RTdlnpᵢ vs.                dGᵢ=RTdlnfᵢ (混合前fᵢ⁰; 混合後fᵢ)

Gᵢ=Gᵢ⁰+RTlnfᵢ let fᵢ⁰=1 atm, ∆Gᵢ=Gᵢ-Gᵢ⁰=RTlnfᵢ, ∆Gₘᵢₓ=RT∑Xᵢln(fᵢ/f)