Peter Atkins • Julio de Paula Atkins’ Physical Chemistry

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Presentation transcript:

Peter Atkins • Julio de Paula Atkins’ Physical Chemistry Eighth Edition Chapter 3 – Lecture 3 The Second Law Copyright © 2006 by Peter Atkins and Julio de Paula

Concentrating on the System Consider a system in thermal equilibrium with its surroundings at constant T When heat is transferred between system and surroundings, the Clausius inequality reads: We can use this to develop expressions for sponteneity under constant volume and under constant pressure

Constant volume conditions dqV = dU so becomes which rearranges to Constant pressure conditions dqP = dH so becomes

Concentrating on the System We introduce two new thermo functions: Helmholtz energy (constant V) A ≡ U – TS → dA = dU - TdS → ΔAT,V = ΔU - TΔS Gibbs energy (constant P) G ≡ H – TS → dG = dH - TdS → ΔGT,P = ΔH - TΔS

The Gibbs Energy Criterion for a spontaneous process: dGT,P ≤ O based on the system alone “Free energy”: max additional (non-PV) work obtainable dwadd,max = dG or wadd,max = ΔG

Standard Molar Gibbs Energies where: and: Or:

Combining First and Second Laws Major objectives: Find relationships between various thermo properties Derive expressions for G(T,P) The Fundamental Equation From 1st Law: dU = dq + dw From 2nd Law: Substituting: dw = -PΔV and dq = TdS Gives: dU = TdS - PdV

Properties of the Internal Energy dU = TdS – PdV indicates that U = f (S,T) So the full differential of U: Indicates that: The first partial is a pure thermo definition of temperature

A Math Lesson An infinitesimal change in the function f(x,y) is: df = M dx + N dy where M and N are functions of x and y Criterion for df to be exact is: Euler Reciprocity relationship (p 968)

A Math Lesson (cont’d) dU = TdS – PdV must pass this test if it indeed is exact Applying the Euler Reciprocity: This equation is one of the four Maxwell relations df = M dx + N dy

The Maxwell Relations Table 3.5 Natural Variables Relation U(S,V) H(S,P) A(V,T) G(P,T)

Variation of Internal Energy with Volume Internal pressure Beginning with dU = TdS – PdV Dividing both sides by dV and imposing constant T Maxwell #3 here... A thermo equation of state

Properties of Gibbs Energy dG = dH – TdS So: G(H,T,S) and expanding: dG = dH – d(TS) = dH – TdS – SdT = dU + d(PV) – TdS – SdT = dU + PdV + VdP – TdS – SdT = TdS - PdV + PdV + VdP – TdS – SdT dG = VdP - SdT

Fig 3.18 Variation of Gibbs energy with T and with P dG = VdP - SdT

Fig 3.19 Variation of Gibbs energy with T depends on S

Fig 3.20 Variation of Gibbs energy with P depends on V