{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch7 - Advanced Topics in Equilibrium Statistical Mechanics...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Advanced Topics in Equilibrium Statistical Mechanics Glenn Fredrickson 7. Critical Phenomena The subject of critical phenomena , i.e., the study of the equilibrium and nonequi- librium properties of systems near a critical point (e.g., a gas-liquid critical point), has a long history, but many notable recent advances (culminating in a Nobel prize to Robert Wilson in 1982). A. Mean-Field Theory: The VDW fluid The simplest theories to construct of critical phenomena are so-called mean-field theories —approaches that average the environment around a given molecule, thus decoupling that molecule from its neighbors and making the statistical mechanics calculation simple, like that of an ideal gas . Indeed the Van der Waals EOS for a fluid can be “derived” using this sort of reasoning: p + a N V 2 [ V Nb ] = Nk B T and is one of the simplest MF theories of the gas-liquid critical point. Let’s begin by rewriting the equation in terms of ρ = N V rather than V : ( p + 2 )(1 ) = ρk B T. Qualitatively, we have: plane: pT plane: 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
plane: Let’s now locate the critical point in the VDW model. Clearly ∂p ∂ρ T = 0 , 2 p ∂ρ 2 T = 0 , VDW equation give three equations that can give p c , T c , ρ c in terms of a, b . Simpler is to write out the VDW equation as abρ 3 + 2 ( bp + k B T ) ρ + p = 0 or ρ 3 1 b ρ 2 + p a + k B T ab ρ p ab = 0 . Above T c , this has 1 real, 2 imaginary roots; these merge to 3 equal real roots at T c . Thus, at ( T c , p c ): ρ 3 1 b ρ 2 + p c a + k B T c ab ρ p c ab = 0 = ( p p c ) 3 = p 3 3 ρ 2 ρ c + 3 ρρ 2 c ρ 3 c ρ c = 1 3 b , p c = a 27 b 2 , k B T c = 8 a 27 b , Z c = β c p c ρ c = 3 8 Next, we introduce reduced variables: ρ r = ρ/ρ c , p r = p/p c , T r = T/T c which allows the VDW equation to be rewritten as: ( p r + 3 ρ 2 r )(3 ρ r ) = 8 T r ρ r . The absence of any material-dependent parameters is the basis for the familiar Law of Corresponding States ”, which you have undoubtedly studied in your thermo classes. Next, let’s zoom in on the critical region by further changing variables and expanding the VDW equation: t T T c T c = T r 1 “temp difference” ψ ρ ρ c ρ c = ρ r 1 “order parameter” (density difference) π p p c p c = p r 1 “pressure difference” 2
Image of page 2
[ π + 1 + 3( ψ + 1) 2 ][3 ( ψ + 1)] = 8( t + 1)( ψ + 1) This reduces to: π = 4 t 1+ ψ 1 1 2 ψ + 3 2 ψ 3 1 1 1 2 ψ π 4 t ( 1 + 3 2 ψ + . . . ) + 3 2 ψ 3 + . . . for | t | 1, | ψ | 1. Next, we do a thermodynamic integration to get the Helmholtz f.e.: write A V A V , A V = A V ( ρ, T ) Helmholtz f.e. per unit volume (intensive) p = ∂A ∂V N,T = A V V ∂A V ∂V N,T = A V V ∂ρ ∂V N ∂A V ∂ρ T = A V + ρ ∂A V ∂ρ T p c ( π + 1)= A V + (1 + ψ ) ∂A V ∂ψ t or in terms of a dimensionless f.e. density F A V /p c : F + (1 + ψ ) ∂F ∂ψ t = 1 + π ( t, ψ ) Integrate this ODE subject to F ( t, 0) F 0 ( t ) .
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern