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Mathematical Methods Physics 313 Professor Lee Carkner Lecture 20
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Mathematical Thermodynamics Experiment or theory often produce relationships in forms that are inconvenient for the problem at hand Many differential equations are hard to compute
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Legendre Differential Transformation For an equation of the form: we can define, and get: We use a Legendre transform when f is not a convenient variable and we want xdu instead of udx e.g. replace PdV with -VdP
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Characteristic Functions The internal energy can be written: dU = -PdV +T dS We can use the Legendre transformation to find other expressions relating P, V, T and S We will deal specifically with the hydrostatic thermodynamic potential functions, which are all energies
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Enthalpy From dU = -PdV + T dS we can define: H is the enthalpy Since work is done in an isobaric process, enthalpy measures the energy needed to do work in this case Change in internal energy is a measure of the energy needed for a temperature change
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Using Enthalpy Enthalpy can be written: For isobaric reactions H = Q which is ~C T For an adiabatic process, H = V dP Energy carried by flowing fluid Can look up the flow work or heat needed to do isobaric work, etc.
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Helmholtz Function From dU = T dS - PdV we can define: A is called the Helmholtz function Used when T and V are convenient variables Can be related to the partition function
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Gibbs Function If we start with the enthalpy, dH = T dS +V dP, we can define: G is called the Gibbs function For isothermal and isobaric processes, G remains constant chemical reactions
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A PDE Theorem The characteristic functions are all equations of the form: or dz = M dx + N dy ( M/ y) x = ( N/ x) y
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Maxwell’s Relations We can apply the previous theorem to the four characteristic equations to get: ( T/ V) S = ( T/ P) S = ( S/ V) T = ( S/ P) T = We can also replace V and S (the extensive coordinates) with v and s per unit mass
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König - Born Diagram Example: What is expression for dU? dU = TdS-PdV ( T/ V) S =-( P/ S) V H G A U S P VT
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Using Maxwell’s Relations Example: finding entropy Using the last two Maxwell relations we can find the change in S by taking the derivative of P or V Example: Can read off “straddling” values on a table
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Key Equations We can use the characteristic equations and Maxwell’s relations to find key relations involving:
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Entropy Equations T dS = C V dT + T ( P/ T) V dV T dS = C P dT - T( V/ T) P dP Examples: Since = (1/V) ( V/ T) P, the second equation can be integrated to find the heat
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Internal Energy Equations ( U/ P) T = -T ( V/ T) P - P( V/ P) T Example:
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Heat Capacity Equations C P - C V = -T( V/ T) P 2 ( P/ V) T Examples: Volume always increases with T and pressure always decreases with V ( V/ T) P = 0 (when volume is at minima or maxima)
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