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# Solutions5 - PHYS 333 Winter 2009 Assignment 5 1 A question...

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Unformatted text preview: PHYS 333 Winter 2009 Assignment 5 1. A question concerning the mathematics of the T dS equation. In class, and in the textbook, we have seen the T dS equation: T dS = dU + P dV In many traditional textbooks, this equation is derived from thermodynamics, and then used to derive the two identities ∂S ∂U V = 1 T and ∂S ∂V U = P T (a) There is one other identity that can be obtained directly from the T dS equation. What is it? Write dU = T dS- P dV so that ∂U ∂S V = T and ∂U ∂V S =- P The first of these is familiar, but the second is new. It could also be obtained by setting S constant, so that 0 = dU- ∂U ∂V S dV ≡ dU + P dV In class, and in the textbook, we introduced the enthalpy, H = U + P V . (b) There is a T dS equation that involves H and P , not U and V . What is it? We write dH = dU + P dV + V dP , and then substitute for dU in the T dS equation: T dS = dH- V dP (c) Use this equation to derive three more thermodynamic identities. They are ∂S ∂H P = 1 T ∂S ∂P H =- V T and ∂H ∂P S = V 1 The heat capacity at constant volume is defined as C V = ∂U ∂T V and that at constant pressure is C P = ∂H ∂T P (d) Show that C V = T ∂S ∂T V and that C P = T ∂S ∂T P The derivative ∂S ∂T V is not one of those that appears directly from the T dS equation. However, it can be written as ∂S ∂T V = ∂S ∂U V ∂U ∂T V so that ∂S ∂T V = C V T Similarly: ∂S ∂T P = ∂S ∂H P ∂H ∂T P so that ∂S ∂T P = C P T 2. Para-magnetism and adiabatic demagnetisation. Starting with Schroeder 3.23. The entropy of a 2-state para-magnet (as discussed in class, see also Schroeder, pages 98 – 107), is related to the multiplicity Ω given by Ω = N ! /N ↑ ! N ↓ ! The up-spins N ↑ and the down-spins N ↓ have, respectively, energies- μB and + μB in a magnetic field B . Thus the total magnetic energy is U M = μB ( N ↓- N ↑ ) and Schroeder shows on page 104 that U M =- NμB tanh ( μB/k B T ) However, the magnetisation is M = μ ( N ↑- N ↓ ) =- U M /B 2 which suggests that the analogy with a P-V system is U M → H not U M → U . [When there is no field, there is no energy: this is like saying that U = 0.] To complete the analogy, we note that the work done on a magnet by a field which increases the magnetisation by dM is ¯ dW = BdM so that B → P and M → - V The analogue of the T dS equation is therefore (see question 1) T dS = dU M + MdB (a) Show that the entropy can be written as S = Nk B [ln (2 cosh x )- x tanh x ] where x = μB/k B T ....
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Solutions5 - PHYS 333 Winter 2009 Assignment 5 1 A question...

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