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PHYS 427_HE02_A_ClosedBook_Solutions

# PHYS 427_HE02_A_ClosedBook_Solutions - PHYS 427 2nd Midterm...

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Unformatted text preview: PHYS 427 2nd Midterm Exam‐Part A Solutions Fall 2008 PHYS 427 Thermal and Statistical Physics 2nd Midterm Exam-Part A Solutions Zuanyi Li Problem 1 The occupied density of orbitals is the product of the density of states D3 (ε ) and the Fermi distribution function f (ε ) . Its curves for both T=0 and T finite is shown below: Density of orbitals T=0 T is finite μ 0.0 0.5 ε/εF 1.0 1.5 Problem 2 The P-V diagram for a Carnot cycle is shown below: Where A and C are isothermal processes with temperature τ h , τ l , respectively; B and D are isentropic processes. 1 / 2 PHYS 427 2nd Midterm Exam‐Part A Solutions Fall 2008 Problem 3 One should not try to invent a perpetual motion machine of the second kind because it violates the second law of thermodynamics: “It is impossible for any cyclic process to occur whose sole effect is the extraction of heat from a reservoir and the performance of an equivalent amount of work”. Be specific, entropy ( σ h = Qh / τ h ) enters the system when it extracts heat ( Qh ) from a reservoir, and this entropy must ultimately be removed from the system because an ideal device or machine need to operate in a reversible cycle. However, work does not carry entropy. To prevent the accumulation of entropy there must be some output heat ( Ql = σ lτ l = σ hτ l = Qh (τ l / τ h ) ) to remove this entropy. Therefore, heat cannot be completely converted into work. Problem 4 dF = μ dN − σ dτ − PdV dG = μ dN − σ dτ + VdP F is useful to a system at constant volume, e.g., heating a box of gas. G is useful to a system at constant pressure, e.g., chemical reactions. Problem 5 FN is approximately independent of a magnetic field Ba while FS ∝ Ba2 . 2.5 2.0 Free energy density Fs 1.5 1.0 FN Bac 0.5 0.0 0.00 0.25 0.50 0.75 1.00 1.25 1.50 Ba/Bac 2 / 2 ...
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