KALMAN KNIZHNIK - KITTEL AND KROEMER SOLUTIONS K&K 2.1 Part a Suppose g(U ) = CU 3N/2 , where C is a constant and N is the number of particles. Show that U = 32 N τ . We use the definition of temperature as
∂σ 1 = ∂U N τ
(1)
So, let’s calculate σ. σ = ln(g) =
3N ln(U ) + ln(C) 2
(2)
Therefore, ∂σ 3N 1 = = ∂U 2U τ
(3)
3 U = Nτ 2 2
(4)
So finally,
Part b Show that ∂ 2 σ/∂U 2 |N is negative. This form of g(U ) actually applies to an ideal gas. ∂σ Referring to the definition of ∂U in equation 3 and taking another derivative: ∂2σ 3N =− 2
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