Kittel and Kroemer Thermal Physics

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