 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2016/paper-2/section-a/a1/solution

S = k_{B} ln Ω = k_{B} ln Ω_{0} + \frac{5 N}{2} k_{B} ln (E / E_{0}) .

The thermodynamic definition of temperature gives

\frac{1}{T} = (\frac{\partial S}{\partial E})_{N, \dots} = \frac{5 N k _{B}}{2 E}, T = \frac{2 E}{5 N k _{B}} .

Holding the other system parameters fixed,

E = (5 N /2) k_{B} T

, so its heat capacity is $C = 5 N k_{B} /2$ . The constants

Ω_{0}

and

E_{0}

set an additive reference for entropy; neither affects this derivative or the resulting heat capacity.

 New to topics? Read the docs here!