Calculate the canonical partition function, mean encrgy \(U\) and entropy \(S\), for a system having just two energy levels 0 and \(E\). If \(E=E(a)\) for a paramcter \(a\), calculate the force \(A\) and verify the thermodynamic relation \(\mathrm{dS}=\frac{1}{r}(\mathrm{~d} U+A \mathrm{da})\).

Short Answer

Expert verified
Canonical partition function, Z, is \(1 + e^{-E/k_BT}\). Mean energy, U, is \(E/(1+e^{\beta E})\). Using these values for Z and U, entropy can be calculated. To calculate A, write E as a function of parameter a and take derivative w.r.t a. Substitute these values in the given thermodynamic relation to check for its validity.

Step by step solution

01

Calculation of the canonical partition function

The canonical partition function, denoted as Z, is the sum of the exponentials of the energy divided by the product of Boltzmann constant \( k_B \) and temperature \( T \). In this case, there are two energy levels, 0 and E. So, \(Z = e^{0/k_BT} + e^{-E/k_BT} = 1 + e^{-E/k_BT}\)
02

Calculation of the mean energy, U

\(-\partial lnZ/\partial \beta\) where \( \beta = 1/k_BT\). \(\partial lnZ/\partial \beta = -Ee^{-\beta E}/Z\). Substituting the Z from step1, the mean energy \(U = -\partial lnZ/\partial \beta = E/(1+e^{\beta E})\).
03

Calculation of the entropy, S

With the partition function, Z, and the mean energy, U, we can calculate the entropy, S. Using the Thermodynamics definition of entropy, \( S = k_B*(lnZ + \beta U)\). Substitute Z and U from the previous steps and simplify to get the entropy.
04

Calculation of 'A' and verification of the Thermodynamic relation

Write the expression of Energy as function of paramter \(a\), say \(E= E(a)\). Force \(A\) is then defined as \( A = -dE/da\). Now take derivatives and substitute values of U, S and A in the given thermodynamic relation and check if both the sides are equal. If they are, then the thermodynamical relation is verified.

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