- #1
stunner5000pt
- 1,461
- 2
[itex] \tau = k_{B} T [/itex]
a) Find the expression for the free energy as a function of the temperature of the system with two states - one iwth eneryg zero and one with energy [itex] \epsilon_{0} [/itex]
b) From the free energy find the expressions for the energy and entropy of the system
c) Plot the average energy and the entropy as a function of tau. [itex] \tau = k_{B} T [/itex]
Ok for a) wek now that
[tex] F = U - T \sigma [/tex]
Partition fun ction [itex] Z = \sum_{s} \exp(-\epsilon_{s}/\tau) = 1 + \exp(-\epsilon_{0}/\tau) [/itex]
so then
[tex] U = \frac{\epsilon_{0} \exp(-\epsilon_{s}/\tau)}{1 + \epsilon_{0} \exp(-\epsilon_{s}/\tau)} [/tex]
but I am not quite sure how to proceed with the calculation of the entropy, sigma ...
for b)
for entropy use this
[tex] \sigma = \left(\frac{\partial F}{\partial \tau}\right)_{V} [/tex]
but not sure about how to find the nergy for hte system... is it simply the expression wh9ich doesn't involve tau??
I was thiking a bit more
isnt helmholtz free enryg given by simple
[tex] F- F(0) = -\tau \log Z [/tex]??
a) Find the expression for the free energy as a function of the temperature of the system with two states - one iwth eneryg zero and one with energy [itex] \epsilon_{0} [/itex]
b) From the free energy find the expressions for the energy and entropy of the system
c) Plot the average energy and the entropy as a function of tau. [itex] \tau = k_{B} T [/itex]
Ok for a) wek now that
[tex] F = U - T \sigma [/tex]
Partition fun ction [itex] Z = \sum_{s} \exp(-\epsilon_{s}/\tau) = 1 + \exp(-\epsilon_{0}/\tau) [/itex]
so then
[tex] U = \frac{\epsilon_{0} \exp(-\epsilon_{s}/\tau)}{1 + \epsilon_{0} \exp(-\epsilon_{s}/\tau)} [/tex]
but I am not quite sure how to proceed with the calculation of the entropy, sigma ...
for b)
for entropy use this
[tex] \sigma = \left(\frac{\partial F}{\partial \tau}\right)_{V} [/tex]
but not sure about how to find the nergy for hte system... is it simply the expression wh9ich doesn't involve tau??
I was thiking a bit more
isnt helmholtz free enryg given by simple
[tex] F- F(0) = -\tau \log Z [/tex]??
Last edited: