- #1
NewtonApple
- 45
- 0
Homework Statement
A system of N distinguishable particles is arranged such that each particle can exist in one of the two states: one has energy \epsilon_{1}, the other has energy \epsilon_{2}. The populations of these states are n_{1} and n_{2} respectively, (N = n_{1}+n_{2}). The system is placed in contact with a heat bath at temperature T. A simple quantum process occurs in which the populations change: n_{2}\rightarrow n_{2} - 1 and n_{1}\rightarrow n_{1} + 1 with the energy released going into the heat bath.
(a) Calculate the change in the entropy of the two level system.
(b) Calculate the change in the entropy of the heat bath.
(c) If the process is reversible, what is the ratio of n_{2} to n_{1}?
Homework Equations
Boltzmann's Hypothesis - Entropy (S) is S=k_{B}ln(W)
Stirling's approximation for large factorials ln N! = N\,ln\,N - N
The Attempt at a Solution
The number of ways for initial state
W_{i}=\frac{N!}{n_{1}!\, n_{2}!}
The number of ways for final state
W_{f}=\frac{N!}{({n_{2}-1)!\, (n_{1}+1)!}}
Using S=k_{B}ln(W) Change in Entropy is
\Delta S_{2LS} = S_{f}-S_{i} = k_{B}\Big [ln W_{f} - ln W_{i}\Big ] = k_{B}\Big [\frac {ln W_{f}}{ln W_{i}}\Big ]
Substituting values from above and simplifying
\Delta S_{2LS} = S_{f}-S_{i} =k_{B} ln \Big[\frac{n_{1}!\, n_{2}!} {(n_{2}-1)!\, (n_{1}+1)!} \Big ]
Using Stirling's approximation ln N! = N\,ln\,N - N \Delta S_{2LS}=k_{B} \Big[ \frac{ \big(n_{1}\, ln(n_{1})-n_{1}))(n_{2}\, ln(n_{2})-n_{2})\big)}{ \big( (n_{2}-1)ln(n_{2}-1)-(n_{2}-1)\big ) \big((n_{1}+1)ln(n_{1}+1)-(n_{1}+1)\big) } \Big]
and simplifying
\Delta S_{2LS}=k_{B} [n_{1} \, ln(n_1) + n_{2} \, ln(n_2) - (n_{2}-1) \, ln(n_{2}-1) - (n_{1}+1) \, ln(n_{1} + 1)]
.
.
Suppose to get following Solution
\Delta S_{2LS}=k_{B}\, ln(n_{1}/n_{2})
No idea how to get it used all simplification techniques. Please give me some hints.
Last edited: