# Solutions 14

- Page ID
- 204098

## Q1

a). The normalization constant is the inverse of the partition function, Z.

\[ \dfrac{1}{Z} = \dfrac{1}{e^{-E_0/kT} + e^{-E_1/kT} + e^{-E_2/kT}} = \dfrac{1}{e^0 + e^{-1} + e^{-2} } = 0.6652\]

b). We would have to explicitly include the degeneracy of 5 next to the \(E_1\) term:

\[ \dfrac{1}{Z} = \dfrac{1}{e^{-E_0/kT} + 5 e^{-E_1/kT} + e^{-E_2/kT}} = \dfrac{1}{e^0 + e^{-1} + e^{-2} } = 0.3362\]

c). \[P(E_1) = \dfrac{\rho(E_1) e^{-1}}{Z}\] where \(\rho(E_i)\) is the degeneracy or the weight of energy state, \(E_i\). First case: 0.2447 Second case: 0.6185

## Q2

Calculating the weights of the configurations involves correct interpretation of the multinomial distribution formula. We distribute the 5 particles into the 4 different energy bins (0, E, 2E, 3E, and 4E). Using the given formula \( W(a_1,a_2) = \dfrac{A!}{a_1!a_2!}\), we have \(A =5\) and each \(a_i\) represent the *population* *of each energy bin*. For example, in the first configuration, the population of the first bin is 5 while all other populations are 0.

Total Energy (a-f) |
Weight (W) |
---|---|

0 | \( \dfrac{5!}{5!0!0!0!0!} = 1\) |

E | \(\dfrac{5!}{4!1!0!0!0!} = 5\) |

2E | \(\dfrac{5!}{3!2!0!0!0!} = 10\) |

8E | \(\dfrac{5!}{1!0!4!0!0!}= 5\) |

20E | \(\dfrac{5!}{0!0!0!0!5!}=1\) |

9E | \(\dfrac{5!}{1!1!1!1!1!} = 120\) |

The state where the particles are spread out to different energies has the greatest weight. I would interpret this result as an indication for how molecules tend to distribute among energy states.

## Q3

From the weight \(W\) of each configuration, we can calculate the associated entropy \(S\) via

\[S= k\ln W\]

Weight |
\(S= k\ln W\) |
---|---|

\(\dfrac{5!}{5!0!0!0!0!} = 1\) | 0 |

\(\dfrac{5!}{4!1!0!0!0!} = 5\) | 0.2222 |

\(\dfrac{5!}{3!2!0!0!0!} = 10\) | 0.3179 |

\(\dfrac{5!}{1!0!4!0!0!}= 5\) | 0.2222 |

\(\dfrac{5!}{0!0!0!0!5!}=1\) | 0 |

\(\dfrac{5!}{1!1!1!1!1!} = 120\) | 0.6610 |

Notice that the zero entropy configuration is where all particles in the the lowest energy state. That is mathematical representation of the the third law of thermodynamics. Although technically if all the particles in an ensemble were in the **same **excited state instead, that would have zero entropy too (although with increased total energy). This emphasizes that entropy is not related to energy, but to dispersal of states.

## Q4

If \(kT = 2E\), and only listed configurations were possible, the partition function would be:

\[Z = \Sigma weight_{E_i} e^{E_i/kT} = e^{0} +5 e^{-.5} + 10e^{-1} + 5e^{-4} + e^{-10} + 120 e^{-4.5} = 9.1362 \]

Total Energy (a-f) |
Weights |
Probability: \(\dfrac{weight*e^{-E_i/kT}}{Z}\) |
---|---|---|

0 | \(\dfrac{5!}{5!0!0!0!0!} = 1\) | 0.1095 |

E | \(\dfrac{5!}{4!1!0!0!0!} = 5\) | 0.3319 |

2E | \(\dfrac{5!}{3!2!0!0!0!} = 10\) | 0.4027 |

8E | \(\dfrac{5!}{1!0!4!0!0!}= 5\) | 0.01 |

20E | \(\dfrac{5!}{0!0!0!0!5!}=1\) | 4.979e-6 |

9E | \(\dfrac{5!}{1!1!1!1!1!} = 120\) | 0.1459 |

## Q5

We are asked to calculate the partition function for different values of kT. We simply need to remember to include the degeneracy of each energy state (the weights).

\(kT\) | \(Z = \Sigma weight_ie^{-E_i/kT}\) |
---|---|

\(E\) | 4.2091 |

\(2E\) | 9.1362 |

\(3E\) | 16.04 |