Consider \(N=1000\) hypothetical molecules "M" each of which has three heme sites that can bind an oxygen molecule \(\mathrm{O}_{2}\). The binding energies \(E_{n}\) when \(n \mathrm{O}_{2}\) are bound are \(E_{0}=0, E_{1}=\) \(-0.49 \mathrm{eV}, E_{2}=-1.02 \mathrm{eV}\), and \(E_{3}=-1.51 \mathrm{eV}\). Assume that the " \(\mathrm{M}\) " molecules are in equilibrium with air at \(T=310 \mathrm{~K}\) and the partial pressure of \(\mathrm{O}_{2}\) in air is \(P_{\mathrm{O}_{2}}=0.2\) bar. Also assume that the " \(\mathrm{M}\), molecules don't interact with each other and air can be treated as an ideal gas. Of the \(N=1000\) "PP" molecules present, how many will have (a) zero \(\mathrm{O}_{2}\) molecules bound to them; (b) one \(\mathrm{O}_{2}\) molecule bound to them; (c) two \(\mathrm{O}_{2}\) molecules bound to them; (d) three \(\mathrm{O}_{2}\) molecules bound to them?

Step by step solution

01

- Understanding Equilibrium and Partition Function

First, determine the partition function to find the probability of each state. The partition function, denoted as Z, is the sum of the Boltzmann factor for each state. The Boltzmann factor is given by \(\text{e}^{-E_n/k_BT}\) where \(E_n\) is the energy of state n, \(k_B\) is the Boltzmann constant \(\approx 8.617 \times 10^{-5}\) eV/K, and T is the temperature.

02

- Calculate Boltzmann Factors

Calculate the Boltzmann factors for each state \(E_n\): \[E_0 = 0 \rightarrow \text{e}^{-E_0/k_BT} = 1\]\[E_1 = -0.49 \text{ eV} \rightarrow \text{e}^{-E_1/k_BT} = \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)}\]\[E_2 = -1.02 \text{ eV} \rightarrow \text{e}^{-E_2/k_BT} = \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)}\]\[E_3 = -1.51 \text{ eV} \rightarrow \text{e}^{-E_3/k_BT} = \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)}\]

03

- Calculate Partition Function (Z)

Sum the Boltzmann factors to calculate the partition function Z: \[Z = \text{e}^{-E_0/k_BT} + \text{e}^{-E_1/k_BT} + \text{e}^{-E_2/k_BT} + \text{e}^{-E_3/k_BT} = 1 + \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)} + \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)} + \text{e}^{0.49 / (8.617 \times 10^{-5} \times 310)} \]

04

- Compute Probabilities

Using the partition function, find the probability of each state n: \[P_n = \frac{\text{e}^{-E_n/k_BT}}{Z}\] (a) For zero \(O_2\) molecules: \[P_0 = \frac{\text{e}^{-E_0/k_BT}}{Z} = \frac{1}{Z}\](b) For one \(O_2\) molecule: \[P_1 = \frac{\text{e}^{-E_1/k_BT}}{Z}\](c) For two \(O_2\) molecules: \[P_2 = \frac{\text{e}^{-E_2/k_BT}}{Z}\](d) For three \(O_2\) molecules: \[P_3 = \frac{\text{e}^{-E_3/k_BT}}{Z}\]

05

- Calculate Number of Molecules

Multiply the probability of each state by the total number of molecules \(N = 1000\) to find how many molecules have that state:(a) For zero \(O_2\) molecules: \[ 1000 \times P_0\](b) For one \(O_2\) molecule: \[ 1000 \times P_1 \](c) For two \(O_2\) molecules: \[ 1000 \times P_2 \](d) For three \(O_2\) molecules: \[ 1000 \times P_3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partition Function

In statistical physics, the partition function is a vital concept. It helps us understand how particles distribute themselves among various energy states at a particular temperature. The partition function, denoted by Z, is essentially the sum of the Boltzmann factors of all possible states. The Boltzmann factor is given by \(e^{-E_n/(k_B \, T)}\), where \(E_n\) is the energy of the state, \(k_B\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin. This aggregate measure tells us how the molecules are spread across different energy levels. Calculating Z allows us to determine the probability of finding a molecule in a specific state, crucial for solving problems like the given exercise.

Boltzmann Factor

The Boltzmann factor is a term that appears frequently in statistical mechanics. It is an indication of the relative probability of a system being in a state with energy \(E_n\). The Boltzmann factor is expressed as \(e^{-E_n/(k_B \, T)}\). Here's what the terms mean:

• \(E_n\) is the energy of a particular state
• \(k_B\) is the Boltzmann constant, approximately \(8.617 \times 10^{-5}\) eV/K
• \(T\) is the temperature in Kelvin.

In practice, a high Boltzmann factor (closer to 1) means the state is more probable, while a low factor (closer to 0) means it is less probable. Each state's Boltzmann factor is crucial for calculating the partition function and subsequently the probabilities.

Binding Energy

Binding energy is the energy required to disassemble a whole system into separate parts. It tells us how strongly particles are bound together. In the context of our problem, binding energies are given for molecules with different numbers of bound oxygen molecules (O_2). Here's their significance:

• \(E_0 = 0\) eV: No O_2 molecules bound
• \(E_1 = -0.49\) eV: One O_2 molecule bound
• \(E_2 = -1.02\) eV: Two O_2 molecules bound
• \(E_3 = -1.51\) eV: Three O_2 molecules bound

Negative values indicate bound states because energy must be added to break the bond. In our problem, these binding energies are utilized to calculate the Boltzmann factors for each state.

Probability Distribution

The probability distribution in statistical physics helps determine the likelihood of a system existing in various energy states at thermal equilibrium. This is derived from the partition function, which we already discussed. The probability \(P_n\) of the system being in a state with energy \(E_n\) of interest in our problem is given by: \[ P_n = \frac{e^{-E_n/(k_B \, T)}}{Z} \]Here, the Boltzmann factor and partition function (Z) come into play. Let's break it down:

• The numerator \(e^{-E_n/(k_B \, T)}\) quantifies the relative likelihood of state \(n\).
• The denominator Z makes sure all probabilities sum up to 1 by acting as a normalizing factor.

This form of calculation helps us determine the fraction of molecules in each possible state.

Thermodynamic Equilibrium

Thermodynamic equilibrium is a state where all macroscopic flows of matter and energy have ceased, and all parts of the system are at the same temperature. This is a stable condition where the properties of the system do not change over time.
In our problem, equilibrium implies that the molecules are stabilized at a given temperature \(T\) and partial pressure. At this point, the number of molecules in each state remains constant over time. Given that the molecules don't interact and the system behaves like an ideal gas, the steady-state condition simplifies calculations through well-established principles like the Boltzmann distribution and partition function.