Activities
Problem
In carbon monoxide poisoning the CO replaces the \(\textsf{O}_{2}\) adsorbed on hemoglobin (\(\text{Hb}\)) molecules in the blood. To show the effect, consider a model for which each adsorption site on a heme may be vacant or may be occupied either with energy \(\varepsilon_A\) by one molecule \(\textsf{O}_{2}\) or with energy \(\varepsilon_B\) by one molecule CO. Let \(N\) fixed heme sites be in equilibrium with \(\textsf{O}_{2}\) and CO in the gas phases at concentrations such that the activities are \(\lambda(\text{O}_2) = 1\times 10^{-5}\) and \(\lambda(\text{CO}) = 1\times 10^{-7}\), all at body temperature \(37^\circ\text{C}\). Neglect any spin multiplicity factors.
First consider the system in the absence of CO. Evaluate \(\varepsilon_A\) such that 90 percent of the \(\text{Hb}\) sites are occupied by \(\textsf{O}_{2}\). Express the answer in eV per \(\textsf{O}_{2}\).
Now admit the CO under the specified conditions. Fine \(\varepsilon_B\) such that only 10% of the Hb sites are occupied by \(\textsf{O}_{2}\).
Problem
Suppose that a system of \(N\) atoms of type \(A\) is placed in diffusive contact with a system of \(N\) atoms of type \(B\) at the same temperature and volume.
Show that after diffusive equilibrium is reached the total entropy is increased by \(2Nk\ln 2\). The entropy increase \(2Nk\ln 2\) is known as the entropy of mixing.
If the atoms are identical (\(A=B\)), show that there is no increase in entropy when diffusive contact is established. The difference has been called the Gibbs paradox.
Since the Helmholtz free energy is lower for the mixed \(AB\) than for the separated \(A\) and \(B\), it should be possible to extract work from the mixing process. Construct a process that could extract work as the two gasses are mixed at fixed temperature. You will probably need to use walls that are permeable to one gas but not the other.
- Note
This course has not yet covered work, but it was covered in Energy and Entropy, so you may need to stretch your memory to finish part (c).