91Ó°ÊÓ

Gibb's Factor is given by formula:

$Gibb\text{'}sFactor=e^{\left[\frac{-(\mathrm{Î\mu }-n\mathrm{μ})}{kT}\right]}$

where, $k$is Boltzmann's constant, $T$is temperature, $\mathrm{μ}$is chemical potential, and $\mathrm{Î\mu }$is energy of state .

As, in hemoglobin molecule, oxygen can have four different states given by:

localid="1647144603783" $$\begin{gathered} {\mathrm{Î\mu }}_o=0eV \\ {\mathrm{Î\mu }}_1=-0.55eV(two\sin glyoccupied) \\ {\mathrm{Î\mu }}_2=-1.3eV \end{gathered}$$

The average number of oxygen molecules is given by,

$$\begin{gathered} \overline{N}=\underset{S}{∑}N\left(S\right) \\ =\frac{2e^{{\displaystyle \frac{-({\mathrm{Î\mu }}_1-\mathrm{μ})}{kT}}}}{Z}+\frac{2e^{{\displaystyle \frac{-({\mathrm{Î\mu }}_2-2\mathrm{μ})}{kT}}}}{Z} \end{gathered}$$

Occupancy of system is given by:

$$\begin{gathered} \overline{n}=\frac{\overline{N}}{Z} \\ =\frac{1}{Z}\left(e^{\frac{-({\mathrm{Î\mu }}_1-\mathrm{μ})}{kT}+\frac{-({\mathrm{Î\mu }}_2-2\mathrm{μ})}{kT}}\right) \end{gathered}$$

Chemical Potential formula is :

$\mathrm{μ}=-kT×\ln \left(\frac{V{Z}_{in}}{N{\mathrm{ν}}_Q}\right)$

Substitute $\frac{V}{N}=\frac{kT}{P}$in above formula,

$\mathrm{μ}=-kT×\ln \left(\frac{kT{Z}_{in}}{P{\mathrm{ν}}_Q}\right)$

$e^{\frac{\mathrm{μ}}{kT}}=\frac{P{\mathrm{ν}}_Q}{kT{Z}_{in}}$

Gibb's Factor can be written as:

$$\begin{gathered} e^{\left[\frac{-({\mathrm{Î\mu }}_1-\mathrm{μ})}{kT}\right]}=e^{\frac{-{\mathrm{Î\mu }}_1}{kT}}×\frac{P{\mathrm{ν}}_Q}{kT{Z}_{in}} \\ =\frac{P{\mathrm{ν}}_Q}{kT{Z}_{in}{e}^{\frac{{\mathrm{Î\mu }}_1}{kT}}} \\ =\frac{P}{P_o} \end{gathered}$$

Oxygen molecule quantum volume is:

${\mathrm{ν}}_Q={\left(\frac{h}{\sqrt{2\mathrm{Ï€³¾°ì°Õ}}}\right)}^3$

Substitute the values of the variables in above formula,

$$\begin{gathered} {\mathrm{ν}}_Q={\left(\frac{6.63×{10}^{-34}J·s}{\sqrt{2\mathrm{π}(32×1.66×{10}^{-27}kg)(1.38×{10}^{-23}J/K)(310K)}}\right)}^3 \\ =5.3×{10}^{-33}{m}^3 \end{gathered}$$

In equation, $\overline{n}=\frac{1}{Z}\left(e^{\frac{-({\mathrm{Î\mu }}_1-\mathrm{μ})}{kT}+\frac{-({\mathrm{Î\mu }}_2-2\mathrm{μ})}{kT}}\right)$

Substitute $e^{\frac{\mathrm{Î\mu }}{KT}}=1790$ and $P_o=2.03bar$at room temperature,

$\overline{n}=\frac{\left({\displaystyle \frac{P}{2.03}}+1790{\left({\displaystyle \frac{P}{2.03}}\right)}^2\right)}{\left(1+\frac{P}{2.03}+1790{\left(\frac{P}{2.03}\right)}^2\right)}$

Occupancy in this model is nearly $100\%$near lungs given $P=0.20bar$

The table is created between Pressure $P$and fraction of occupied sites:

The graph between fraction of occupied sites and oxygen partial pressure is created as,

We see that the occupancy rises gradually for smaller values of $P$. This shows that with cooperative binding, hemoglobin reacts with oxygen much fast in cells when $P$is small.