Near the cells where oxygen is used, its chemical potential is significantly lower than near the lungs. Even though there is no gaseous oxygen near these cells, it is customary to express the abundance of oxygen in terms of the partial pressure of gaseous oxygen that would be in equilibrium with the blood. Using the independent-site model just presented, with only oxygen present, calculate and plot the fraction of occupied heme sites as a function of the partial pressure of oxygen. This curve is called the Langmuir adsorption isotherm ("isotherm" because it's for a fixed temperature). Experiments show that adsorption by myosin follows the shape of this curve quite accurately.

Gibb's Factor is given by formula:

$Gibb\text{'}sFactor=e^{-\left[\frac{\left(E\right(s)-\mu N(s\left)\right)}{kT}\right]}$

Where, T is temperature, $k$is Boltzmann's constant,$\mu$is chemical potential, $N\left(s\right)$is number of state $s$atoms, $E\left(s\right)$is state$s$energy.

We consider the system as single donor atom. So, three cases are possible:

(1) Unoccupied state

Here, state energy and number of atoms are both equal to zero.

So, in formula (1) we put $N=0$and $E=0$.

So, localid="1647110725971" $Gibb\text{'}sFactor=e^{-\left[\frac{(0-\mu \times 0)}{kT}\right]}$

localid="1647110803077" $=e^0$

localid="1647110809090" $=1$

Second case possible is:

(2) States with two ionization

Here, state energy is $-1$and number of atoms is $1$.

In formula (1) we put $E=-1$and $N=1$.

$Gibb\text{'}sFactor=e^{-\left[\frac{(-1-\mu )}{kT}\right]}$

$=e^{\left[\frac{(1+\mu )}{kT}\right]}$

The degeneracy is $2$because there are two independent states of electron.

So, $Gibb\text{'}sFactor=2e^{\left[\frac{(1+\mu )}{kT}\right]}$

So, grand partition function is:

$Z=1+2e^{\left[\frac{(1+\mu )}{kT}\right]}$

We can say that probability of ionization of donor atom is:

$P_{ion}=\frac{1}{Z}$

We put $Z=1+2e^{\left[\frac{(1+\mu )}{kT}\right]}$in above equation,

$P_{ion}=\frac{1}{1+2e^{\left[\frac{(1+\mu )}{kT}\right]}}$

Formula of chemical potential is:

localid="1647112562133" $\mu =-kT\times \ln \left(\frac{V{Z}_{in}}{N{\nu }_Q}\right)$

We know that ideal gas equation is given by:

$PV=N\times k\times T$

So, $\frac{V}{N}=\frac{kT}{P}$

We put value of $\frac{kT}{P}$as $\frac{V}{N}$in above chemical potential formula:

So, $\mu =-kT\times \ln \left(\frac{kT{Z}_{in}}{P{\nu }_Q}\right)$

So, $e^{\frac{-\mu }{kT}}=\frac{kT{Z}_{in}}{P{\nu }_Q}$

Heme site is occupied by Oxygen $O_2$, the probability is:

$P=\frac{e^{\frac{-(\epsilon -\mu )}{kT}}}{Z}$

In above formula we put localid="1647113657590" $Z=1+e^{\frac{-(\epsilon -\mu )}{kT}}$

localid="1647113747407" $P=\frac{e^{\frac{-(\epsilon -\mu )}{kT}}}{1+e^{\frac{-(\epsilon -\mu )}{kT}}}$

$=\frac{1}{1+e^{\frac{(\epsilon -\mu )}{kT}}}$

$=\frac{1}{1+[e^{\frac{\left(\epsilon \right)}{kT}}\times e^{\frac{(-\mu )}{kT}}]}$

In above equation we put, $e^{\frac{(-\mu )}{kT}}=\frac{kT{Z}_{in}}{P{\nu }_Q}$

localid="1647114360985" $P=\frac{1}{1+[e^{\frac{\left(\epsilon \right)}{kT}}\times \frac{kT{Z}_{in}}{P{\nu }_Q}]}$

$=\frac{1}{1+{\displaystyle \frac{P_o}{P}}}$

We can write $P_o=\frac{kT{Z}_{in}}{{\nu }_Q}\times e^{\frac{\epsilon }{kT}}$

Therefore, Heme site is occupied by Oxygen $O_2$Probability is:

$P=\frac{1}{1+\frac{kT{Z}_{in}}{P{\nu }_Q}\times e^{\frac{\epsilon }{kT}}}$

For a box of width $1cm$, we will find temperature at which translation motion of $O_2$molecule freezes, the formula for quantum length is:

$l_Q=\frac{h}{\sqrt{2\mathrm{\pi mkT}}}$

Putting the values of variables in above expression, the volume is:

${\nu }_Q={\left(\frac{h}{\sqrt{2\mathrm{\pi mkT}}}\right)}^3$

${\nu }_Q={\left(\frac{6.63\times {10}^{-34}J·s}{\sqrt{2\pi (32\times 1.66\times {10}^{-27}kg)(1.38\times {10}^{-23}J/K)(310K)}}\right)}^3$

$=5.38\times {10}^{-33}{m}^3$

Now, we will calculate value of $P_o$

$P_o=\frac{(1.38\times {10}^{-23}J/K)(310K)\left(223\right)}{(5.4\times {10}^{-33}{m}^3)}\times e^{\frac{-0.7eV}{(8.617\times {10}^{-5}eV)(310K)}}$

$=738.33Pa\left(\frac{1atm}{{10}^{5}pa}\right)$

$=0.00738atm$

We make table for Pressure $p$against fraction of pressure $\frac{P}{P_o+P}$

We can now draw graph between fraction of occupied heme sites and oxygen partial pressure.

This graph curve is called Langmuir Adsorption Isotherm.