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Gibb's Factor is given by formula:

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

Where, T is temperature, $k$is Boltzmann's constant,$渭$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" $Gibb\text{'}sFactor=e^{-\left[\frac{(0-渭脳0)}{kT}\right]}$

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 $N=1$.

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

$=e^{\left[\frac{(1+渭)}{kT}\right]}$

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

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

So, grand partition function is:

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

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

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

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

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

Formula of chemical potential is:

localid="1647112562133" $渭=-kT脳\ln \left(\frac{V{Z}_{in}}{N{谓}_Q}\right)$

We know that ideal gas equation is given by:

$PV=N脳k脳T$

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

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

So, $渭=-kT脳\ln \left(\frac{kT{Z}_{in}}{P{谓}_Q}\right)$

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

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

$P=\frac{e^{\frac{-(蔚-渭)}{kT}}}{Z}$

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

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

$=\frac{1}{1+e^{\frac{(蔚-渭)}{kT}}}$

$=\frac{1}{1+[e^{\frac{\left(蔚\right)}{kT}}脳e^{\frac{(-渭)}{kT}}]}$

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

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

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

We can write $P_o=\frac{kT{Z}_{in}}{{谓}_Q}脳e^{\frac{蔚}{kT}}$

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

$P=\frac{1}{1+\frac{kT{Z}_{in}}{P{谓}_Q}脳e^{\frac{蔚}{kT}}}$

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:

$l_Q=\frac{h}{\sqrt{2\mathrm{蟺尘办罢}}}$

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

${谓}_Q={\left(\frac{h}{\sqrt{2\mathrm{蟺尘办罢}}}\right)}^3$

${谓}_Q={\left(\frac{6.63脳{10}^{-34}J路s}{\sqrt{2蟺(32脳1.66脳{10}^{-27}kg)(1.38脳{10}^{-23}J/K)(310K)}}\right)}^3$

$=5.38脳{10}^{-33}{m}^3$

Now, we will calculate value of $P_o$

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

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

$=0.00738atm$

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.