Q. 7.1 - An Introduction to Thermal Physics

Question

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.

Step-by-Step Solution

Answer

Fraction of occupied heme states as a function of partial pressure of oxygen is correctly calculated and plotted in graph to obtain the curve.

Step 1. Formula

Gibb's Factor is given by formula:

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

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, $Gibb\text{'}s Factor =e^{-\left[\frac{(0-\mu \times 0)}{kT}\right]}$

                                $= e^0$

                                $= 1$

Step 2. Case Discussion

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 $N = 1$ .

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

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

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

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

So, grand partition function is:

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

Step 3. Potential Formula

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:

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

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

Step 4. Finding Probability

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

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

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

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

   

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

  

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

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

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

        

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

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:

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

Step 5. Calculation

For a box of width $1 cm$, 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:

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

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

   

  $= 5.38\times {10}^{-33} m^3$

Now, we will calculate value of $P_o$

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

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

       $= 0.00738 atm$

Step 6. Table and Graph

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

P	$\frac{P}{P_{o } + P}$
1	0.9926
2	0.9963
3	0.9975
4	0.9981
5	0.9985
6	0.9987
7	0.9989
8	0.9990
9	0.9991
10	0.9992

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

This graph curve is called Langmuir Adsorption Isotherm.