Answer:
The answers to the questions are;
(a) The data consistent with a Langmuir isotherm
(b) The adsorption equilibrium ratio (Kco) is 0.696 bar⁻¹ and
the volume of CO that is adsorbed at complete coverage, V[tex]_m[/tex] is 123.09 cm³
(c) The total surface area (S_a) of the piece of charcoal in m² per gram of charcoal is 66.16 m²/gm.
Explanation:
(a) Yes the data is consistent with Langmuir isotherm as it satisfies the following equation
[tex]V = \frac{V_m\times K \times P}{1 +K \times P}[/tex]
That is a plot of the data forms a straight line of the form [tex]\frac{P}{V} = \frac{1}{V_m\times K}+\frac{P}{V_m}[/tex] where the intercept is [tex]\frac{1}{V_m\times K}[/tex] and the slope is [tex]\frac{1}{V_m}[/tex]
The values of the plot are as follows;
Pco/V Pco (bar) V (cm³)
0.012943919 0.133322368 10.3
0.013815789 0.266644737 19.3
0.01465081 0.399967105 27.3
0.015638987 0.533289474 34.1
0.016665296 0.666611842 40
0.017580972 0.799934211 45.5
0.019442845 0.933256579 48
Slope 1/V[tex]_m[/tex] = 0.008124326 cm⁻³
V[tex]_m[/tex] = 123.0871369 cm³
1/(V[tex]_m[/tex]×K) = 0.011670999 bars/cm³
K =0.696112311 bar⁻¹
(b) From the graph, the slope is given as = 0.008124326, that is
[tex]\frac{1}{V_m}[/tex] = 0.008124326 and V[tex]_m[/tex] =[tex]\frac{1}{0.008124326 cm^{-3}}[/tex] = 123.09 cm³
The intercept is given as = 0.011670999
.
Therefore, [tex]\frac{1}{V_m\times K}[/tex] = 0.011670999 bars/cm³ and
K[tex]_{co}[/tex] = [tex]\frac{1}{123.09 cm^3*0.011670999 bars/cm^3}[/tex] = 0.696 bar⁻¹.
(c) The CO adsorption site density is given as 5 x 10¹⁵ cm⁻²
Therefore the we have the specific volume given as
[tex]\frac{V}{n} = \overline{\rm V} =\frac{RT}{P}[/tex] = [tex]\frac{(0.0821L \cdot atm/moles \cdot K)(273K)}{1 atm}[/tex] = 22.41 l
With a CO adsorption site density of 5 x 10¹⁵ cm⁻² we have
CO molecule cross sectional area, σ, given as [tex]\frac{1}{5*10^{15} cm^{-2}}[/tex] = 2 x 10⁻¹⁶ cm²
Therefore we have
Area A = [tex]\frac{\sigma \cdot N_A \cdot V_m}{\overline{\rm V}}[/tex] = [tex]\frac{(2\times 10^{-16} cm^2)(6.02\times 10^{23}mole ^{-1})(123.09 cm^3/gm)}{22400 cm^3}[/tex] = 6.62 x 10⁵ cm²/gm
= 66.16 m²/gm.
