Engineering, 18.03.2020 22:36 salvadorjr1226p4zkp3

The data below were obtained for the adsorption of CO on charcoal at 273 K using an instrument that measures the volume (corrected to 1 atm pressure) of gas molecules that become adsorbed onto a surface as a function of its pressure in the surround gas phase. Such measurements are commonly performed in order to quantify the surface area of a material. The charcoal sample had a mass of 0.5 g and a CO adsorption site density of 5x10¹⁵ cm⁻².

Pco (torr) 100 200 300 400 500 600 700
V (cm³) 10.3 19.3 27.3 34.1 40 45.5 48

(a) Is the data consistent with a Langmuir isotherm? Support your answer. Hint: you will need to modify the Langmuir isotherm for a single species to convert θ_a to V.
(b) Determine the adsorption equilibrium ratio (Kco) and the volume of CO that is adsorbed at complete coverage.
(c) What is the total surface area (S_a) of the piece of charcoal in m² per gram of charcoal?

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Answer from: jenorajordan5387

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 is 123.09 cm³

Explanation:

(a) Yes the data is consistent with Langmuir isotherm as it satisfies the following equation

$V = \frac{V_m\times K \times P}{1 +K \times P}$

That is a plot of the data forms a straight line of the form $\frac{P}{V} = \frac{1}{V_m\times K}+\frac{P}{V_m}$ where the intercept is $\frac{1}{V_m\times K}$ and the slope is $\frac{1}{V_m}$

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 = 0.008124326 cm⁻³

V = 123.0871369 cm³

1/(V×K)= 0.011670999 bars/cm³

K =0.696112311 bar⁻¹

(b) From the graph, the slope is given as = 0.008124326, that is

$\frac{1}{V_m}$ = 0.008124326 and V = $\frac{1}{0.008124326 cm^{-3}}$ = 123.09 cm³

The intercept is given as = 0.011670999

Therefore, $\frac{1}{V_m\times K}$ = 0.011670999 bars/cm³ and

K $_{co}$ = $\frac{1}{123.09 cm^3*0.011670999 bars/cm^3}$ = 0.696 bar⁻¹.

Therefore the we have the specific volume given as

$\frac{V}{n} = \overline{\rm V} =\frac{RT}{P}$ = $\frac{(0.0821L \cdot atm/moles \cdot K)(273K)}{1 atm}$ = 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 $\frac{1}{5*10^{15} cm^{-2}}$ = 2 x 10⁻¹⁶ cm²

Therefore we have

Area A = $\frac{\sigma \cdot N_A \cdot V_m}{\overline{\rm V}}$ = $\frac{(2\times 10^{-16} cm^2)(6.02\times 10^{23}mole ^{-1})(123.09 cm^3/gm)}{22400 cm^3}$ = 6.62 x 10⁵ cm²/gm

= ‪66.16 m²/gm.

The data below were obtained for the adsorption of CO on charcoal at 273 K using an instrument that

Answer from: Quest

Ithink i did this class in 7th grade doe i don’t remember

Answer from: Quest

There is no names to check of

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