The Brunauer-Emmett-Teller (BET) surface area measurement data for adsorption of $\mathrm{N_2}$ gas at 77 K on 1.0 g of an adsorbent fits into a straight line, when $\dfrac{z}{(1-z)V}$ is plotted against $z$. The slope and intercept of the straight line are $6 \times 10^{-4}$ mm$^{-3}$ and $4 \times 10^{-6}$ mm$^{-3}$, respectively. The surface area (in m$^2$ g$^{-1}$) of the adsorbent is (rounded off to one decimal place).
(Given: 1 mm$^3$ of $\mathrm{N_2}$ gas corresponds to $2.7 \times 10^{16}$ molecules and each molecule occupies 0.16 nm$^2$. $V$ is the volume of gas adsorbed in mm$^3$, $z$ is $p/p_o$, where $p$ is the pressure of the $\mathrm{N_2}$ gas and $p_o$ is equilibrium vapour pressure of liquid $\mathrm{N_2}$)

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What is the polymer of elastomer?

Answer 1

Instead of finding $V_m$ and the molecule count as two separate steps, this route folds the volume-to-area conversion into a single constant first, then applies it once to $V_m$.

Get $V_m$ from the BET plot: for the linear BET plot $\frac{z}{(1-z)V}$ against $z$, the sum of slope and intercept always equals $1/V_m$ (the $C$ term cancels), so $V_m = \frac{1}{\text{slope}+\text{intercept}} = \frac{1}{6\times10^{-4}+4\times10^{-6}} = \frac{1}{6.04\times10^{-4}} = 1655.63\ \text{mm}^3$.
Build one conversion factor: each mm$^3$ of gas holds $2.7\times10^{16}$ molecules, and each molecule covers $0.16\ \text{nm}^2 = 1.6\times10^{-19}\ \text{m}^2$. So one mm$^3$ of monolayer gas corresponds to a surface area of $2.7\times10^{16}\times1.6\times10^{-19} = 4.32\times10^{-3}\ \text{m}^2$.
Apply the factor to $V_m$: $A = V_m \times 4.32\times10^{-3}\ \text{m}^2/\text{mm}^3 = 1655.63 \times 4.32\times10^{-3} = 7.152\ \text{m}^2$.
Report per gram: the whole measurement used a 1.0 g adsorbent sample, so this area is already the specific surface area, $7.152\ \text{m}^2\,\text{g}^{-1}$.

Let's summarize:

Slope plus intercept of the BET plot always gives $1/V_m$ directly, no need to solve for the BET constant $C$ separately.
Wrapping the molecules-per-volume and area-per-molecule data into one combined conversion factor avoids carrying two large exponents through the last step.

So the BET surface area, rounded to one decimal place, is $\boxed{7.2\ \text{m}^2\,\text{g}^{-1}}$.

Answer 2

Instead of finding $V_m$ and the molecule count as two separate steps, this route folds the volume-to-area conversion into a single constant first, then applies it once to $V_m$.

Get $V_m$ from the BET plot: for the linear BET plot $\frac{z}{(1-z)V}$ against $z$, the sum of slope and intercept always equals $1/V_m$ (the $C$ term cancels), so $V_m = \frac{1}{\text{slope}+\text{intercept}} = \frac{1}{6\times10^{-4}+4\times10^{-6}} = \frac{1}{6.04\times10^{-4}} = 1655.63\ \text{mm}^3$.
Build one conversion factor: each mm$^3$ of gas holds $2.7\times10^{16}$ molecules, and each molecule covers $0.16\ \text{nm}^2 = 1.6\times10^{-19}\ \text{m}^2$. So one mm$^3$ of monolayer gas corresponds to a surface area of $2.7\times10^{16}\times1.6\times10^{-19} = 4.32\times10^{-3}\ \text{m}^2$.
Apply the factor to $V_m$: $A = V_m \times 4.32\times10^{-3}\ \text{m}^2/\text{mm}^3 = 1655.63 \times 4.32\times10^{-3} = 7.152\ \text{m}^2$.
Report per gram: the whole measurement used a 1.0 g adsorbent sample, so this area is already the specific surface area, $7.152\ \text{m}^2\,\text{g}^{-1}$.

Let's summarize:

Slope plus intercept of the BET plot always gives $1/V_m$ directly, no need to solve for the BET constant $C$ separately.
Wrapping the molecules-per-volume and area-per-molecule data into one combined conversion factor avoids carrying two large exponents through the last step.

So the BET surface area, rounded to one decimal place, is $\boxed{7.2\ \text{m}^2\,\text{g}^{-1}}$.

Show Hint

Correct Answer: 7.2

Solution and Explanation

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