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Direct Air Capture of CO 2 with Chemicals

Jun 1, 2011 **...** For illustrative purposes, the benchmark system assumes that the natural-**gas**-
derived CO2 emissions are captured at the kiln and then combined ...

For more information, see Direct Air Capture of CO 2 with Chemicals

I found an answer from www.britannica.com

**Avogadro's law** | Definition, Explanation, & Facts | Britannica

This empirical **relation** can be derived from the kinetic theory of gases under the
... For example, the **molecular** weight of oxygen is 32.00, so that one gram-mole
of ... The **volume** occupied by one gram-mole of gas is about 22.4 litres (0.791
cubic ... **physicist** who showed in what became known as **Avogadro's law** that,
under ...

For more information, see **Avogadro's law** | Definition, Explanation, & Facts | Britannica

Given that

Molar volume of ideal gas( H_2 ) = 22.4 \text{ litters } = 22.4 m^{3}

Diameter of the hydrogen molecule = 1 \text{ angstrom } (A^{^o})

Radius of hydrogen molecule = 0.5 * 10^{-10} meters

Avogadro constant = 6.023*10^{23}\ mol^{-1}

Step 1: Calculating one mole of hydrogen's atomic volume

Volume of hydrogen atom(Sphere) = \frac{4}{3} \pi r^3

=\frac{4}{3}*3.14*(0.5*10^{-10})^3

= 0.524 * 10^{-30} m^3

1 mole of hydrogen contains 6.023 * 10^{23} hydrogen atoms.

Volume of 1 mole of hydrogen atoms = 6.023 * 10^{23} * 0.524 * 10^{-30}

= 3.16* 10^{-7} m^3

Hence, total volume of mole of hydrogen atoms = 3.16* 10^{-7} m^3

Step 2: Determine the ratio of molar volume to the atomic volume

\frac{\text{ molar volume }}{\text{ atomic volume }}=\frac{22.4*10^{-3}}{3.16*10^{-7}}

\frac{\text{ molar volume }}{\text{ atomic volume }} = 7 * 10^{4}

Hence, \text{ molar volume } = \text{ atomic volume } * 7 * 10^{4}

This high ratio is due to the intermolecular gasses being significantly greater than the molecular size.