How can I calculate the root mean square speed of carbon dioxide molecules at STP?

Tuesday, February 7, 2012

How can I calculate the root mean square speed of carbon dioxide molecules at STP?

Hello!

As we know from the kinetic theory of gases, the mean kinetic energy of gas molecules is proportional to its absolute temperature $\displaystyle{T}.$ Because the kinetic energy of a molecule is $\displaystyle\frac{{{m}{{v}}^{{2}}}}{{2}},$ the root mean square is proportional to $\displaystyle\sqrt{{{T}}}.$ The exact formula is

$\displaystyle{v}_{{{r}{m}{s}}}=\sqrt{{\frac{{{3}{R}{T}}}{{M}}}},$

where $\displaystyle{R}\approx{8.3}\frac{{J}}{{{m}{o}{l}\cdot{K}}}$ is the ideal gas constant and $\displaystyle{M}$ is the molar mass of a gas. $\displaystyle{M}$ must be expressed in $\displaystyle\frac{{{k}{g}}}{{{m}{o}{l}}}.$

We know the relative atomic masses of $\displaystyle{O}$ and $\displaystyle{C}.$ They are $\displaystyle{16}$ and $\displaystyle{12},$ and we can compute the molar mass of $\displaystyle{C}{O}_{{2}}:$

$\displaystyle{12}+{2}\cdot{16}={44}{\left(\frac{{g}}{{{m}{o}{l}}}\right)}={4.4}\cdot{{10}}^{{-{2}}}{\left(\frac{{{k}{g}}}{{{m}{o}{l}}}\right)}.$

STP usually means $\displaystyle{0}$ degrees Celsius, or about $\displaystyle{273}{K}.$ Now we can obtain the numerical result:

$\displaystyle{v}_{{{r}{m}{s}}}=\sqrt{{\frac{{{3}\cdot{8.3}\cdot{273}}}{{{4.4}\cdot{{10}}^{{-{2}}}}}}}\approx{393}{\left(\frac{{m}}{{s}}\right)}.$

Ninenotes

Tuesday, February 7, 2012

How can I calculate the root mean square speed of carbon dioxide molecules at STP?

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