Question #3a307

Aug 2, 2017

$\text{Volume of CO you need to inject"=2*10^-2" } m L$

Explanation:

For ideal gases, we can say that the mole fraction, ${x}_{i}$, of each gas component is equal to the following:

${x}_{i} = {n}_{i} / {n}_{T} = {P}_{i} / {P}_{T} = {V}_{i} / {V}_{T}$

${n}_{i} = \text{number of moles of component i}$

${P}_{i} = \text{partial pressure of component i}$

${V}_{i} = \text{volume taken up by component i}$

${n}_{T} , {P}_{T} , {V}_{T} = \text{total moles, pressure, volume}$

Look up partial pressures and mole fractions if you are unsure about this. It is telling us a lot of useful information. Here's a link to the wiki:

https://en.wikipedia.org/wiki/Partial_pressure

And here is a useful example of defining ppm for gases:

https://www.co2meter.com/blogs/news/15164297-co2-gas-concentration-defined

So I will assume that ppm for gases means the following:

$10 \text{ ppm" = 10" ""molecules"/"million molecules"=10" } \frac{\mu L}{L}$

So we have 10 CO molecules for every 999,990 air molecules, or 10 moles of CO for every 999,990 moles of air.

The useful thing about ppm is that temperature and pressure don't matter (they cancel out).

See an explanation on the website below:

http://how-it-looks.blogspot.com.au/2010/07/how-to-convert-to-and-from-parts-per.html

This means that we don't need to worry about the temperature or pressure in our calculation, we can just use the mole fraction to determine what volume the CO will take up.

So using the mole fraction law above we have:

${x}_{i} = {n}_{i} / {n}_{T} = \frac{10}{10} ^ 6 = {10}^{-} 5 = {P}_{i} / {P}_{T} = {V}_{i} / {V}_{T}$

This means that

${10}^{-} 5 = {V}_{i} / {V}_{T}$

We know that ${V}_{T} = 2 \text{ } L$

$\therefore {V}_{i} = {10}^{-} 5 \cdot 2 = 2 \cdot {10}^{-} 5 \text{ "L=2*10^-2" } m L$

So the volume of air will be

$2 - \left(2 \cdot {10}^{-} 5\right) = 1.99998 \text{ } L$

We can also calculate the partial pressure of the CO:

${x}_{i} = {10}^{-} 5 = {P}_{i} / {P}_{T}$

${P}_{T} = 170 \text{ ""bar}$

$\therefore {P}_{i} = {10}^{-} 5 \cdot 170 = 1.7 \cdot {10}^{-} 3 \text{ ""bar}$

And the pressure of the air will be:

$170 - \left(1.7 \cdot {10}^{-} 3\right) = 169.9983 \text{ ""bar}$