# Question 0d4c9

Mar 13, 2016

$\text{476 mmHg}$

#### Explanation:

The idea here is that you need to use the composition of air to find the partial pressure of oxygen that would correspond to a total air pressure of $\text{650. mmHg}$.

This will allow you to find the total cabin pressure at which the oxygen masks are dropped.

So, the mole percent of oxygen in air is approximately 21%. This tells you that out of very $100$ moles of gases that make up air, $21$ will be moles of oxygen.

You can find the partial pressure of oxygen in terms of the total pressure in the cabin bu using the ideal gas law equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} P V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$P$ - the pressure of the gas
$V$ - the volume it occupies
$n$ - the number of moles of gas
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

If you take ${P}_{\text{total}}$ to be the total pressure of the air, and ${n}_{\text{total}}$ to be the total number of moles of gas present in the cabin, you can say that

${P}_{\text{total" = n_"total}} \cdot \frac{R T}{V}$

If you were to isolate the oxygen in the same cabin, i.e. in the same volume $V$, and at the same temperature $T$, you could write, using ${n}_{{O}_{2}}$ as the number of moles of oxygen

${P}_{{O}_{2}} = {n}_{{O}_{2}} \cdot \frac{R T}{V}$

These two equations will be equivalent to

color(purple)(|bar(ul(color(white)(a/a)color(black)(P_"total"/n_"total" = P_(O_2)/n_(O_2) implies P_(O_2) = n_(O_2)/n_"total" * P_"total")color(white)(a/a)|)))" " " "color(red)("(*)")

But you know that the mole fraction of oxygen, ${\chi}_{{O}_{2}}$, is defined as the number of moles of oxygen divided by the total number of moles of gas present in the mixture.

${\chi}_{{O}_{2}} = {n}_{{O}_{2}} / {n}_{\text{total}}$

Moreover, mole percent is simply mole fraction multiplied by $100$

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{mole %" = "mole fraction} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that you have

${n}_{{O}_{2}} / {n}_{\text{total" = "mole %}} / 100 = \frac{21}{100} = 0.21$

Plug this into equation $\textcolor{red}{\text{(*)}}$ to get the partial pressure of oxygen

${P}_{{O}_{2}} = 0.2 \cdot \text{650. mmHg" = "136.5 mmHg}$

Now, when the partial pressure of oxygen drops below $\text{100. mmHg}$, oxygen masks are released. Use the normal partial pressure of oxygen to help you find the total pressure that would correspond to a partial pressure of oxygen of $\text{100. mmHg}$

100color(red)(cancel(color(black)("mmHg O"_2))) * overbrace("650. mmHg total"/(136.5color(red)(cancel(color(black)("mmHg O"_2)))))^(color(brown)("normal partial pressure of O"_2)) = "476.19 mmHg"

Rounded to three sig figs, the number of sig figs you have for your values, the answer will be

P_"total" = color(green)(|bar(ul(color(white)(a/a)"476 mmHg"color(white)(a/a)|)))

ALTERNATIVE APPROACH

Alternatively, you can solve the problem without calculating the normal partial pressure of oxygen.

All you really need to know here is the mole fraction of oxygen in the air. Once you determine that you have

${\chi}_{{O}_{2}} = 0.21$

you can use the fact that this mole fraction is always true, regardless of the total pressure of the air.

Since the partial pressure of oxygen will always be

${P}_{{O}_{2}} = 0.21 \times {P}_{\text{total}}$

you can rearrange to get ${P}_{\text{total}}$ in terms of ${P}_{{O}_{2}}$

${P}_{\text{total}} = \frac{100}{21} \cdot {P}_{{O}_{2}}$

Plug in the value given for the partial pressure of oxygen at which the masks are released to get

P_"total" = 100/21 * "100. mmHg" = color(green)(|bar(ul(color(white)(a/a)"476 mmHg"color(white)(a/a)|)))#