Question #0d4c9
1 Answer
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
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
This tells you that out of very
You can find the partial pressure of oxygen in terms of the total pressure in the cabin bu using the ideal gas law equation
#color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "# , where
If you take
#P_"total" = n_"total" * (RT)/V#
If you were to isolate the oxygen in the same cabin, i.e. in the same volume
#P_(O_2) = n_(O_2) * (RT)/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) = n_(O_2)/n_"total"#
Moreover, mole percent is simply mole fraction multiplied by
#color(blue)(|bar(ul(color(white)(a/a)"mole %" = "mole fraction" xx 100color(white)(a/a)|)))#
This means that you have
#n_(O_2)/n_"total" = "mole %"/100 = 21/100 = 0.21#
Plug this into equation
#P_(O_2) = 0.2 * "650. mmHg" = "136.5 mmHg"#
Now, when the partial pressure of oxygen drops below
#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 xx P_"total"#
you can rearrange to get
#P_"total" = 100/21 * 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)|)))#