Question #48d69
1 Answer
Explanation:
The idea here is that the two gases will contribute to the total pressure of the mixture proportionally to how many moles each has in the mixture - think Dalton's Law of Partial Pressures.
IF you take
color(blue)(|bar(ul(color(white)(a/a)P_"total" = P_(He) + P_(O_2)color(white)(a/a)|)))" " " "color(orange)("(*)")
Now, to get the partial pressure of each gas you must use the ideal gas law equation
color(blue)(|bar(ul(color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" " , where
The trick here is to realize that you can find the partial pressure of each gas by isolating them in same volume and at the same temperature.
![http://www.ck12.org/book/CK-12-Chemistry-Basic/section/14.3/]()
This means that you can write
P_(He) = n_(He) * (RT)/V -> the partial pressure of helium
P_(O_2) = n_(O_2) * (RT)/V -> hte partial pressure of oxygen
Now, the total pressure of the mixture will depend on the total number of moles present.
n_"total" = n_(He) + n_(O_2)
You can thus write
P_"total" = n_"total" * (RT)/V
Use this equation to write
(RT)/V = P_"total"/n_"total"
and plug this into the equations for the partial pressures of the two gases. You will find
P_(He) = n_(He) * P_"total"/n_"total" = n_(He)/n_"total" * P_"total"
and
P_(O_2) = n_(O_2) * P_"total"/n_"total" = n_(O_2)/n_"total" * P_"total"
The ratio between the number of moles of a gas that's part of a gaseous mixture and the total number of moles of gas present in the mixture will give you that gas' mole fraction,
You will thus have
color(purple)(|bar(ul(color(white)(a/a)color(black)(P_(He) = chi_(He) * P_"total")color(white)(a/a)|)))" " and" "color(purple)(|bar(ul(color(white)(a/a)color(black)(P_(O_2) = chi_(O_2) * P_"total")color(white)(a/a)|)))
Finally, plug in your values to get
P_(He) = (2.0 color(red)(cancel(color(black)("moles"))))/((2.0 + 6.0)color(red)(cancel(color(black)("moles")))) * "2400 torr" = color(green)(|bar(ul(color(white)(a/a)6.0 * 10^2"torr"color(white)(a/a)|)))
P_(O_2) = (6.0 color(red)(cancel(color(black)("moles"))))/((2.0 + 6.0)color(red)(cancel(color(black)("moles")))) * "2400 torr" = color(green)(|bar(ul(color(white)(a/a)"1800 torr"color(white)(a/a)|)))
Notice that both values satisfy equation
P_"total" = 6.0 * 10^2"torr" + "1800 torr" = "2400 torr"
The answers are rounded to two sig figs.