Question #956b2
1 Answer
Explanation:
The idea here is that the volume of a gas,
#V prop n#
The balloon is said to contain
Consequently, you can say that the balloon contains twice as many moles of helium gas than moles of air. In other words, if you take
Mathematically, you can show that this is the case because
#V_"He" prop n_"He" " " -># for helium
#V_"air" prop n_"air" " " -># for air
gets you
#V_"He"/V_"air" = n_"He"/n_"air"#
And since you know that
#V_"He" = 2 * V_"air"#
you will end up with
#n_"He" = (2 * color(red)(cancel(color(black)(V_"air"))))/color(red)(cancel(color(black)(V_"air"))) * n_"air"#
#n_"He" = 2 * n_"air"#
Now, the partial pressure of a gas that's part of a gaseous mixture depends on the mole fraction of the gas in the mixture
The mole fraction of helium,
Since we've said that for
#n + 2n = 3n#
This means that you have
#chi_"He" = (color(blue)(cancel(color(black)(n))) color(red)(cancel(color(black)("moles"))))/(3color(blue)(cancel(color(black)(n))) color(red)(cancel(color(black)("moles")))) = 1/3#
The partial pressure of helium in the balloon will be
#P_"He" = chi_ "he" * P_"total"#
Here
Plug in your values to find
#color(darkgreen)(ul(color(black)(chi_"He"))) = 1/3 * "100.1 kPa" = color(radkgreen)(ul(color(black)("33.37 kPa")))#
The answer is rounded to four sig figs, the number of sig figs you have for the total pressure in the balloon.