Question #1cee1

1 Answer
Jun 6, 2016

Answer:

#"7.07 mmHg"#

Explanation:

The idea here is that you can determine the partial pressure of argon by using Dalton's Law of Partial Pressures, which states that the partial pressure of a gas #i#, #P_i#, that's part of a gaseous mixture is equal to

#color(blue)(|bar(ul(color(white)(a/a)P_i = chi_i xx P_"total"color(white)(a/a)|)))#

Here

#chi_i# - the mole fraction of gas #i# in the mixture
#P_"total"# - the total pressure of the mixture

Now, you know that argon makes up #0.93%# by volume of the atmosphere, which is equivalent to saying that for every #"100 L"# of air, you have #"0.93 L"# of argon.

Since volume is directly proportional to number of moles at constant temperature and pressure -- think Avogadro's Law here -- you can say that you have #0.93# moles of argon for every #100# moles of air.

This means that the mole fraction of argon, which is defined as the ratio between the number of moles of argon and the *total number of moles8 of gas present in a given sample of air, will be

#chi_"argon" = (0.93 color(red)(cancel(color(black)("moles"))))/(100color(red)(cancel(color(black)("moles")))) = 0.0093#

The partial pressure of argon in air will be

#P_"argon" = 0.0093 * "760 mmHg" = color(green)(|bar(ul(color(white)(a/a)color(black)("7.07 mmHg")color(white)(a/a)|)))#

I'll leave the answer rounded to three sig figs.