The atmospheric pressure at the summit of Mt. Everest is 255 torr. The percentage of #O_2# and #N_2# is 20% and 79%, respectively. How many #N_2# and #O_2# molecules are there in a volume of 10.0L at 12°C?

1 Answer
Dec 2, 2015

Answer:

Here's what I got.

Explanation:

Your strategy here will be to use the ideal gas law equation to determine the total number of moles of air you have in that #"10.0-L"# sample under those conditions for pressure and temperature.

Once you know how many moles of air you have, you can use the given percentages to find how many moles of nitrogen gas and oxygen gas you have.

Finally, Avogadro's number will allow you to convert the number of moles to number of molecules.

So, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the sample
#V# - its volume
#n# - the number of moles of gas present in the sample
#R# - the universal gas constant, usually given as #0.0821("atm" * "L")/("mol" * "K")#
#T# - the temperature of the gas

So, plug in your values and solve for #n# - do not forget to convert the pressure from torr to atm and the temperature from degrees Celsius to Kelvin

#PV = nRT implies n = (PV)/(RT)#

#n = (255/760 color(red)(cancel(color(black)("atm"))) * 10.0 color(red)(cancel(color(black)("L"))))/(0.0821 * (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + 12)color(red)(cancel(color(black)("K")))) = "0.1433 moles"#

Now, out of these moles of air, you know that #20%# are moles of oxygen and #79%# are moles of nitrogen. This means that you will have

#0.1433 color(red)(cancel(color(black)("moles air"))) * "20 moles O"_2/(100color(red)(cancel(color(black)("moles air")))) = "0.02866 moles O"_2#

#0.1433 color(red)(cancel(color(black)("moles air"))) * "79 moles N"_2/(100color(red)(cancel(color(black)("moles air")))) = "0.1132 moles N"_2#

Now, as you know, in order to have one mole of a substance, you need exactly #6.022 * 10^(23)# molecules of that substance - this is what Avogadro's number is all about.

In your case, the sample will contain

#0.02866 color(red)(cancel(color(black)("moles O"_2))) * (6.022 * 10^(23)"molecules of O"_2)/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(1.7 * 10^(22)"molecules of O"_2#

and

#0.1132 color(red)(cancel(color(black)("moles N"_2))) * (6.022 * 10^(23)"molecules of N"_2)/(1color(red)(cancel(color(black)("mole N"_2)))) = color(green)(6.8 * 10^(22)"molecules of N"_2#

The answers are rounded to two sig figs.