# At a total pressure of #"1 atm"#, if the mol percentages of #"CO"_2#, #"O"_2#, and #"N"_2# are #5%#, #15%#, and #80%# respectively, what are the partial pressures?

##### 1 Answer

We are given that:

#P_(t ot) = "1 atm"# #%"v/v" ("CO"_2) = 5%# #%"v/v" ("O"_2) = 15%# #%"v/v" ("N"_2) = 80%#

By assuming the *appropriate* conditions (low enough pressure, high enough temperature that intermolecular interactions are minimized), all of these gases are ideal enough that we can get partial pressures from:

#bb(P_i = chi_i P_(t ot))# where:

#P_i# is thepartial pressureof gas#i# in the mixture, the pressure that each gas in the mixture exerts on the walls of the container.#chi_i = n_i/(n_1 + . . . + n_N)# is themol fractionof gas#i# in the mixture of#N# different gases.

Each mol fraction is just **the decimal form of the percentage given**, since we are assuming all these gases are ideal, having *the same molar volume* (whatever it may be at the appropriate

(For example,

In other words, we are saying:

#(%"v/v" ("CO"_2))/(100%) ~~ chi_(CO_2)# #(%"v/v" ("O"_2))/(100%) ~~ chi_(O_2)# #(%"v/v" ("N"_2))/(100%) ~~ chi_(N_2)#

As a result, the partial pressures are:

#color(blue)(P_(CO_2)) = 0.05P_(t ot) = color(blue)("0.05 atm")#

#color(blue)(P_(O_2)) = 0.15P_(t ot) = color(blue)("0.15 atm")#

#color(blue)(P_(N_2)) = 0.80P_(t ot) = color(blue)("0.80 atm")#