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 the partial pressure of 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 the mol fraction of 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")#
