# A mixture of gases contains 0.75 mol N2, 0.30 mol O2, and 0.15 mol CO2. If the total pressure of the mixture is 1.56 atm, what is the partial pressure of each component?

Dec 22, 2015

Here's what I got.

#### Explanation:

To make this problem a little interesting, let's assume that you're not familiar with Dalton's Law of partial pressures, which tells you that the partial pressure of a gas that's part of a gaseous mixture is proportional to that gas' mole fraction.

Here's how you can think about what's going on, Let's assume that the mixture is at a pressure ${P}_{\text{total}}$, a temperature $T$, and occupies a volume $V$.

If you were to isolate the nitrogen gas in the same volume, you could write, using the ideal gas law equation

${P}_{{N}_{2}} \cdot V = {n}_{{N}_{2}} \cdot R T \implies {P}_{{N}_{2}} = {n}_{{N}_{2}} \cdot \frac{R T}{V}$

Here ${P}_{{N}_{2}}$ is the pressure exerted by the nitrogen gas when alone in the same volume as the mixture.

Now do the same for oxygen and carbon dioxide.

${P}_{{O}_{2}} \cdot V = {n}_{{O}_{2}} \cdot R T \implies {P}_{{O}_{2}} = {n}_{{O}_{2}} \cdot \frac{R T}{V}$

${P}_{C {O}_{2}} \cdot V = {n}_{C {O}_{2}} \cdot R T \implies {P}_{C {O}_{2}} = {n}_{C {O}_{2}} \cdot \frac{R T}{V}$

Now, what would happen if you were to have the nitrogen gas, the oxygen gas, and the carbon dioxide in the same volume? The pressure would change to ${P}_{\text{total}}$, and the total number of moles in the mixture would be

${n}_{\text{total}} = {n}_{{N}_{2}} + {n}_{{O}_{2}} + {n}_{C {O}_{2}}$

This means that you could write

${P}_{\text{total" * V = n_"total}} \cdot R T$

${P}_{\text{total" = n_"total}} \cdot \frac{R T}{V}$

This is equivalent to

${P}_{\text{total" = [n_(N_2) + n_(O_2) + n_(CO_2)] * (RT)/V" " " }} \textcolor{p u r p \le}{\left(1\right)}$

${P}_{\text{total}} = {\overbrace{{n}_{{N}_{2}} \cdot \frac{R T}{V}}}^{\textcolor{b l u e}{{P}_{{N}_{2}}}} + {\overbrace{{n}_{{O}_{2}} \cdot \frac{R T}{V}}}^{\textcolor{red}{{P}_{{O}_{2}}}} + {\overbrace{{n}_{C {O}_{2}} \cdot \frac{R T}{V}}}^{\textcolor{g r e e n}{{P}_{C {O}_{2}}}}$

Therefore,

${P}_{\text{total}} = {P}_{{N}_{2}} + {P}_{{O}_{2}} + {P}_{C {O}_{2}}$

Now, to get the partial pressure of, let's say nitrogen gas, you would use equation $\textcolor{p u r p \le}{\left(1\right)}$

$\frac{R T}{V} = {P}_{\text{total}} / \left({n}_{{N}_{2}} + {n}_{{O}_{2}} + {n}_{C {O}_{2}}\right)$

This will give you

${P}_{{N}_{2}} = {n}_{{N}_{2}} \cdot {P}_{\text{total}} / \left({n}_{{N}_{2}} + {n}_{{O}_{2}} + {n}_{C {O}_{2}}\right)$

P_(N_2) = overbrace(n_(N_2)/(n_(N_2) + n_(O_2) + n_(CO_2)))^(color(blue)("mole fraction of N"_2)) * P_"total"

Therefore,

${P}_{{N}_{2}} = {\chi}_{{N}_{2}} \cdot {P}_{\text{total}}$

Similarly,

${P}_{{O}_{2}} = {\chi}_{{O}_{2}} \cdot {P}_{\text{total}}$

${P}_{C {O}_{2}} = {\chi}_{C {O}_{2}} \cdot {P}_{\text{total}}$

In your case, the total number of moles will be

${n}_{\text{total" = 0.75 + 0.30 + 0.15 = "1.2 moles}}$

This means that you have

P_(N_2) = (0.75 color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) * "1.56 atm" = color(green)("0.98 atm")

P_(O_2) = (0.30 color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) * "1.56 atm" = color(green)("0.39 atm")

P_(CO_2) = (0.15 color(red)(cancel(color(black)("moles"))))/(1.2color(red)(cancel(color(black)("moles")))) * "1.56 atm" = color(green)("0.20 atm")

The values don';t add up to give $\text{1.56 atm}$ because they must be rounded to two sig figs, the number of sig figs you have for the moles of each gas.