# A tank contains a mixture of 54.5 g of oxygen gas and 68.9 g of carbon dioxide gas at 38°C. The total pressure in the tank is 10.00 atm. How do you calculate the partial pressure (in atm) of each gas in the mixture?

##### 1 Answer
Apr 25, 2016

${P}_{{O}_{2}} = \text{5.21 atm}$
${P}_{C {O}_{2}} = \text{4.79 atm}$

#### Explanation:

You can solve this problem by using Dalton's Law of Partial Pressures, which states that the partial pressure of a component of a gaseous mixture depends on the mole ratio of said component and the total pressure of the gaseous mixture.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {P}_{i} = {\chi}_{i} \times {P}_{\text{total}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$, where

${P}_{i}$ - the partial pressure of component $i$
${\chi}_{i}$ - its mole fraction in the mixture
${P}_{\text{total}}$ - the total pressure of the mixture

Since you already know the total pressure of the mixture, focus on finding how many moles of each gas you get in the mixture.

To do that, use the molar masses of the two gases. You will have

54.5 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "1.703 moles O"_2

68.9 color(red)(cancel(color(black)("g"))) * "1 mol CO"_2/(44.01color(red)(cancel(color(black)("g")))) = "1.566 moles CO"_2

The total number of moles present in the mixture will be

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

${n}_{\text{total" = "1.703 moles" + "1.566 moles" = "3.269 moles}}$

Now, to get the mole fraction of a gas $i$ that's part of a mixture, simply divide the number of moles of that gas by the total number of moles present in the mixture

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} {\chi}_{i} = \text{number of moles of i"/"total number of moles} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the partial pressures of the two gases will be equal to

P_(O_2) = (1.703 color(red)(cancel(color(black)("moles"))))/(3.269color(red)(cancel(color(black)("moles")))) * "10.00 atm" = color(green)(|bar(ul(color(white)(a/a)"5.21 atm"color(white)(a/a)|)))

P_(CO_2) = (1.566 color(red)(cancel(color(black)("moles"))))/(3.269color(red)(cancel(color(black)("moles")))) * "10.00 atm" = color(green)(|bar(ul(color(white)(a/a)"4.79 atm"color(white)(a/a)|)))

The answers are rounded to three sig figs.

Notice that the answers check out because the total pressure of the gaseous mixture is equal to the sum of the partial pressures of its components

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