If 5.0 moles of O_2 and 3.0 moles of N_2 are placed in a 30.0 L tank at a temperature of 25°C, what will the pressure of the resulting mixture of gases be?

Feb 22, 2016

$\text{6.5 atm}$

Explanation:

In order to find the pressure of the mixture, you need to know the total number of moles present in the mixture.

The problem provides you with the volume of the container and the temperature at which the gases are kept, which means that you can use the total number of moles in the ideal gas law to find the *pressure8 of the mixture.

So, the total number of moles present in the container will be

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

${n}_{\text{total" = "5.0 moles" + "3.0 moles" = "8.0 moles}}$

Now, the ideal gas law equation looks like this

$\textcolor{b l u e}{P V = n R T} \text{ }$, where

$P$ - the total pressure of the mixture
$V$ - the volume it occupies
$n$ - the total number of moles of gas in the mixture
$R$ - the universal gas constant, usually given as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the gas

Rearrange to solve for $P$ and plug in your values to find

$P V = n R T \implies P = \frac{n R T}{V}$

$P = \left(8.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * overbrace((273.15 + 25)color(red)(cancel(color(black)("K"))))^(color(purple)("absolute temperature")))/(30.0color(red)(cancel(color(black)("L}}}}\right)$

$P = \textcolor{g r e e n}{\text{6.5 atm}} \to$ rounded to two sig figs

$\textcolor{w h i t e}{a}$

ALTERNATIVE APPROACH

You can check the result by using Raoult's Law, which tells you that the total pressure of a gaseous mixture will be equal to the sum of the partial pressures of each component of that mixture.

$\textcolor{b l u e}{{P}_{\text{mix}} = {P}_{{O}_{2}} + {P}_{{N}_{2}}}$

To get the partial pressures of the two gases, assume that each gas occupies the given volume alone.

For oxygen gas, you will have

${P}_{{O}_{2}} = \left(5.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(30.0color(red)(cancel(color(black)("L}}}}\right)$

${P}_{{O}_{2}} = \text{4.08 atm}$

For nitrogen gas, you will have

${P}_{{N}_{2}} = \left(3.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K"))))/(30.0color(red)(cancel(color(black)("L}}}}\right)$

${P}_{{N}_{2}} = \text{2.45 atm}$

The total pressure of the mixture will once again be

P_"mix" = "4.08 atm" + "2.45 atm" = color(green)("6.5 atm") -> rounded to two sig figs