# 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?

##### 1 Answer

#### Answer:

#### 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_"total" = n_(O_2) + n_(N_2)#

#n_"total" = "5.0 moles" + "3.0 moles" = "8.0 moles"#

Now, the ideal gas law equation looks like this

#color(blue)(PV = nRT)" "# , where

*universal gas constant*, usually given as

**absolute temperature** of the gas

Rearrange to solve for

#PV = nRT implies P = (nRT)/V#

#P = (8.0 color(red)(cancel(color(black)("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"))))#

#P = color(green)("6.5 atm") -># rounded to twosig figs

**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.

#color(blue)(P_"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) = (5.0 color(red)(cancel(color(black)("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"))))#

#P_(O_2) = "4.08 atm"#

For nitrogen gas, you will have

#P_(N_2) = (3.0 color(red)(cancel(color(black)("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"))))#

#P_(N_2) = "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