Example 12.3-1

A mixture of H₂ at 2.33 atm and N₂ at 0.77 atm is in a container. What is the total pressure in the container?

Solution

Dalton's law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:

P_tot = 2.33 atm + 0.77 atm = 3.10 atm

Exercise

Air can be thought of as a mixture of N₂ and O₂. In 760 torr of air, the partial pressure of N₂ is 608 torr. What is the partial pressure of O₂?

Answer

152 torr

Example 12.3-2

A 2.00 L container with 2.50 atm of H₂ is connected to a 5.00 L container with 1.90 atm of O₂ inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?

Solution

Because gases act independently of each other, we can determine the resulting final pressures using Boyle's law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle's law to determine the final pressure of H₂:

Create a table of values from the problem.

From Boyle's Law, since the volume is increasing, the pressure must be decreasing. Therefore the volume factor must be less than one. In other words, the smaller volume must go on top of the bigger volume; otherwise the volume factor would be more than one.

$$P_2=(2.50\text{atm})\left(\frac{2.00\cancel{\text{L}}}{7.00\cancel{\text{L}}}\right)=0.714\text{atm}$$

Now we do that same thing for the O₂:

Create a table of values from the problem.

From Boyle's Law, since the volume is increasing, the pressure must be decreasing. Therefore the volume factor must be less than one. In other words, the smaller volume must go on top of the bigger volume; otherwise the volume factor would be more than one.

$$P_2=(1.90\text{atm})\left(\frac{5.00\cancel{\text{L}}}{7.0\cancel{\text{L}}}\right)=1.36\text{atm}$$

The total pressure is the sum of the two resulting partial pressures:

P_tot = 0.714 atm + 1.36 atm = 2.07 atm

Exercise

If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure?

Answer

0.51 atm

Example 12.3-3

Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H₂ gas were generated?

Solution

We need to take into account that the total pressure includes the vapor pressure of water. According to Table 12.3(1) , the vapor pressure of water at 22°C is 19.84 torr. According to Dalton's law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so

$$\text{733 torr}=P_{{\text{H}}_2}+P_{{\text{H}}_{\text{2}}\text{O}}=P_{{\text{H}}_{\text{2}}}+\text{19}.\text{84 torr}$$

We solve by subtracting:

$$P_{{\text{H}}_{\text{2}}}=\text{713 torr}$$

Since we are heading to the ideal gas law, we need to convert the pressure units to atm.

$\left(\text{713 torr}\right)\left(\frac{\mathrm{atm}}{760\mathrm{torr}}\right)=0.938\mathrm{atm}$

Now we can use the ideal gas law to determine the number of moles (remembering to convert temperature to kelvins, making it 295 K):

$$\text{(0.938 atm)(2}\text{.00 L)}=n\left(0.08206\frac{\text{L}·\text{atm}}{\text{mol}·\text{K}}\right)(\text{295 K)}$$

All the units cancel except for mol, which is what we are looking for. So

n = 0.0775 mol H₂ collected

Exercise

CO₂, generated by the decomposition of CaCO₃, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO₂ were generated?

Answer

0.129 mol