Pressure of several gases in the same container

You have probably studied the behavior of a single ideal gas in a container, how its pressure, temperature and volume change when one or another is manipulated.

For example, Charles law, $P_1V_1=P_2V_2$, tells us that the pressure of a system will increase as the volume is decreased. That is, if $V_2 \lt V_1$, then $P_2 \gt P_1$.

But what about mixtures of gases? For example, how do we characterize the pressure of a mixture of argon (Ar) and helium (He), and can they be treated separately?

Dalton's law of partial pressures, due to John Dalton (1766 - 1844), says that the total pressure of a system of non-interacting gases is just the sum of the partial pressures of each of the components.

The basic assumption behind Dalton's law of partial pressures is that ideal gas particles are so far apart, except at extremely-high pressures, that collisions between them are not as important as collisions with the walls of the container. We treat them as if they are completely independent.

Example 1

11.0 g of N₂ and 5.0 g of O₂ are let into an evacuated container of volume V = 12.5 liters. Calculate the partial pressures of each gas and the total pressure in the container if the temperature is maintained at T = 300K.

Solution: Aiming for calculating each pressure with the ideal gas law, $PV = nRT$, first we'll need to calculate the number of moles of each gas. The number of moles of N₂ is:

$$ \require{cancel} 11 \cancel{g \, N_2} \left( \frac{1 \, mole \, N_2}{28 \cancel{g \, N_2}} \right) = 0.393 \, mol$$

and O₂:

$$5 \cancel{g \, O_2} \left( \frac{1 \, mole \, O_2}{32 \cancel{g \, O_2}} \right) = 0.156 \, mol$$

Now we use the ideal gas equation to calculate the pressures:

$$P_{N_2} = \frac{n_{N_2} RT}{V}$$

Example 2

A container holds two gases, helium and argon. Helium constitutes 30% of the volume. Calculate the partial pressures of He and Ar if the total pressure in the container is 4.2 atm.

Solution: All other things (numbers of moles, temperature) remaining constant, then the pressures are proportional to the volumes. If $x$ is the total volume, then

$$ \begin{align} V_{Ar} &= 0.7 x \\[5pt] V_{He} &= 0.3 x \end{align}$$

Example 3

Containers A, B and C have the following volumes and contain gases at the pressures given:

If all of these gases are put into container D with volume 2.85 L, what will the pressure in container D be? Related question: Is it possible to transfer all of the gases in containers A, B and C into D?

Solution: All other things (numbers of moles, temperature) remaining constant, then the pressure-volume products are constant before and after the gases are transferred to container D:

$$P_1 V_1 = P_2 V_2 \: \color{magenta}{\longrightarrow} \: P_2 = \frac{P_1 V_1}{V_2}$$

$$ \begin{align} P_{A \rightarrow D} &= \frac{P_A V_A}{V_D} = \frac{3.25(1.25)}{2.85} = 1.425 \, atm \\[5pt] P_{B \rightarrow D} &= \frac{P_B V_B}{V_D} = \frac{2.85(1.05)}{2.85} = 1.05 \, atm \\[5pt] P_{C \rightarrow D} &= \frac{P_C V_C}{V_D} = \frac{1.20(1.50)}{2.85} = 0.632 \, atm \end{align}$$

Example 4

A gas mixture contains oxygen (O₂), argon (Ar) and nitrogen (N₂). The O₂ has a partial pressure of 99 torr and the partial pressure of N₂ is 0.335 atm. The total pressure is 675 torr. Calculate the total pressure of this system if all of the oxygen is removed from it.

Practice problems

1. What volume will a mixture of 5.2 moles of argon (Ar) gas and 6.4 moles of nitrogen (N₂) gas occupy at 50˚C and 0.80 atm?

   Solution

   
   

   The total number of moles of gas is $5.2+6.4=11.6$ moles. Then we just apply the ideal gas law:

   $$PV = nRT \longrightarrow V = \frac{nRT}{P}$$

   $$ \begin{align} V &= \frac{nRT}{P} \\[5pt] &= \frac{11.6 \; \cancel{mol} (0.0821 \; \frac{L \, \cancel{atm}}{\cancel{mol} \, \cancel{K}}) (273+50)K}{0.80 \; \cancel{atm}} \\[5pt] &= 385 \, L \end{align}$$




2. A 30 L gas tank has a total pressure of 760 torr at 0˚C. It contains nitrogen gas (P_N2 = 240 torr), argon gas (P_Ar = 150 torr) and helium gas (He). Calculate the number of moles of He atoms in the tank.

   Solution

   
   

   The total pressure must add to 760 torr, so the partial pressure of He is

   $$P_{He} = 760-240-150=370 \; torr$$

   Converting that pressure to atmospheres gives

   $$370 \, \cancel{torr} \left( \frac{1 \, atm}{760 \, \cancel{torr}} \right) = 0.4868 \, atm$$

   Now we just rearrange the ideal gas law and calculate the number of moles of He:

   $$ \begin{align} PV &= nRT \longrightarrow n = \frac{PV}{RT} \\[5pt] n &= \frac{04868 \, \cancel{atm} (30 \, \cancel{L})}{0.0821 \,\frac{\cancel{L} \,\cancel{atm}}{mol \,\cancel{K}}(273.15 \, \cancel{K})} \\[5pt] n &= 0.652 \; \text{moles} \end{align}$$




3. The vapor pressure of water at 20˚C is 17.5 torr. If a sample of oxygen gas (O₂) is collected over water at 20˚C and 1.0 atm, what is the partial pressure of O₂?

   Solution

   
   

   The temperature is $273.15+20=293.15 \; K$

   $$P_{H_2O} = 17.5 \, \cancel{torr} \left( \frac{1 \; atm}{760 \, \cancel{torr}} \right) = 0.023 \; atm$$

   Now the total pressure is 1.0 atm, so the pressure of oxygen is

   $$P_{O_2} = 1.0 - 0.023 = 0.977 \; atm$$




4. The partial pressure of fluorine gas (F₂) in a mixture of gases that has a total pressure of 1.10 atm is 300 torr. Calculate the mole fraction of F₂.

   Solution

   
   

   The partial pressure of F₂ is

   $$P_{F_2} = 300 \, \cancel{torr} \left( \frac{1 \; atm}{760 \, \cancel{torr}} \right) = 0.3947 \; atm$$

   Now we have

   $$n_{F_2} = \frac{P_{F_2} V}{RT} \phantom{00} \text{and} \phantom{00} n_{tot} = \frac{P_{tot}V}{RT}$$

   where volume (V), temperature (T) and R (a constant) are the same in both cases. Therefore the ratio of the two, which is the mole fraction, is just

   $$\frac{\frac{P_{F_2} \cancel{V}}{\cancel{RT}}}{\frac{P_{tot}\cancel{V}}{\cancel{RT}}} = \frac{P_{F_2}}{P_{tot}}$$

   Then the mole fraction is

   $$\chi_{F_2} = \frac{P_{F_2}}{P_{tot}} = \frac{0.3947}{1.10} = 0.3588$$




5. A container of volume 6.5 L holds 72 g of chlorine gas (Cl₂) and 5.0 g of helium gas (He). If the total pressure is 740 mm Hg, calculate the partial pressures of each gas.

   Solution

   
   

   The total pressure in atmospheres is

   $$P_{tot} = \frac{740}{760} = 0.9737 \; atm$$

   The numbers of moles of Cl₂ and He are

   $$ \begin{align} 72 \; \cancel {g \; Cl_2} \left( \frac{1 \; mol \; Cl_2}{70.7 \;\cancel{ g \; Cl_2}} \right) &= 1.0184 \; mol \; Cl_2 \\[5pt] 5 \; \cancel {g \; He} \left( \frac{1 \; mol \; He}{4.0 \;\cancel{ g \; He}} \right) &= 1.25 \; mol \; He \end{align}$$

   Now we need the temperature, which we can calculate from the total pressure and the other known conditions:

   $$ \begin{align} T &= \frac{PV}{nR} = \frac{0.9737 \; \cancel{atm}(6.5 \; \cancel{L})}{2.2684 \; \cancel{mol}(0.0821 \; \frac{\cancel{L} \, \cancel{atm}}{\cancel{mol} \, K})} \\[5pt] T &= 33.9841 \; K \end{align}$$

   Now we can calculate the partial pressures:

   $$ \begin{align} P_{Cl_2} &= \frac{1.0184 \; \cancel{mol}(0.0821 \; \frac{\cancel{L} \; atm}{\cancel{mol} \; \cancel{K}})(33.9841 \; \cancel{K})}{6.5 \; \cancel{L}} \\[5pt] &= 0.4371 \; atm \\[5pt] P_{He} &= \frac{1.25 \; \cancel{mol}(0.0821 \; \frac{\cancel{L} \; atm}{\cancel{mol} \; \cancel{K}})(33.9841 \; \cancel{K})}{6.5 \; \cancel{L}} \\[5pt] &= 0.5372 \; atm \end{align}$$

   These pressures sum to $0.9743 \; atm$, which is just a little off of our total pressure ($P = 0.9737 \; atm$) from above. We can probably attribute the small difference (0.07%) to some rounding.