14.4: Gas Mixtures and Molecular Speeds
Lesson Objectives
 Use Dalton’s law and mole fraction to calculate the partial pressure of a gas in a mixture.
 Calculate the pressure of a gas that has been collected by water displacement in order to determine the volume of the dry gas.
 Define diffusion and effusion.
 Use Graham’s law to calculate the velocity ratios of two gases based on their molar masses.
Lesson Vocabulary
 Dalton’s law of partial pressures
 diffusion
 effusion
 mole fraction
 partial pressure
Check Your Understanding
Recalling Prior Knowledge
 Do gas molecules interact with one another as they exert pressure?
 How is the vapor pressure of water related to its temperature?
 How is kinetic energy related to the mass and velocity of a moving object?
The gas laws that have been introduced so far have applied to a pure gas. In this lesson, you will learn about Dalton’s Law, which is important for dealing with mixtures of gases. In addition, the velocities of gas molecules will be studied by use of Graham’s Law.
Dalton’s Law
Gas pressure results from collisions between gas particles and the walls of their container. If more gas is added to a rigid container, the gas pressure increases. This is true whether the newly added gas is the same as the original gas or a different one; the identities of the gases do not matter. John Dalton, the English chemist who proposed the atomic theory, also studied mixtures of gases. He found that each gas in a mixture exerts a pressure independently of every other gas in the mixture. For example, our atmosphere is composed of about 78% nitrogen and 21% oxygen, with smaller amounts of several other gases making up the rest. Since nitrogen makes up 78% of the gas particles in a given sample of air, it exerts 78% of the pressure. If the overall atmospheric pressure is 1.00 atm, then the pressure of just the nitrogen in the air is 0.78 atm. The pressure of the oxygen in the air is 0.21 atm.
The partial pressure of a gas is the contribution that gas makes to the total pressure when the gas is part of a mixture. The partial pressure of a gas is indicated by a P with a subscript that is the symbol or formula of that gas. The partial pressure of nitrogen is represented by \begin{align*}\text{P}_{\text{N}_2}\end{align*}


 P_{Total} = P_{1} + P_{2} + P_{3} ...

Figure below shows two gases that are in separate, equalsized containers at the same temperature and pressure. Each exerts a different pressure, P_{1} and P_{2}, reflective of the number of particles in the container. On the right, the two gases are combined into the same container, with no volume change. The total pressure of the gas mixture is equal to the sum of the individual pressures. If P_{1} = 300 mmHg and P_{2} = 500 mmHg, then P_{Total} = 800 mmHg.
Dalton’s law says that the total pressure of a gas mixture is equal to the sum of the partial pressures contributed by each component of the mixture.
Mole Fraction
One way to express relative amounts of substances in a mixture is with the mole fraction. Mole fraction (X) is the ratio of moles of one substance in a mixture to the total number of moles of all substances. For a mixture of two substances, A and B, the mole fractions of each would be written as follows:



\begin{align*}\mathrm{X_A=\dfrac{mol \ A}{mol \ A+mol \ B}}\end{align*}
XA=mol Amol A+mol B and \begin{align*}\mathrm{X_B=\dfrac{mol \ B}{mol \ A+mol \ B}}\end{align*}XB=mol Bmol A+mol B

\begin{align*}\mathrm{X_A=\dfrac{mol \ A}{mol \ A+mol \ B}}\end{align*}

If a mixture consists of 0.50 mol A and 1.00 mol B, then the mole fraction of A would be X_{A} = 0.5/1.5 = 0.33. Similarly, the mole fraction of B would be X_{B} = 1.0/1.5 = 0.67.
Mole fraction is a useful quantity for analyzing gas mixtures in conjunction with Dalton’s law of partial pressures. Consider the following situation: A 20.0 liter vessel contains 1.0 mol of hydrogen gas at a pressure of 600 mmHg. Another 20.0 liter vessel contains 3.0 mol of helium at a pressure of 1800 mmHg. These two gases are mixed together in an identical 20.0 liter vessel. Because each will exert its own pressure according to Dalton’s law, we can express the partial pressures as follows:



\begin{align*}\mathrm{P_{H_2}=X_{H_2} \times P_{Total}}\end{align*}
PH2=XH2×PTotal and \begin{align*}\mathrm{P_{He}=X_{He} \times P_{Total}}\end{align*}PHe=XHe×PTotal

\begin{align*}\mathrm{P_{H_2}=X_{H_2} \times P_{Total}}\end{align*}

The partial pressure of a gas in a mixture is equal to its mole fraction multiplied by the total pressure. For our mixture of hydrogen and helium:



\begin{align*}\mathrm{X_{H_2}=\dfrac{1.0 \ mol}{1.0 \ mol+3.0 \ mol}=0.25}\end{align*}
XH2=1.0 mol1.0 mol+3.0 mol=0.25 and \begin{align*}\mathrm{X_{He}=\dfrac{3.0 \ mol}{1.0 \ mol+3.0 \ mol}=0.75}\end{align*}XHe=3.0 mol1.0 mol+3.0 mol=0.75

\begin{align*}\mathrm{X_{H_2}=\dfrac{1.0 \ mol}{1.0 \ mol+3.0 \ mol}=0.25}\end{align*}

The total pressure according to Dalton’s law is 600 mmHg + 1800 mmHg = 2400 mmHg. So, each partial pressure will be:



\begin{align*}\mathrm{P_{H_2}=0.25 \times 2400 \ mmHg=600 \ mmHg}\end{align*}
PH2=0.25×2400 mmHg=600 mmHg

\begin{align*}\mathrm{P_{H_2}=0.25 \times 2400 \ mmHg=600 \ mmHg}\end{align*}




\begin{align*}\mathrm{P_{He}=0.75 \times 2400 \ mmHg=1800 \ mmHg}\end{align*}
PHe=0.75×2400 mmHg=1800 mmHg

\begin{align*}\mathrm{P_{He}=0.75 \times 2400 \ mmHg=1800 \ mmHg}\end{align*}

The partial pressures of each gas in the mixture do not change, since they were mixed into the same size vessel and the temperature was not altered.
Sample Problem 14.8: Dalton’s Law
A flask contains a mixture of 1.24 moles of hydrogen gas and 2.91 moles of oxygen gas. If the total pressure is 104 kPa, what is the partial pressure of each gas?
Step 1: List the known quantities and plan the problem.
Known
 1.24 mol H_{2}
 2.91 mol O_{2}
 P_{Total} = 104 kPa
Unknown

\begin{align*}\text{P}_{\text{H}_2}\end{align*}
PH2 = ? kPa 
\begin{align*}\text{P}_{\text{O}_2}\end{align*}
PO2 = ? kPa
First, the mole fraction of each gas can be determined. Then, the partial pressure can be calculated by multiplying the mole fraction by the total pressure.
Step 2: Solve.



\begin{align*}\mathrm{X_{H_2}=\dfrac{1.24 \ mol}{1.24 \ mol+2.91 \ mol}=0.299}\end{align*}
XH2=1.24 mol1.24 mol+2.91 mol=0.299

\begin{align*}\mathrm{X_{H_2}=\dfrac{1.24 \ mol}{1.24 \ mol+2.91 \ mol}=0.299}\end{align*}




\begin{align*}\mathrm{X_{O_2}=\dfrac{2.91 \ mol}{1.24 \ mol+2.91 \ mol}=0.701}\end{align*}
XO2=2.91 mol1.24 mol+2.91 mol=0.701

\begin{align*}\mathrm{X_{O_2}=\dfrac{2.91 \ mol}{1.24 \ mol+2.91 \ mol}=0.701}\end{align*}




\begin{align*}\mathrm{P_{H_2}=0.299 \times 104 \ kPa=31.1 \ kPa}\end{align*}
PH2=0.299×104 kPa=31.1 kPa

\begin{align*}\mathrm{P_{H_2}=0.299 \times 104 \ kPa=31.1 \ kPa}\end{align*}




\begin{align*}\mathrm{P_{O_2}=0.701 \times 104 \ kPa=72.9 \ kPa}\end{align*}
PO2=0.701×104 kPa=72.9 kPa

\begin{align*}\mathrm{P_{O_2}=0.701 \times 104 \ kPa=72.9 \ kPa}\end{align*}

Step 3: Think about your result.
The hydrogen is slightly less than one third of the mixture, so it exerts slightly less than one third of the total pressure.
 A mixture of 0.249 mol He, 0.812 mol Ne, and 0.429 mol Ar are mixed into a container at a total pressure of 1.340 atm. Calculate the partial pressure of each gas.
Collecting a Gas by Water Displacement
Gases that are produced in laboratory experiments are often collected by a technique called water displacement (Figure below). A bottle is filled with water and placed upsidedown in a pan of water. The reaction flask is fitted with rubber tubing that is fed under the bottle of water. As the gas is produced in the reaction flask, it exits through the rubber tubing and displaces the water in the bottle. When the bottle is full of the gas, it can be sealed with a lid.
Note that if the water level inside the bottle is the same as the water level outside the bottle, the pressure of the gas in the bottle must be equal to the ambient atmospheric pressure outside the bottle. If the levels are different, the difference in height can be used to determine the difference in pressure between the enclosed sample and the external atmosphere. In this book, we will assume that the bottle is adjusted so that the inner and outer water levels are equal.
A gas produced in a chemical reaction can be collected by water displacement.
However, the collected gas will not be pure. Because it is collected over water, the gas will be mixed with water vapor due to evaporation. Dalton’s law can be used to calculate the amount of the desired gas by subtracting the contribution of the water vapor.



\begin{align*}\text{P}_{\text{Total}} = \text{P}_\text{g} + \text{P}_{\text{H}_2\text{O}}\end{align*}
PTotal=Pg+PH2O

\begin{align*}\text{P}_{\text{Total}} = \text{P}_\text{g} + \text{P}_{\text{H}_2\text{O}}\end{align*}





\begin{align*}\text{P}_\text{g}\end{align*}
Pg is the pressure of the desired gas.

\begin{align*}\text{P}_\text{g}\end{align*}





\begin{align*}\text{P}_\text{g} = \text{P}_{\text{Total}}  \text{P}_{\text{H}_2\text{O}}\end{align*}
Pg=PTotal−PH2O

\begin{align*}\text{P}_\text{g} = \text{P}_{\text{Total}}  \text{P}_{\text{H}_2\text{O}}\end{align*}

In order to solve a problem, it is necessary to know the vapor pressure of water at the temperature of the collection vessel (Table below). Sample Problem 14.9 illustrates the use of Dalton’s law when a gas is collected over water.
Temperature (°C)  Vapor Pressure (mmHg)  Temperature (°C)  Vapor Pressure (mmHg) 

0  4.58  40  55.32 
5  6.54  45  71.88 
10  9.21  50  92.51 
15  12.79  55  118.04 
20  17.54  60  149.38 
25  23.76  65  187.54 
30  31.82  70  233.7 
35  42.18 
Sample Problem 14.9: Gas Collected by Water Displacement
A certain experiment generates 2.58 L of hydrogen gas, which is collected over water. The temperature is 20°C and the atmospheric pressure is 98.60 kPa. Find the volume that the dry hydrogen would occupy at STP.
Step 1: List the known quantities and plan the problem.
Known
 V_{Total} = 2.58 L
 T = 20°C = 293 K
 P_{Total} = 98.60 kPa = 739.7 mmHg
Unknown

\begin{align*}\text{V}_{\text{H}_2}\end{align*}
VH2 at STP = ? L
The atmospheric pressure is converted from kPa to mmHg in order to match units with Table above. The sum of the pressures of the hydrogen and the water vapor is equal to the atmospheric pressure. The pressure of the hydrogen is found by subtraction. Then, the volume of the gas at STP can be calculated by using the combined gas law.
Step 2: Solve.



\begin{align*}\text{P}_{\text{H}_2} = \text{P}_{\text{Total}}  \text{P}_{\text{H}_2\text{O}} = 739.7 \ \text{mmHg}  17.54 \ \text{mmHg} = 722.2 \ \text{mmHg}\end{align*}
PH2=PTotal−PH2O=739.7 mmHg−17.54 mmHg=722.2 mmHg

\begin{align*}\text{P}_{\text{H}_2} = \text{P}_{\text{Total}}  \text{P}_{\text{H}_2\text{O}} = 739.7 \ \text{mmHg}  17.54 \ \text{mmHg} = 722.2 \ \text{mmHg}\end{align*}

Now the combined gas law is used, solving for V_{2}, the volume of hydrogen at STP.



\begin{align*}\mathrm{V_2=\dfrac{P_1 \times V_1 \times T_2}{P_2 \times T_1}=\dfrac{722.2 \ mmHg \times 2.58 \ L \times 273 \ K}{760 \ mmHg \times 293 \ K}=2.28 \ L \ H_2}\end{align*}
V2=P1×V1×T2P2×T1=722.2 mmHg×2.58 L×273 K760 mmHg×293 K=2.28 L H2

\begin{align*}\mathrm{V_2=\dfrac{P_1 \times V_1 \times T_2}{P_2 \times T_1}=\dfrac{722.2 \ mmHg \times 2.58 \ L \times 273 \ K}{760 \ mmHg \times 293 \ K}=2.28 \ L \ H_2}\end{align*}

Step 3: Think about your result.
If the hydrogen gas were to be collected and STP and without the presence of the water vapor, its volume would be 2.28 L. This is less than the actual collected volume because some of that is water vapor. The conversion to its dry volume at STP is useful for stoichiometry purposes.
 845 mL of oxygen gas is collected over water at 25°C. The atmospheric pressure is 754.3 mmHg. Calculate the volume of the dry oxygen gas at STP.
Graham’s Law
When a person opens a bottle of perfume in one corner of a large room, it doesn’t take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. Diffusion is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly (Figure below). Solids essentially do not diffuse.
Two liquids, yellow and green, diffuse into one another.
Another related process is effusion. Effusion is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a heliumfilled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.
Scottish chemist Thomas Graham (18051869) studied the rates of effusion and diffusion for various gases. Graham’s Law states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham’s Law can be understood by comparing two gases (A and B) at the same temperature, meaning the gases have the same average kinetic energy. The kinetic energy of a moving object is given by the equation \begin{align*}\text{KE} = \frac{1}{2}\text{mv}^2\end{align*}



\begin{align*}\frac{1}{2}\text{m}_\text{A}\text{v}_\text{A}^2 = \frac{1}{2}\text{m}_\text{B}\text{v}_\text{B}^2\end{align*}
12mAv2A=12mBv2B

\begin{align*}\frac{1}{2}\text{m}_\text{A}\text{v}_\text{A}^2 = \frac{1}{2}\text{m}_\text{B}\text{v}_\text{B}^2\end{align*}

The equation can be rearranged to solve for the ratio of the velocity of gas A to the velocity of gas B (\begin{align*}\text{v}_\text{A}/\text{v}_\text{B}\end{align*}



\begin{align*}\mathrm{\dfrac{v_A^2}{v_B^2}=\dfrac{m_B}{m_A}}\end{align*}
v2Av2B=mBmA which becomes \begin{align*}\mathrm{\dfrac{v_A}{v_B}=\sqrt{\dfrac{m_B}{m_A}}}\end{align*}vAvB=mBmA−−−−√

\begin{align*}\mathrm{\dfrac{v_A^2}{v_B^2}=\dfrac{m_B}{m_A}}\end{align*}

For the purposes of comparing the rates of effusion or diffusion for two gases at the same temperature, the molar masses of each gas can be used in the equation for \begin{align*}m\end{align*}
Sample Problem 14.10: Graham’s Law
Calculate the ratio of diffusion rates for ammonia gas (NH_{3}) and hydrogen chloride (HCl) at the same temperature and pressure.
Step 1: List the known quantities and plan the problem.
Known
 molar mass of NH_{3} = 17.04 g/mol
 molar mass of HCl = 36.46 g/mol
Unknown
 velocity ratio \begin{align*}\text{v}_{\text{NH}_3}/\text{v}_\text{HCl}\end{align*}
vNH3/vHCl
Substitute the molar masses of the gases into Graham’s law and solve for the ratio.
Step 2: Solve.


 \begin{align*}\mathrm{\dfrac{v_{NH_3}}{v_{HCl}}=\sqrt{\dfrac{36.46 \ g/mol}{17.04 \ g/mol}}=1.46}\end{align*}

The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.
Step 3: Think about your result.
Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.
 Determine the ratio of effusion rates for hydrogen gas and argon gas at the same temperature and pressure.
 A certain gas is observed to diffuse 3.32 times slower than helium. Calculate the molar mass of the gas.
The relative rates of diffusion for ammonia and hydrogen chloride from Sample Problem 14.10 can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride (NH_{4}Cl), a white solid that appears in the glass tube as a ring (Figure below).
Graham’s law can be demonstrated by a glass tube capped on either end with cotton balls that have been soaked in ammonia or hydrochloric acid. The NH_{3} vapor travels faster down the tube than the HCl vapor, as evidenced by the location of the ring of ammonium chloride.
The ring of ammonium chloride appears closer to the HCl end of the tube, indicating that the ammonia vapor has diffused faster than the hydrogen chloride vapor.
Lesson Summary
 According to Dalton’s law, the sum of the individual partial pressures of the gases in a mixture is equal to the total gas pressure.
 The partial pressure of a gas in a mixture can be calculated by multiplying its mole fraction by the total pressure.
 The pressure of a gas that has been collected by water displacement is determined by subtracting the vapor pressure of water at that temperature from the total pressure of the sample.
 Diffusion is the net movement of molecules from an area of high concentration to an area of lower concentration. Effusion is the process of a gas escaping from a tiny hole in its container. Graham’s Law allows the relative rates of diffusion or effusion for various gases to be calculated from the molar masses of the gases. Lighter gases move at a faster rate than heavier gases, resulting in faster diffusion and effusion rates.
Lesson Review Questions
Reviewing Concepts
 Explain what is meant by the term partial pressure.
 What is a mole fraction? What must be the sum of the mole fractions for all substances in a mixture?
 Why does a gas displace water when bubbled into it?
Problems
 Hydrogen, nitrogen, and oxygen gases are mixed into a container with a total pressure of 767 mmHg. If the partial pressure of the hydrogen is 85 mmHg, and the partial pressure of the nitrogen is 517 mmHg, what is the partial pressure of the oxygen?
 On the summit of Mount McKinley, the highest mountain in North America, the atmospheric pressure on a certain day is about 338 mmHg. Given that the mole fraction of oxygen in Earth’s atmosphere is 0.2095, what is the partial pressure of oxygen on the summit of Mount McKinley?
 Rank the following gases by their average velocities at a given temperature, from slowest to fastest: Ar, SO_{2}, N_{2}, CH_{4}, Kr.
 A mixture is made of 32.5 g of helium gas and 38.7 g of neon gas.
 Calculate the mole fraction of each of the two gases.
 The total pressure in the container is 278 kPa. Calculate the partial pressure of each gas.
 2.20 liters of gas is collected over water at a temperature of 30°C and a total pressure of 98.05 kPa. Find the volume that the dry form of this gas would occupy at STP.
 What is the ratio of the effusion rates for oxygen gas and xenon gas when both gases are at the same temperature and pressure?
 At a certain temperature and pressure, the average velocity of N_{2} molecules is 524 m/s. What is the average velocity of O_{2} molecules at the same temperature and pressure?
 Water can be made to decompose into hydrogen and oxygen gases by passing an electrical current through it. \begin{align*}2\text{H}_2\text{O}{(l)} \rightarrow 2\text{H}_2{(g)}+\text{O}_2{(g)}\end{align*} A 50.0 g sample of water completely decomposes in a sealed 10.0 L container that was previously under vacuum. The temperature is 25°C. Determine the partial pressure (in atm) of both gases produced in the reaction.
Further Reading / Supplemental Links
 Dalton’s Law of Partial Pressure and Mole Fractions, (http://www.kentchemistry.com/links/GasLaws/dalton.htm)
 Graham’s Law of Effusion, (http://www.kentchemistry.com/links/GasLaws/GrahamsLaw.htm)
Points to Consider
Many chemical reactions take place between reactants that have been dissolved in water. A thorough understanding of the properties of water is essential to an analysis of reactions that occur in aqueous solution.
 How does the shape of a water molecule affect its polarity and its properties?
 How do the properties of water change when it freezes to ice?
Notes/Highlights Having trouble? Report an issue.
Color  Highlighted Text  Notes  

Please Sign In to create your own Highlights / Notes  
Show More 