Gas Laws
| Boyle's Law | Amonton's Law | Charles' Law | Gay-Lussac's Law |
| Avogadro's Hypothesis | The Ideal Gas Equation | Ideal Gas Calculations | Dalton's Law of Partial Pressures |
Torricelli's experiment did more than just show that air has weight; it also provided away of creating a vacuum because the space above the column of mercury at the top of a barometer is almost completely empty. (It is free ofair or other gases except a negligible amount of mercury vapor.) Torricelli's work with avacuum soon caught the eye of the British scientist Robert Boyle.
Boyle's most famous experiments with gases dealt with what he called the "springof air." These experiments were based on the observation that gases are elastic.(They return to their original size and shape after being stretched or squeezed.) Boylestudied the elasticity of gases in a J-tube similar to the apparatus shown in the figurebelow. By adding mercury to the open end of the tube, he trapped a small volume of air inthe sealed end.
Boyle studied what happened to the volume of the gas in the sealed end of the tube ashe added mercury to the open end.
Boyle noticed that the product of the pressure times the volume for any measurement inthis table was equal to the product of the pressure times the volume for any othermeasurement, within experimental error.
P1V1 = P2V2
This expression, or its equivalent,
is now known as Boyle's Law.
Calculate the pressure in atmospheres in a motorcycle engine at the end of the compression stroke. Assume that at the start of the stroke, the pressure of the mixture of gasoline and air in the cylinder is 745.8 mm Hg and the volume of each cylinder is 246.8 mL. Assume that the volume of the cylinder is 24.2 mL at the end of the compression stroke.
Toward the end of the 1600s, the French physicist Guillaume Amontons built athermometer based on the fact that the pressure of a gas is directly proportional to itstemperature. The relationship between the pressure and the temperature of a gas istherefore known as Amontons' law.
P 
T
Amontons' law explains why car manufacturers recommend adjusting the pressure of yourtires before you start on a trip. The flexing of the tire as you drive inevitably raisesthe temperature of the air in the tire. When this happens, the pressure of the gas insidethe tires increases.
Amontons' law can be demonstrated with the apparatus shown in the figure below, whichconsists of a pressure gauge connected to a metal sphere of constant volume, which isimmersed in solutions that have different temperatures.
 |
| The apparatus for demonstrating Amonton's law consists of . |
The following data were obtained with this apparatus.
In 1779 Joseph Lambert proposed a definition for absolute zero on the temperature scalethat was based on the straight-line relationship between the temperature and pressure of agas shown in the figure above.
He defined absolute zero as the temperature at which the pressure of agas becomes zero when a plot of pressure versus temperature for a gas is extrapolated. Thepressure of a gas approaches zero when the temperature is about -270°C. When moreaccurate measurements are made, the pressure of a gas extrapolates to zero when thetemperature is -273.15°C. Absolute zero on the Celsius scale is therefore -273.15°C.
The relationship between temperature and pressure can be greatly simplified byconverting the temperatures from the Celsius to the Kelvin scale.
TK = ToC +273.15
When this is done, a plot of the temperature versus the pressure of a gas gives astraight line that passes through the origin. Any two points along the line therefore fitthe following equation.
It is important to remember that this equation is only valid if thetemperatures are converted from the Celsius to the Kelvin scale before calculations aredone.
Assume that the pressure in the tires of your car is 32 lb/in2 at 20°C. What is the pressure when the gas in these tires heats up to a temperature of 40°C?
On 5 June 1783, Joseph and Etienne Montgolfier used a fire to inflate a sphericalballoon about 30 feet in diameter that traveled about a mile and one-half before it cameback to earth. News of this remarkable achievement spread throughout France, andJacques-Alexandre-Cesar Charles immediately tried to duplicate this performance. As aresult of his work with balloons, Charles noticed that the volume of a gas is directlyproportional to its temperature.
V T
This relationship between the temperature and volume of a gas, which became known as Charles'law, provides an explanation of how hot-air balloons work. Ever since the thirdcentury B.C., it has been known that an object floats when it weighs less than the fluidit displaces. If a gas expands when heated, then a given weight of hot air occupies alarger volume than the same weight of cold air. Hot air is therefore less dense than coldair. Once the air in a balloon gets hot enough, the net weight of the balloon plus thishot air is less than the weight of an equivalent volume of cold air, and the balloonstarts to rise. When the gas in the balloon is allowed to cool, the balloon returns to theground.
Charles' law can be demonstrated with the apparatus shown in the figure below. A 30-mLsyringe and a thermometer are inserted through a rubber stopper into a flask that has beencooled to 0ºC. The ice bath is then removed and the flask is immersed in a warm-waterbath. The gas in the flask expands as it warms, slowly pushing the piston out of thesyringe. The total volume of the gas in the system is equal to the volume of the flaskplus the volume of the syringe.
The figure below shows a plot of the typical data obtained from this experiment.
This graph provides us with another way of defining absolute zero on the temperaturescale. Absolute zero is the temperature at which the volume of a gasbecomes zero when the a plot of the volume versus temperature for a gas are extrapolated.As expected, the value of absolute zero obtained by extrapolating the data is essentiallythe same as the value obtained from the graph of pressure versus temperature in thepreceding section. Absolute zero can therefore be more accurately defined as thetemperature at which the pressure and the volume of a gas extrapolate to zero.
A plot of the volume versus the temperature of a gas (when the temperatures obtainedare converted from Celsius to the Kelvin scale) becomes a straight line that passesthrough the origin. Any two points along this line can therefore be used to construct thefollowing equation, which is known as Charles' law.
Before using this equation, it is important to remember thattemperatures must be converted from ºC to K.
Assume that the volume of a balloon filled with H2 is 1.00 L at 25°C. Calculate the volume of the balloon when it is cooled to -78°C in a low-temperature bath made by adding dry ice to acetone.
Joseph Louis Gay-Lussac (1778-1850) began his career in 1801 by very carefully showingthe validity of Charles' law for a number of different gases. Gay-Lussac's most importantcontributions to the study of gases, however, were experiments he performed on the ratioof the volumes of gases involved in a chemical reaction.
Gay-Lussac studied the volume of gases consumed or produced in a chemical reactionbecause he was interested in the reaction between hydrogen and oxygen to form water. Heargued that measurements of the weights of hydrogen and oxygen consumed in thisreaction could be influenced by the moisture present in the reaction flask, but thismoisture would not affect the volumes of hydrogen and oxygen gases consumed inthe reaction.
Much to his surprise, Gay-Lussac found that 199.89 parts by volume of hydrogen wereconsumed for every 100 parts by volume of oxygen. Thus, hydrogen and oxygen seemed tocombine in a simple 2:1 ratio by volume.
| hydrogen | + | oxygen |  | water |
| 2 volumes | 1 volume |
Gay-Lussac found similar whole-number ratios for the reaction between other pairs ofgases. The compound we now know as hydrogen chloride (HCl) combined with ammonia (NH3)in a simple 1:1 ratio by volume:
| hydrogen chloride | + | ammonia | | ammonium chloride |
| 1 volume | 1 volume |
Carbon monoxide combined with oxygen in a 2:1 ratio by volume:
| carbon monoxide | + | oxygen | | carbon dioxide |
| 2 volumes | 1 volume |
Gay-Lussac obtained similar results when he analyzed the volumes of gases given offwhen compounds decomposed. Ammonia, for example, decomposed to give three times as muchhydrogen by volume as nitrogen:
| ammonia | | nitrogen | + | hydrogen |
| 1 volume | 3 volumes |
On 31 December 1808, Gay-Lussac announced his law ofcombining volumes to a meeting of the Societ Philomatique in Paris. At that time,he summarized the law as follows: Gases combine among themselves in very simpleproportions. Today, Gay-Lussac's law is stated as follows: The ratio of thevolumes of gases consumed or produced in a chemical reaction is equal to the ratio ofsimple whole numbers.
Use the following balanced chemical equations to explain the results of Gay-Lussac's experiments:
2 H2(g) + O2(g) 2H2O(g)
HCl(g) + NH3(g) NH4Cl(s)
2 CO(g) + O2(g) 2 CO2(g)
2 NH3(g) N2 + 3 H2(g)
Gay-Lussac's law of combining volumes was announced only a few years after John Daltonproposed his atomic theory. The link between these two ideas was first recognized by theItalian physicist Amadeo Avogadro three years later, in 1811. Avogadro argued thatGay-Lussac's law of combining volumes could be explained by assuming that equal volumes ofdifferent gases collected under similar conditions contain the same number of particles.
HCl and NH3 therefore combine in a 1:1 ratio by volume because one moleculeof HCl is consumed for every molecule of NH3 in this reaction and equal volumesof these gases contain the same number of molecules.
| NH3(g) | + | HCl(g) | | NH4Cl(s) |
Anyone who has blown up a balloon should accept the notion that the volume of a gas isproportional to the number of particles in the gas.
V n
The more air you add to a balloon, the bigger it gets. Unfortunately this example doesnot test Avogadro's hypothesis that equal volumes of different gasescontain the same number of particles. The best way to probe the validity of thishypothesis is to measure the number of molecules in a given volume of different gases,which can be done with the apparatus shown in the figure below.
A small hole is drilled through the plunger of a 50-mL plastic syringe. The plunger isthen pushed into the syringe and the syringe is sealed with a syringe cap. The plunger isthen pulled out of the syringe until the volume reads 50 mL and a nail is inserted throughthe hole in the plunger so that the plunger is not sucked back into the barrel of thesyringe. The "empty" syringe is then weighed, the syringe is filled with 50 mLof a gas, and the syringe is reweighed. The difference between these measurements is themass of 50 mL of the gas.
The results of experiments with six gases are given in the table below.
Experimental Data for the Mass of 50-mL Samples of Different Gases
| Compound | Mass of 50 mL of Gas (g) | Molecular Weight of Gas | Number of Gas Molecules | |||
| H2 | 0.005 | 2.02 | 1 x 1021 | |||
| N2 | 0.055 | 28.01 | 1.2 x 1021 | |||
| O2 | 0.061 | 32.00 | 1.1 x 1021 | |||
| CO2 | 0.088 | 44.01 | 1.2 x 1021 | |||
| C4H10 | 0.111 | 58.12 | 1.15 x 1021 | |||
| CCl2F2 | 0.228 | 120.91 | 1.14 x 1021 |
The number of molecules in a 50-mL sample of any one of these gases can be calculatedfrom the mass of the sample, the molecular weight of the gas, and the number of moleculesin a mole. Consider the following calculation of the number of H2 molecules in50 mL of hydrogen gas, for example.
The last column in the table above summarizes the results obtained when thiscalculation is repeated for each gas. The number of significant figures in the answerchanges from one calculation to the next. But the number of molecules in each sample isthe same, within experimental error. We therefore conclude that equal volumes of differentgases collected under the same conditions of temperature and pressure do in fact containthe same number of particles.
Gases can described in terms of four variables: pressure (P), volume (V),temperature (T), and the amount of gas (n). There are five relationshipsbetween pairs of these variables in which two of the variables were allowed to cahngewhile the other two were held constant.
| P | | n | (T and V constant) | ||||
| Boyle's law: | P | | 1/V | (T and n constant) | |||
| Amontons' law: | P | | T | (V and n constant) | |||
| Charles' law: | V | | T | (P and n constant) | |||
| Avogadro's hypothesis: | V | | n | (P and T constant) |
Each of these relationships is a special case of a more general relationship known asthe ideal gas equation.
PV = nRT
In this equation, R is a proportionality constant known as the ideal gasconstant and T is the absolute temperature. The value of R dependson the units used to express the four variables P, V, n, and T.By convention, most chemists use the following set of units.
Calculate the value of the ideal gas constant, R, if exactly 1 mole of an ideal gas occupies a volume of 22.414 liters at 0°C and 1 atmosphere pressure.
The ideal gas equation can be used to predict thevalue of any one of the variables that describe a gas from known values of the otherthree.
Practice Problem 8:
Many gases are available for use inthe laboratory in compressed gas cylinders, in which they are stored at high pressures. Let's calculate the mass of O2 that can be stored at 21ºC and 170 atm in a cylinder with a volume of 60.0 L.
The key to solving ideal gas problems ofteninvolves recognizing what is known and deciding how to use this information.
Let's calculate the mass of the air in a hot-air balloon that has a volume of 4.00 x 105 liters when the temperature of the gas is 30ºC and the presure is 748 mmHg. Let's assume the average molar mass of air is 29.0 grams per mole.
The ideal gas equation can be applied to problems that don't seemto ask for one of the variables in this equation.
Let's calculate the molecular weight of butane if 0.5813 gram of this gas fills a 250.0-mL flask at a temperature of 24.4ºC and a pressure of 742.6 mmHg.
The ideal gas equation can even be used tosolve problems that don't seem to contain enough information.
Let's calculate the density in grams per liter of O2 gas at 0ºC and 1.00 atm.
Gas law problems often ask you to predict what happens when one or more changes aremade in the variables that describe the gas. There are two ways of working these problems.A powerful approach is based on the fact that the ideal gas constant is in fact aconstant.
We start by solving the ideal gas equation for the ideal gas constant.
We then note that the ratio of PV/nT at any time must be equal tothis ratio at any other time.
We then substitute the known values of pressure, temperature, volume, and amount of gasinto this equation and solve for the appropriate unknown. This approach has twoadvantages. First, only one equation has to be remembered. Second, it can be used tohandle problems in which more than one variable changes at a time.
Dalton's Law of Partial Pressures
The CRC Handbook of Chemistry and Physics describes the atmosphere as 78.084%N2, 20.946% O2, 0.934% Ar, and 0.033% CO2 by volume whenthe water vapor has been removed. What image does this description evoke in your mind? Doyou believe that only 20.463% of the room in which you are sitting contains O2?Or do you believe that the atmosphere in your room is a more or less hom*ogeneous mixtureof these gases?
Gases expand to fill their containers. The volume of O2 in your room istherefore the same as the volume of N2. (Both gases expand to fill the room.)When we describe the atmosphere as 20.946% O2 by volume, we mean that thevolume of the atmosphere would shrink by 20.946% if the O2 is removed.
What about the pressure of the different gases in your room? Is the pressure of the O2in the atmosphere the same as the pressure of the N2? We can answer thisquestion by rearranging the ideal gas equation as follows.
According to this equation, the pressure of a gas is proportional to the number ofmoles of gas, if the temperature and volume are held constant. Because the temperature andvolume of the O2 and N2 in the atmosphere are the same, the pressureof each gas must be proportional to the number of the moles of the gas. Because there ismore N2 in the atmosphere than O2, the contribution to the totalpressure of the atmosphere from N2 is larger than the contribution from O2.
John Dalton was the first to recognize that the total pressure of a mixture of gases isthe sum of the contributions of the individual components of the mixture. By convention,the part of the total pressure of a mixture that results from one component is called the partialpressure of that component. Dalton's law of partial pressuresstates that the total pressure of a mixture of gases is the sum of the partial pressuresof the various components.
Calculate the total pressure of a mixture that contains 1.00 g H2 and 1.00 g of He in a 5.00 L container at 21° C.
Dalton derived the law of partial pressures from his work on the amount of water vaporthat could be absorbed by air at different temperatures. It is therefore fitting that thislaw is used most often to correct for the amount of water vapor picked up when a gas iscollected by displacing water. Suppose, for example, that we want to collect a sample of O2prepared by heating potassium chlorate until it decomposes.
| 2 KClO3(s) | | 2 KCl(s) | + | 3 O2(g) |
The gas given off in this reaction can be collected by filling a flask with water,inverting the flask in a trough, and then letting the gas bubble into the flask as shownin the figure below.
Because some of the water in the flask will evaporate during the experiment, the gasthat collects in this flask is going to be a mixture of O2 and water vapor. Thetotal pressure of this gas is the sum of the partial pressures of these two components.
PT = Poxygen + Pwater
The total pressure of this mixture must be equal to atmospheric pressure. (If it wasany greater, the gas would push water out of the container. If it was any less, waterwould be forced into the container.) If we had some way to estimate the partial pressureof the water in this system, we could therefore calculate the partial pressure of theoxygen gas.
By convention, the partial pressure of the gas that collects in aclosed container above a liquid is known as the vapor pressure of theliquid. If we know the temperature at which a gas is collected by displacing water, and weassume that the gas is saturated with water vapor at this temperature, we can calculatethe partial pressure of the gas by subtracting the vapor pressure of water from the totalpressure of the mixture of gases collected in the experiment.
Let's calculate the number of grams of O2 that can be collected by displacing water from a 250-mL flask at 21ºC and 746.2 mmHg.
