Gases and Gas Laws

Introduction The properties of gases will be introduced along with five ways of predicting the behavior of gases: Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, Avogadro’s Law and the Ideal Gas Law.

Properties of Gases You can predict the behavior of gases based on the following properties: Pressure Volume Amount (moles) Temperature Lets review each of these briefly…

Physical Characteristics of Gases Although gases have different chemical properties, gases have remarkably similar physical properties.  Gases always fill their containers (recall solids and liquids). No definite shape and volume  Gases are highly compressible: Volume decreases as pressure increases. Volume increases as pressure decreases.  Gases diffuse (move spontaneously throughout any available space).  Temperature affects either the volume or the pressure of a gas, or both.

Definition of a Gas Therefore a definition for gas is: a substance that fills and assumes the shape of its container, diffuses rapidly, and mixes readily with other gases.

Pressure Automobile tires, basketballs, balloons, and soda bottles.

Pressure Evangelista Torricelli invented the –BAROMETER: device to measure pressure –At sea level the height of the column of mercury is 760 mm. Pressure exerted by the atmospheric gases on the surface of the mercury in the dish keeps the mercury in the tube.

Measuring pressure of a confined gas Manometer Atmospheric pressure > gas Atmospheric pressure < gas

Pressure Units of Pressure -1mm Hg = 1 torr Standard atmosphere (atm) 1 atm. = 760.0mm Hg = torr

The height of mercury in a mercury barometer is measured to be 732 mm Hg. Represent this pressure in atm and torr. Problem 1 atm. = 760.0mm Hg = torr

Answer: atm; 732 torr The height of mercury in a mercury barometer is measured to be 732 mm Hg. Represent this pressure in atm and torr. Problem 1 atm. = 760.0mm Hg = torr

Kelvin Temperature Scale Kelvin Temp Scale: based on absolute zero — all kinetic motion stops Formulas °C = K K= °C+273 0°C = 273K 30°C =303 K -20°C = 253 K

Various gas laws describe the relationships among four of the important physical properties of gases: 1.Volume (liters) 2.Pressure (usually in atms) 3. Temperature (Kelvin) 4. Amount (moles)

Five Gas Laws Boyle’s Law Charles Law Gay-Lussac’s Law Avogadro’s Law Ideal Gas Law

Boyle’s Law Describes the pressure-volume relationship of gases if the temperature and amount are kept constant. P 1 V 1 = P 2 V 2 P 1 V 1 = initial pressure and volume P 2 V 2 = final pressure and volume

Boyle’s Law Volume and pressure are inversely proportional

Illustration of Boyle’s law. P 1 V 1 = P 2 V 2

Pressure and Volume: Boyle’s Law A sample of neon gas has a pressure of 7.43 atm in a container with a volume of 45.1 L. This sample is transferred to a container with a volume of 18.4 L. What is the new pressure of the neon gas? P 1 V 1 = P 2 V atm

Pressure and Volume: Boyle’s Law A steel tank of oxygen gas has a volume of 2.00L. If all of the oxygen is transferred to a new tank with a volume of 5.50 L, the pressure is measured to be 6.75 atm. What was the original pressure of the oxygen gas? 18.6 atm

Charles’s Law Describes the temperature-volume relationship of gases if the pressure and amount are kept constant. the volume of a gas increases proportionally as the temperature of the gas increases. V1 = V2 T1 T2

Charles’s Law shows that the volume of a gas (at constant pressure) increases with the temperature. Charles’s Law is used to explain what happens to a balloon when placed in a freezer.

Volume and Temperature: Charles’s Law A sample of methane gas is collected at 285 K and cooled to 245K. At 245 K the volume of the gas is 75.0 L. Calculate the volume of the methane gas at 285K. V1 = V2 T1 T2 V1 = 87.2 L

Volume and Temperature: Charles’s Law A 2.45 L sample of nitrogen gas is collected at 273 K and heated to 325K. Calculate the volume of the nitrogen gas at 325 K. V1 = V2 T1 T2 V2 = 2.92 L

Volume and Temperature: Charles’s Law Consider a gas with a volume of 5.65 L at 27 C and 1 atm pressure. At what temperature will this gas have a volume of 6.69 L and 1 atm pressure. V1 = V2 T1 T2 T2 = 82 o C (355K)

Volume and Temperature: Charles’s Law Consider a gas with a volume of 9.25L at 47 o C and 1 atm pressure. At what temperature does this gas have a volume of 3.50 L and 1 atm pressure. T2 = -152 o C (121K)

Gay-Lussac’s Law Describes the temperature-pressure relationship of gases if the volume and amount are kept constant. the pressure of a gas increases proportionally as the temperature of the gas increases. P1 = P2 T1 T2

Pressure and Temperature: Gay- Lussac’s Law If you have a tank of gas at 800 torr pressure and a temperature of 250 Kelvin, and it is heated to 400 Kelvin, what is the new pressure? P1 = P2 T1 T2 P2 = 1,280 torr

Sample Problem 13.3 Page 448 Classwork Page 448, Problems 8 and 9

Combined Gas Law This is when all variables (T,P, and V) are changing P1 V1 = P2 V2 T1 T2

The Combined Gas Law Consider a sample of helium gas at 23 o C with a volume of 5.60 L at a pressure of 2.45 atm. The pressure is changed to 8.75 atm and the gas is cooled to 15 o C. Calculate the new volume of the gas using the combined gas law equation. P1 V1 = P2 V2 T1 T2 V2 = 1.53 L

The Combined Gas Law Consider a sample of helium gas at 28 o C with a volume of 3.80 L at a pressure of 3.15 atm. The gas expands to a volume of 9.50 L and the gas is heated to 43 o C. Calculate the new pressure of the gas using the combined gas law equation. P2 = 1.32 atm

Avogadro’s Law Describes the amount-volume relationship of gases if the pressure and temperature are kept constant. Equal volume of gases at the same temperature and pressure contain equal number of moles of gas. the volume of a gas is directly proportional to the number of moles of gas V1 = V2 n1 n2

Avogadro’s Law V 1 = V 2 n 1 n 2

Volume and Moles: Avogadro’s Law If 2.55 mol of helium gas occupies a volume of 59.5 L at a particular temperature and pressure, what volume does 7.83 mol of helium occupy under the same conditions? V1 = V2 n1 n2 V2 = 183 L

Volume and Moles: Avogadro’s Law If 4.35 g of neon gas occupies a volume of 15.0 L at a particular temperature and pressure, what volume does 2.00 g of neon gas occupy under the same conditions? V2 = 6.90 L

Avogadro’s Law A very useful consequence of Avogadro’s law is that the volume of a mole gas can be calculated at any temperature and pressure. An extremely useful form to know when calculating the volume of a mole of gas is 1 mole of any gas at STP occupies 22.4 liters. STP stands for standard temperature and pressure. Standard Pressure: 1.00 atm (760 torr or 760 mm Hg) Standard Temperature: 273K

Molar Volume When gases are at STP:  1 mole of any gas = 22.4 L/mol  We will return to this concept with gas stoichiometry problems.

Ideal Gas Equation PV=nRT

The Ideal Gas Law PV=nRT R=Universal gas constant (proportionality constant) R= L atm/ K Pressure is in atm. Volume is in liters Temperature is in Kelvin (K)

The Ideal Gas Law A sample of neon gas has a volume of 3.45 L at 25 o C and a pressure of 565 torr. Calculate the number of moles of neon present in the gas sample. PV=nRT R= L atm/K Convert pressure to atm and temperature to K n = mol

The Ideal Gas Law A mol sample of argon gas has a volume of 9.00 L at a pressure of 875 mm Hg. What is the temperature (in K) of the gas? PV=nRT T = 505K

Dalton’s Law of Partial Pressures Objectives: To understand the relationship between the partial and total pressures of a gas mixture, and to use this relationship in calculations.

Dalton’s Law of Partial Pressures John Dalton: For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present. Partial pressure: pressure that each gas would exert if it were alone in the container. The pressure that each gas exerts in the mixture is independent of that exerted by other gases present

Dalton’s law of partial pressures: P total = P 1 + P 2 + P 3 P 1, P 2 and P 3 are the partial pressure of component gases 1, 2 and 3.

P total =n total (RT/V) The volume of the individual gas particles is not important.

Gases Collected by Water Displacement Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water, which is more dense. You can apply Dalton’s law of partial pressures in calculating the pressure of gases collected in this way.

The production of oxygen by thermal decomposition.

A gas collected by water displacement is not pure but is always mixed with water vapor. Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure, known as water-vapor pressure. P atm = P gas + P H20 P atm is read from a barometer in the lab and is equal to P total P H20 varies with temperature and is in a reference table

Dalton’s Law of Partial Pressures A sample of solid potassium chlorate KClO 3, was heated in a test tube and decomposed according to the reaction: 2KClO 3 (s) 2KCl(s) + 3O 2 The oxygen produced was collected by displacement of water. The barometric pressure and the temperature during the experiment were torr and 20 o C, respectively. The vapor pressure of water at 20 o C is 17.5 torr. What is the partial pressure of oxygen collected?

Dalton’s Law of Partial Pressures P total = P atm = P O2 + P H2O P H2O = 17.5 torr (vapor pressure of water at 20 o C) torr = P O torr P O2 = torr

Dalton’s Law of Partial Pressures A sample of solid potassium chlorate KClO 3, was heated in a test tube and decomposed according to the reaction: 2KClO 3 (s) 2KCl(s) + 3O 2 The oxygen produced was collected by displacement of water at 22 o C. The resulting mixture of O2 and H2O vapor had a total pressure of 754 torr and a volume of 0.650L. Calculate the partial pressure of O2 in the gas collected and the number of moles of O2 present. The vapor pressure of water at 22 o C is 21 torr.

Dalton’s Law of Partial Pressures P total =P O2 + P H2O 754=P O P O2 = 733 torr n O2 =P O2 V733/760=0.964 atm RT n O2 = (0.964 atm)(0.650L) ( ) (295 K) n =.026 moles

Dalton’s Law of Partial Pressures A sample of oxygen gas is saturated with water vapor at 30.0 o C. The total pressure is 753 torr and the vapor pressure of water at 30.0 o C is torr. What is the partial pressure of the oxygen gas in atm? atm

Gas Stoichiometry Objectives: 1) To understand the molar volume of an ideal gas. 2) To review the definition of STP 3) To use these concepts and the ideal gas equation.

Gas Stoichiometry For 1 mole of an ideal gas at 0 o C (273K) and 1 atm, the volume will be. V= nRT/P = (1.0 mol)( )(273K) = 22.4L 1 atm L is called the molar volume at standard temperature and pressure (abbreviated STP). Contains 1 mole of an ideal gas at STP.

Gas Stoichiometry When magnesium reacts with hydrochloric acid, hydrogen gas is produced: Mg(s) + 2HCl MgCl 2 (aq) + H 2 (g) Calculate the volume of hydrogen gas produced at STP by reacting 5.0g Mg and an excess of HCl (aq)

Volume = 4.66 L = moles H g Mg 0.21 Units match (1.0 atm)(V) = 0.21(0.082)(273K) PV=nRT

Volume = 4.66 L 5.0 g Mg 1 mol H L

Gas Stoichiometry When subjected to an electric current, water decomposes to hydrogen and oxygen gas: 2H 2 O(l) 2H 2 (g) + O 2 (g) If 25.0g of water is decomposed, what volume of oxygen gas is produced at STP?

Volume = 15.5 L = moles O g H Units match (1.0 atm)(V) = 0.69(0.082)(273K) PV=nRT

= L O g H 2 O 1 mol H 2 O 18.0 g H 2 O 1 mol O 2 2 mol H 2 O 22.4 L O 2 1 mol O Method 2 1 mole of a gas occupies 22.4 liters of volume

How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = L O g KClO 3 1 mol KClO g KClO 3 3 mol O 2 2 mol KClO L O 2 1 mol O