Gases Chapter 13

13.1 The Gas Laws The gas laws apply to ideal gases, which are described by the kinetic theory in the following five statements. Gas particles do not attract or repel each other Gas particles are much smaller than the spaces between them. Gas particles are in constant, random motion. No kinetic energy is lost when gas particles collide with each other or within the walls of their container. All gases have the same kinetic energy at a given temperature.

Gases are made up of Two Types of Particles Molecules (2 or more atoms bound together) - Examples: CH4 , O2 , N2 2. Atoms (Noble gases: He, Ne, Ar) Gas particles are widely spread out and moving around:

Measuring Gases Four variables we use to measure a gas and talk about its properties Amount Volume Temperature Pressure (force exerted on the walls of its container)

Amount of Gas Quantity of gas is expressed in units of MOLES Moles allow us to convert to grams of gas, liters of gas, or the number of gas particles we have Don’t forget the three conversion factors… grams 22.4 L 6.02 x 1023 items 1 mole 1 mole 1 mole

Volume Volume of a gas = volume of container Measured in units of LITERS or unit of length3 1 L = 1000 cm3 = .001 m3

Temperature The temperature is often measured in degrees Celsius ALWAYS CONVERT TO KELVIN when doing calculations To Convert: K = °C + 273

Pressure Due to gas particles colliding w/ container (exert force on an area) UNITS: ATMOSPHERES (atm): close to the average atmospheric pressure at sea level What’s atmospheric pressure? The pressure on earth’s surface due to the mass of air in the atmosphere Pascal (Pa) or Kilopascal (kPa) 1 atm = 101,325 Pa = 101.3 kPa mm Hg (millimeters of mercury) 1 atm = 760 mm Hg Torr 1 atm = 760 torr

UNITS In other words… 1 atm = 101.3 kPa = 101325 Pa = 760 mm Hg = 760 torr

STP Stands for: Standard Temp. and Pressure B/c gas behavior depends on Temperature and Pressure, we compare properties of different gases at a standard temperature and pressure STP = 1 atm, 0 ° C

GAS LAWS The Mathematical representations of relationships between the four variables used to describe a gas…

1. Boyle’s Law (P-V relationship) As pressure DECREASES, volume INCREASES As pressure INCREASES, volume DECREASES (inverse relationship) Boyle's Law Video

Boyle’s Law (Equation) AT CONSTANT TEMPERATURE EQUATION: P1V1 = P2V2

Practice Problem (Boyle’s Law) Suppose I have a 2.5 L balloon filled with He gas. The pressure in the balloon is 1.04 atm. What will the volume of the balloon be if I increase the pressure to 1.54 atm? Boyle's Law DEMOS

Solution P1 = 1.04 atm P2 = 1.54 atm V1 = 2.5 L V2 = ? P1V1 = P2V2 (1.04 atm)(2.5 L) = (1.54 atm)(V2) (1.04 atm)(2.5 L) = V2 (1.54 atm) 1.69 L = V2 * NOTE: It is important for V to be in LITERS. Convert to L if you are given mL or some other unit of volume.

2. Charles’s Law (T-V Relationship) Consider a gas in a cylinder w/ a moveable piston... If we DECREASE the temperature, we DECREASE the volume Heating a Balloon Video HIGH TEMP   LOW TEMP

Charles’s Law AT CONSTANT PRESSURE… As temperature increases, so does the volume of gas. As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

Charles’s Law: Equation AT CONSTANT PRESSURE… Equation : V1 = V2 T1 T2 As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

Practice Problem…(You try!) Say I have a balloon that contains 5.1 L of gas at room temperature (25 ° C). What will be the new volume of the gas if I put it in the freezer (15° C)? FIRST convert the temperature values to Kelvin! K = ° C + 273

Solution V1 = 5.1 L V2 = ? T1 = 25°C + 273 K = 298 K T2 = 15°C + 273 K = 288 K Use the equation V1 = V2 T1 T2 5.1 L = V2 298 K 288 K (288 K)(5.1 L) = V2 298 K V2 = 4.93 L

Practice Problem…(You try!) Say I have a balloon that contains 5.1 L of gas at room temperature (25 ° C). What will be the new volume of the gas if I put it in the freezer (15° C)? FIRST convert the temperature values to Kelvin! K = ° C + 273

3. Gay-Lussac’s Law (P-T Relationship) The pressure of a fixed amount of gas varies directly with temperature when the volume remains constant. If we INCREASE the temperature, we INCREASE the pressure Candle-Water Demo Egg in a Bottle Demo If volume is a constant, then there is a direct relationship between temp. and pressure

Gay-Lussac’s Law AT CONSTANT VOLUME… As temperature increases, so does the pressure of gas. As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

Gay-Lussac’s Law: Equation AT CONSTANT VOLUME… Equation : P1 = P2 T1 T2 As temperature increases, so does the volume of gas when the amount of gas and pressure do not change.

Practice Problem…(You try!) The pressure of oxygen gas inside a canister is 5.00 atm at 25.0°C. The canister is located at a camp high on Mount Everest. If the temperature there falls to -10.0°C, what is the new pressure inside the canister? REMEMBER convert the temperature values to Kelvin! K = ° C + 273

Solution T1 T2 P1 = P2 P1 = 5.00 atm, T1 = 25.0°C + 273 = 298 K P2 = ????, T2 = -10.0°C + 273 = 263 K 5.00 atm = P2 298 K 263 K P2 = 4.41 atm

4. Avogadro’s Principle (V-n Relationship) Equal volumes of gases at the same temperature and pressure contain equal numbers of particles. If volume is a constant, then there is a direct relationship between temp. and pressure 28

Avogadro’s Law AT CONSTANT TEMP and PRESSURE… The volume of a gas is completely dependent on how many moles of gas are in the container. As temperature increases, so does the volume of gas when the amount of gas and pressure do not change. 29

Avogadro’s Law: Equation AT CONSTANT TEMP and PRESSURE… Equation : V1 = V2 n1 n2 (n= number of moles) Remember 1 mol = 22.4 L at STP As temperature increases, so does the volume of gas when the amount of gas and pressure do not change. 30

Practice Problem…(You try!) How many moles of acetylene (C2H2) gas occupy a volume of 3.25 L at STP?

Solution n1 n2 V1 = V2 V1 = 3.25 L, n1 = ???? mol 3.25 L = 22.4L n1 1 mol n1 = 0.145 mol

5. Dalton’s Law of Partial Pressures… In a mixture of gases, each gas exerts a pressure as if it were on its own The total pressure of a mixture of gases is the sum of the pressure due to each gas PT = P1 + P2 + P3 + … The total pressure (PT) is the sum of the pressures from each gas (P1 = pressure of gas 1, P2 = pressure of gas 2, etc)

Practice Problem What is the atmospheric pressure if the partial pressures of nitrogen, oxygen, and argon (the components of air) are 604.5 mm Hg, 162.8 mm Hg, and 0.5 mm Hg, respectively?

Solution Use Dalton’s Law… PT = P1 + P2 + P3 + … P = 767.8 mm Hg

Gas Law Summary P1V1= P2V2 V1= V2 T1 T2 P1= P2 n1 n2 PT= P1 + P2 + P3 Equations Constant Boyle’s P1V1= P2V2 Temperature & amount Charles’s V1= V2 T1 T2 Pressure & amount Gay-Lussac’s P1= P2 Volume & amount Avogadro’s n1 n2 Temperature & pressure Dalton’s Partial Pressure PT= P1 + P2 + P3

Putting the Laws Together… Combined gas law: P1V1 = P2V2 n1T1 n2T2

Combining the Laws into one equation… The Ideal Gas Law Combining the Laws into one equation…

THE IDEAL GAS LAW: Describes the behavior of an ideal gas in terms of P, T, and n Ideal Gas: gas described by kinetic molecular theory Real gases behave like ideal gases under ordinary conditions (doesn’t apply to VERY LOW T and VERY HIGH P)

The Value of “R” Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 R is called the “gas constant” The value of R depends on the units you need to use to solve a problem Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 mole –K

Units: Which value of “R” do I use?? Depends on the units of P and V P can be in atm or Pa V can be in L or m3 T is ALWAYS in K

Practice Problem 1: PV = nRT How many moles of gas are contained in a 2.0 L container with a pressure of 1.5 atm at 100°C? Write out the variables given. Convert T to Kelvin if needed. P = V = T = What variable is missing? Which value of R do you need to use? Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 mole –K J____ mole- K

Practice Problem 1: PV = nRT How many moles of gas are contained in a 2.0 L container with a pressure of 1.5 atm at 100°C? Write out the variables given. Convert T to Kelvin if needed. P = 1.5 atm V = 2.0 L T = 100°C + 273 = 373 K What variable is missing? n Which value of R do you need to use? 0.0821 atm-L mole-K Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 mole –K J____ mole- K

Practice Problem 1: Plug the values into the equation and solve for the unknown variable… PV = nRT (1.5 atm)(2.0L) = n (0.0821atm-L/mole-K)(373 K) (1.5 atm)(2.0L)__________ = 0.098 moles (0.0821atm-L/mole-K)(373 K)

If the gas is carbon dioxide, how many grams of gas are produced? Carbon dioxide = CO2 0.098 moles x 44 grams = 4.312 grams 1 mole

Practice Problem 2: PV = nRT Detemine the kelvin temperature required for 0.0140 mol of gas to fill a balloon to 1.20 L under 0.988 atm pressure. Write out the variables given. P = V = n = What variable is missing? Which value of R do you need to use? Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 mole –K J____ mole- K

Practice Problem 2: PV = nRT Detemine the kelvin temperature required for 0.0140 mol of gas to fill a balloon to 1.20 L under 0.988 atm pressure. Write out the variables given. P = 0.988 atm V = 1.20 L n = 0.0140 mol What variable is missing? T Which value of R do you need to use? 0.0821 Value of R Units 0.0821 atm-L mole-K 8.314 Pa-m3 mole –K J____ mole- K

Practice Problem 2: Plug the values into the equation and solve for the unknown variable… PV = nRT T = PV nR T = (0.988 atm)(1.20L) = 307 K (0.0140 )(0.0821atm-L/mole-K)