Properties of Gases. There are 5 important properties of gases:

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1 Chapter 11 Gases Properties of Gases There are 5 important properties of gases: Confined gases exerts pressure on the wall of a container uniformly Gases have low densities Gases can be compressed Gases can expand to fill their contained uniformly Gases mix completely with other gases in the same container Chapter 11 2

2 Kinetic Molecular Theory of Gases The Kinetic Molecular Theory of Gases is the model used to explain the behavior of gases in nature. 1. Gas particles (atoms or molecules) move constantly in random directions (in straight lines) with rapid velocities. 2. Gases are made up of very tiny atoms or molecules meaning that gases are mostly empty space! This fact is what allows for the compression of a gas. 3. Gas molecules or atoms have no attraction for one another. This is the result of the distance between the particles! They undergo elastic collisions with one another 4. The average kinetic energy of a gas is proportional to the temperature in Kelvins. Chapter 11 3 Properties that Describe a Gas These properties are all related to one another. When one variable changes, it causes the other three to react in a predictable manner. Chapter 11 4

3 Gas Pressure (P) Gas pressure (P) is the result of constantly moving gas molecules striking the inside surface of their container. Chapter 11 5 Atmospheric Pressure Atmospheric pressure is the pressure exerted by the air on the earth. Evangelista Torricelli invented the barometer in 1643 to measure atmospheric pressure. Atmospheric pressure is 760 mm of mercury or 1 atmosphere (atm) at sea level. What happens to the atmospheric pressure as you go up in elevation? Chapter 11 6

4 Units of Pressure Standard pressure is the atmospheric pressure at sea level, 760 mm of mercury. Here is standard pressure expressed in other units: Chapter 11 7 Gas Pressure Conversions The barometric pressure is 697 torr. What is the barometric pressure in atmospheres? Step 1: Determine what you have: 697 torr Step 2: Determine what you want:??? atm Step 3: Write out your plan to convert torr to atm Step 4: Select conversion factor(s): 1 atm = 760 torr 697 torr 1 atm 760 torr = atm Chapter 11 8

5 Properties that Describe a Gas These properties are all related to one another. When one variable changes, it causes the other three to react in a predictable manner. Chapter 11 9 Gas Law Problems Chapter 11 10

6 Boyle s Law (V and P) Boyle s Law states that the volume of a gas is inversely proportional to the pressure at constant temperature. Mathematically, we write: P α 1. V For a before and after situation: P 1 V 1 = P 2 V 2 Chapter Boyle s Law Problem A 1.50 L sample of methane gas exerts a pressure of 1650 mm Hg. What is the final pressure if the volume changes to 7.00 L? Step 1: Organize the data in a table with initial (1) and final (2) conditions. Step 2: Rearrange the gas law to solve for the unknown (here, P 2 ) Step 3: Plug your numbers into Boyle s Law and solve for P 2 P 1 V 1 = P 2 V 2 rearranges to P 1 V 1 V 2 = P 2 (1650 mm Hg )(1.50 L) 7.00 L = 354 mm Hg Chapter 11 12

7 Charles Law (V and T) In 1783, Jacques Charles discovered (while hot air ballooning) that the volume of a gas is directly proportional to the temperature in Kelvin. Mathematically, we write: T α V For a before and after situation: V 1 V = 2 T 1 T 2 Chapter Charles Law Problem A 275 L helium balloon is heated from 20 C to 40 C. What is the final volume at constant P? Step 1: Temperatures must be in Kelvin. Convert from C to K if needed: 20 C = 293 K and 40 C = 313 K Step 2: Organize the data in a table with initial (1) and final (2) conditions. Step 3: Rearrange the gas law to solve for the unknown (here, V 2 ) Step 4: Plug your numbers into Charles Law and solve for V 2 V 1 V = 2 rearranges to T 1 T 2 (275 L)(313 K) = 294 L 293 K V 1 T 2 T 1 = V 2 Chapter 11 14

8 Gay-Lussac s Law (P and T) In 1802, Joseph Gay-Lussac discovered that the pressure of a gas is directly proportional to the temperature in Kelvin. Mathematically, we write: T α P For a before and after situation: P 1 P = 2 T 1 T 2 Chapter Gay-Lussac s Law Problem A steel container of nitrous oxide at 15.0 atm is cooled from 25 C to 40 C. What is the final volume at constant V? Step 1: Temperatures must be in Kelvin. Convert from C to K if needed: 25 C = 298 K and -40 C = 233 K Step 2: Organize the data in a table with initial (1) and final (2) conditions. Step 3: Rearrange the gas law to solve for the unknown (here, P 2 ) Step 4: Plug your numbers into Gay-Lussac s Law and solve for P 2 P 1 P = 2 rearranges to T 1 T 2 P 1 T 2 T 1 = P 2 (15.0 atm)(298 K) = 11.7 atm 233 K Chapter 11 16

9 Combined Gas Law When we introduced Boyle s, Charles, and Gay- Lussac s Laws, we assumed that one of the variables remained constant. Experimentally, all three (temperature, pressure, and volume) usually change. By combining all three laws, we obtain the combined gas law: P 1 V 1 P 2 V = 2 T 1 T 2 Chapter Combined Gas Law Problem Lets apply the combined gas law to 10.0L of carbon dioxide gas at 300 K and 1.00 atm. If the volume and Kelvin temperature double, what is the new pressure? Conditions P V T Initial 1.00 atm 10.0 L 300 K Final P L 600 K Chapter 11 18

10 Avogadro s Law In the previous laws, the amount of gas was always constant. However, the amount of a gas (n) is directly proportional to the volume of the gas, meaning that as the amount of gas increases, so does the volume. Mathematically, we write: n α V For a before and after situation: V 1 V = 2 n 1 n 2 Chapter Avogadro s Law Problem A steel container contains 2.6 mol of nitrous oxide with a volume 15.0 L. If the amount of nitrous oxide is increased to 8.4 mol, what is the final volume at constant T and P? Step 1: Organize the data in a table with initial (1) and final (2) conditions. Step 2: Rearrange the gas law to solve for the unknown (here, V 2 ) Step 3: Plug your numbers into Avogardro s Law and solve for V 2 V 1 V = 2 rearranges to n 1 n 2 V 1 n 2 n 1 = V 2 (15.0 L)(8.4 mol) 2.6 mol = 48.5 L Chapter 11 20

11 Gas Laws Summary Chapter Molar Volume and STP Standard temperature and pressure (STP) are defined as 0 C and 1 atm. At standard temperature and pressure, one mole of any gas occupies 22.4 L. The volume occupied by one mole of gas (22.4 L) is called the molar volume. 1 mole Gas = 22.4 L Chapter 11 22

12 Molar Volume Problem Chapter Molar Volume Calculation Volume to Moles A sample of methane, CH 4, occupies 4.50 L at STP. How many moles of methane are present? Step 1: Determine what you have: 4.50 L Step 2: Determine what you want:??? mol Step 3: Write out your plan to convert L to mol Step 4: Select conversion factor(s) needed for plan Volume Molar Volume Moles 1 mol CH 4.50 L CH 4 4 = mol CH 22.4 L CH 4 4 Chapter 11 24

13 Mole Unit Factors We now have three interpretations for the mole: 1 mol = particles 1 mol = molar mass 1 mol = 22.4 L (at STP for a gas) This gives us 3 unit factors to use to convert between moles, particles, mass, and volume. Chapter Mole Calculation - Grams to Volume What is the mass of 3.36 L of ozone gas, O 3, at STP? Step 1: Determine what you have: 3.36 L Step 2: Determine what you want:??? g O 3 Step 3: Write out your plan to convert L to grams Step 4: Select conversion factor(s) needed for plan Grams Molar Mass Moles Molar Volume Volume 1 mol O 3.36 L O g O L O 3 1 mol O 3 = 7.20 g O 3 Chapter 11 26

14 Mole Calculation Molecules to Volume How many molecules of hydrogen gas, H 2, occupy L at STP? Step 1: Determine what you have: L H 2 Step 2: Determine what you want:??? molecule H 2 Step 3: Write out your plan to convert L to molecules Step 4: Select conversion factor(s) needed for plan Volume Molar Volume Moles Avogadro s Number Atoms L H 2 1 mol H molecules H L H 2 1 mole H molecules H 2 Chapter Gas Density The density of a gas is much less than that of a liquid. We can calculate the density of any gas at STP easily. The formula for gas density at STP is: molar mass in grams (MM) molar volume in liters (MV) = density, g/l Chapter 11 28

15 Calculating Gas Density What is the density of ammonia gas, NH 3, at STP? Step 1: Determine what you know: NH 3 and the molar volume Step 2: Determine what you want: density of NH 3 Step 3: Determine what you need to calculate density: MM of NH 3 Step 4: Calculate what you need, then use the formula to solve for the unknown. Calculate the molar mass for ammonia: (N) + 3(1.01) (H) = g/mol NH g/mol 22.4 L/mol = g/l Chapter Molar Mass of a Gas We can also use molar volume to calculate the molar mass of an unknown gas. We can calculate the Molar Mass of any gas at STP easily. The formula for Molar Mass at STP is: Molar Mass = Density (g/l) Molar Volume (L/mol) MM = D MV Chapter 11 30

16 Molar Mass of a Gas 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass? Step 1: Determine what you have: 1.00 L and 1.96 g unknown gas Step 2: Determine what you want:??? g/mol unknown gas Step 3: Write out your plan to convert L and grams to g/mol Step 4: Use the formula for Molar Mass of a gas at STP and solve. Molar Mass = Density (g/l) x Molar Volume (L/mol) 1.96 g 1.00 L 22.4 L = 43.9 g/mol 1 mole Chapter The Ideal Gas Law The four properties used in the measurement of a gas (Pressure, Volume, Temperature and moles) can be combined into a single gas law: PV = nrt Here, R is the ideal gas constant and has a value of: atm L/mol K Note the units of R. When working problems with the Ideal Gas Law, your units of P, V, T and n must match those in the constant! Chapter 11 32

17 Ideal Gas Law Problem How many moles of hydrogen gas occupy L at STP? Step 1: Determine what you have: L of H 2 gas at STP (so 1.00 atm and 273 K) Step 2: Determine what you want:??? moles of H 2 gas (n H2 ) Step 3: Determine what formula you can use to calculate n H2 Step 4: Rearrange the formula for n and solve. n = PV. RT n = (1 atm)(0.500 L). ( atm L/mol K)(273K) n = moles Chapter Gases in Chemical Reactions Gases are involved as reactants in numerous chemical reactions. Typically, the information given for a gas in a reaction is its Pressure (P), volume (V) and temperature (T). We use this information and the Ideal Gas Law to determine the moles of the gas (n). Once we have this information, we can proceed with the problem as we would any other stoichiometry problem. A (g) + X (s) B (s) + Y (l) Chapter 11 34

18 Reaction with a Gas Hydrogen gas is formed when zinc metal reacts with hydrochloric acid. How many liters of hydrogen gas at STP are produced when 15.8 g of zinc reacts? Zn (s) + 2HCl (aq) H 2 (g) + ZnCl 2 (aq) Step 1: Determine what you know: 15.8 g Zn reacts at STP Step 2: Determine what you want:??? L of H 2 are produced Step 3: Write out your plan to convert g Zn to L H 2 Step 4: Select conversion factor(s) and/or formulas needed for plan Molar Mole Molar Mass of Zn Ratio Volume Grams of Zn Moles of Zn Moles of H h Liters of H 2 Chapter Reaction with a Gas Hydrogen gas is formed when zinc metal reacts with hydrochloric acid. How many liters of hydrogen gas at a pressure of 755 atm and 35 C are produced when 15.8 g of zinc reacts? Zn (s) + 2HCl (aq) H 2 (g) + ZnCl 2 (aq) Step 1: Determine what you know: 15.8 g Zn reacts at 755 atm and 35 C Step 2: Determine what you want:??? L of H 2 are produced Step 3: Write out your plan to convert g Zn to L H 2 Step 4: Select conversion factor(s) and/or formulas needed for plan Molar Mole Ideal Gas Mass of Zn Ratio Law Grams of Zn Moles of Zn Moles of H h Liters of H 2 Chapter 11 36