CHAPTER 12 GASES 2 The Gas Laws Describe HOW gases behave. Can be predicted by the theory. Amount of change can be calculated with mathematical equations. The effect of adding gas. When we blow up a balloon we are adding gas molecules. Doubling the the number of gas particles doubles the pressure. (of the same volume at the same temperature). Pressure and the number of molecules are directly related More molecules means more collisions. Fewer molecules means fewer collisions. Gases naturally move from areas of high pressure to low pressure because there is empty space to move in. If you double the number of molecules 1 atm If you double the number of molecules You double the pressure. 2 atm 4 atm As you remove molecules from a container 2 atm As you remove molecules from a container the pressure decreases 1 atm As you remove molecules from a container the pressure decreases Until the pressure inside equals the pressure outside Molecules naturally move from high to low pressure Changing the size of the container In a smaller container molecules have less room to move. Hit the sides of the container more often. As volume decreases pressure increases. 1 atm As the pressure on a gas increases 4 Liters As 2 atm 2 Liters the pressure on a gas increases the volume decreases Pressure and volume are inversely related Temperature Raising the temperature of a gas increases the pressure if the volume is held constant. The molecules hit the walls harder. The only way to increase the temperature at constant pressure is to increase the volume. 300 K If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K one of 2 things happens 600 K 300 K Either the volume will increase to 2 liters at 1 atm 300 K •Or the pressure will increase to 2 atm. •Or someplace in between 600 K Kinetic Molecular Theory Particles • • • • • in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight-line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature. 18 Real Gases Particles in a REAL gas… • have their own volume • attract each other Gas behavior is most ideal… • at low pressures • at high temperatures • in nonpolar atoms/molecules 19 Characteristics of Gases Gases expand to fill any container. • random motion, no attraction Gases are fluids (like liquids). • no attraction Gases have very low densities. • no volume = lots of empty space 20 Characteristics of Gases Gases can be compressed. • no volume = lots of empty space Gases undergo diffusion & effusion. • random motion 21 Temperature Always use absolute temperature (Kelvin) when working with gases. ºF -459 ºC -273 K 0 C 59 F 32 32 212 0 100 273 373 K = ºC + 273 22 Pressure force pressure area Which shoes create the most pressure? 23 Barometer Pressure • measures atmospheric pressure Aneroid Barometer Mercury Barometer 24 Pressure Manometer • measures contained gas pressure U-tube Manometer Bourdon-tube gauge 25 Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr N kPa 2 m 14.7 psi 26 STP STP Standard Temperature & Pressure 0°C 1 atm 273 K -OR- 101.325 kPa 27 Boyle’s Law P Volume (mL) Pressure (torr) P·V (mL·torr) 10.0 20.0 30.0 40.0 760.0 379.6 253.2 191.0 7.60 x 103 7.59 x 103 7.60 x 103 7.64 x 103 PV = k V 28 Boyle’s Law The pressure and volume of a gas are inversely related • at constant mass & temp P PV = k V 29 Charles’ Law V T Volume (mL) Temperature (K) V/T (mL/K) 40.0 44.0 47.7 51.3 273.2 298.2 323.2 348.2 0.146 0.148 0.148 0.147 V k T 30 Charles’ Law The volume and absolute temperature (K) of a gas are directly related • at constant mass & pressure V T V k T 31 Gay-Lussac’s Law Temperature (K) Pressure (torr) P/T (torr/K) 248 273 298 373 691.6 760.0 828.4 1,041.2 2.79 2.78 2.78 2.79 P k T P T 32 Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume P k T P T 33 Combined Gas Law P V PV PV = k T P1V1 P2V2 = T1 T2 P1V1T2 = P2V2T1 34 Gas Law Problems gas occupies 473 cm3 at 36°C. Find its volume at 94°C. A CHARLES’ LAW GIVEN: T V V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3 35 Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: P V V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL 36 Gas Law Problems A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: P T V WORK: V1 = 7.84 cm3 P1V1T2 = P2V2T1 P1 = 71.8 kPa (71.8 kPa)(7.84 cm3)(273 K) T1 = 25°C = 298 K =(101.325 kPa) V2 (298 K) V2 = ? P2 = 101.325 kPa V2 = 5.09 cm3 T2 = 273 K 37 Gas Law Problems A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P T P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C 38 Your Turn A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is change to 1.5 atm what is the new volume? A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change to volume to 43 L? Your Turn What is the temperature of a gas that is expanded from 2.5 L at 25ºC to 4.1L at constant pressure. What is the final volume of a gas that starts at 8.3 L and 17ºC and is heated to 96ºC? Your Turn What is the pressure inside a 0.250 L can of deodorant that starts at 25ºC and 1.2 atm if the temperature is raised to 100ºC? At what temperature will the can above have a pressure of 2.2 atm? Your Turn A 15 L cylinder of gas at 4.8 atm pressure at 25ºC is heated to 75ºC and compressed to 17 atm. What is the new volume? If 6.2 L of gas at 723 mm Hg at 21ºC is compressed to 2.2 L at 4117 mm Hg, what is the temperature of the gas? Avogadro’s Principle Equal volumes of gases contain equal numbers of moles • at constant temp & pressure • true for any gas V k n V n 43 Ideal Gas Law Merge the Combined Gas Law with Avogadro’s Principle: PV V k =R nT T n UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values! 44 Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm3kPa/molK You don’t need to memorize these values! 45 The Ideal Gas Law Pressure times Volume equals the number of moles times the Ideal Gas Constant (R) times the temperature in Kelvin. R does not depend on anything, it is really constant R = 0.0821 (L atm)/(mol K) R = 62.4 (L mm Hg)/(K mol) Ideal Gas Law Problems Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. GIVEN: WORK: P = ? atm PV = nRT n = 0.412 mol P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K T = 16°C = 289 K V = 3.25 L P = 3.01 atm R = 0.0821Latm/molK 47 Ideal Gas Law Problems Find the volume of 85 g of O2 at 25°C and 104.5 kPa. GIVEN: WORK: V=? 85 g 1 mol = 2.7 mol n = 85 g = 2.7 mol 32.00 g T = 25°C = 298 K PV = nRT P = 104.5 kPa (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K R = 8.315 dm3kPa/molK 48 3 V = 64 dm Examples How many moles of air are there in a 2.0 L bottle at 19ºC and 747 mm Hg? What is the pressure exerted by 1.8 g of H2 gas exert in a 4.3 L balloon at 27ºC? Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. Ptotal = P1 + P2 + ... When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor. 50 Dalton’s Law Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor. GIVEN: PH2 = ? Ptotal = 94.4 kPa PH2O = 2.72 kPa Look up water-vapor pressure for 22.5°C. WORK: Ptotal = PH2 + PH2O 94.4 kPa = PH2 + 2.72 kPa PH2 = 91.7 kPa Sig Figs: Round to least number of decimal places. 51 Dalton’s Law A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor. GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr Look up water-vapor pressure for 35.0°C. WORK: Ptotal = Pgas + PH2O 742.0 torr = PH2 + 42.2 torr Pgas = 699.8 torr Sig Figs: Round to least number of decimal places. 52 Density The Molar mass of a gas can be determined by the density of the gas. D= mass = m Volume V Molar mass = mass = m Moles n n = PV RT Molar Mass = m (PV/RT) Molar mass = m RT V P Molar mass = DRT P You might need to review Chapter 8 At STP At STP determining the amount of gas required or produced is easy. 22.4 L = 1 mole For example How many liters of O2 at STP are required to produce 20.3 g of H2O? Not At STP Chemical reactions happen in MOLES. If you know how much gas - change it to moles Use the Ideal Gas Law n = PV/RT If you want to find how much gas use moles to figure out volume V = nRT/P Example #1 HCl(g) can be formed by the following reaction 2NaCl(aq) + H2SO4 (aq) 2HCl(g) + Na2SO4(aq) What mass of NaCl is needed to produce 340 mL of HCl at 1.51 atm at 20ºC? Example #2 2NaCl(aq) + H2SO4 (aq) 2HCl(g) + Na2SO4 (aq) What volume of HCl gas at 25ºC and 715 mm Hg will be generated if 10.2 g of NaCl react? Ideal Gases don’t exist Molecules do take up space There are attractive forces otherwise there would be no liquids Real Gases behave like Ideal Gases When the molecules are far apart The molecules do not take up as big a percentage of the space We can ignore their volume. This is at low pressure Real Gases behave like Ideal gases when When molecules are moving fast. Collisions are harder and faster. Molecules are not next to each other very long. Attractive forces can’t play a role. Diffusion Molecules moving from areas of high concentration to low concentration. Perfume molecules spreading across the room. Effusion Gas escaping through a tiny hole in a container. Depends on the speed of the molecule. Graham’s Law The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules. Kinetic energy = 1/2 mv2 m is the mass v is the velocity. Chem Express Graham’s Law bigger molecules move slower at the same temp. (by Square root) Bigger molecules effuse and diffuse slower Helium effuses and diffuses faster than air - escapes from balloon. Graham’s Law Diffusion • Spreading of gas molecules throughout a container until evenly distributed. Effusion • Passing of gas molecules through a tiny opening in a container 67 Graham’s Law Speed of diffusion/effusion • Kinetic energy is determined by the temperature of the gas. • At the same temp & KE, heavier molecules move more slowly. – Larger m smaller v KE = 2 ½mv 68 Graham’s Law Graham’s Law • Rate of diffusion of a gas is inversely related to the square root of its molar mass. • The equation shows the ratio of Gas A’s speed to Gas B’s speed. vA vB mB mA 69 Graham’s Law Determine the relative rate of diffusion for krypton and bromine. The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”. vA vB v Kr v Br2 m Br2 m Kr mB mA 159.80g/mol 1.381 83.80g/mol Kr diffuses 1.381 times faster than Br702. Graham’s Law A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? vA vB mB mA vH 2 12.3 m/s 32.00 g/mol 2.02 g/mol vH 2 vH 2 vO2 mO2 mH 2 Put the gas with the unknown speed as “Gas A”. 12.3 m/s 3.980 vH2 49.0m/s 71 Graham’s Law An unknown gas diffuses 4.0 times faster than O2. Find its molar mass. The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0. vA vB vA v O2 mB mA mO2 mA Square both sides to get rid of the square root sign. 32.00 g/mol g/mol 32.00 4.0 m A A 32.00 g/mol 16 mA 32.00 g/mol mA 2.0 g/mol 72 16 2 Gas Stoichiometry Liters of a Gas: • STP - use 22.4 L/mol • Non-STP - use ideal gas law Moles Non-STP • Given liters of gas? – start with ideal gas law • Looking for liters of gas? – start with stoichiometry conv. 73 Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 5.25 g CaO + Looking for liters: Start with stoich and calculate moles of CO2. 5.25 g CaCO3 1 mol CaCO3 1 mol CO2 100.09g CaCO3 1 mol CaCO3 CO2 ?L non-STP = 1.26 mol CO2 Plug this into the Ideal Gas Law to find liters. 74 Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: WORK: P = 103 kPa V=? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm3kPa/molK PV = nRT (103 kPa)V =(1mol)(8.315dm3kPa/molK)(298K) V = 1.26 dm3 CO2 75 Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + 3 O2 15.0 L non-STP 2 Al2O3 ?g GIVEN: WORK: P = 97.3 kPa V = 15.0 L n=? T = 21°C = 294 K R = 8.315 dm3kPa/molK PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3kPa/molK) (294K) Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT n = 0.597 mol O2 76 Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? 4 Al + Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 3 O2 15.0L non-STP 2 mol Al2O3 101.96 g Al2O3 3 mol O2 1 mol Al2O3 2 Al2O3 ?g = 40.6 g Al2O3 77