Gas Laws PowerPoint Notes PDF

Ch. 11 - Gases I. Physical Properties A. 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 te...

Ch. 11 - Gases I. Physical Properties A. 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. B. 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 C. 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 C. Characteristics of Gases Gases can be compressed. no volume = lots of empty space Gases undergo diffusion & effusion. random motion D. Temperature Always use absolute temperature (Kelvin) when working with gases. ºF -459 32 212 ºC -273 0 100 K 0 273 373 C = 9 5 (F − 32) K = ºC + 273 E. Pressure force pressure = area Which shoes create the most pressure? E. Pressure Barometer measures atmospheric pressure Aneroid Barometer Mercury Barometer E. Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg N 760 torr kPa = 2 m 14.7 psi F. STP STP Standard Temperature & Pressure 0°C 273 K -OR- 1 atm 101.325 kPa  Gas Laws II. The Gas V T Laws BOYLES P CHARLES GAY- LUSSAC A. Boyle’s Law Volume Pressure P·V (mL) (torr) (mL·torr) 10.0 760.0 7.60 x 103 20.0 379.6 7.59 x 103 30.0 253.2 7.60 x 103 40.0 191.0 7.64 x 103 P PV = k V A. Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp P PV = k V B. Charles’ Law Volume Temperature V/T (mL) (K) (mL/K) 40.0 273.2 0.146 44.0 298.2 0.148 47.7 323.2 0.148 51.3 348.2 0.147 V V =k T T B. Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V V =k T T C. Gay-Lussac’s Law Temperature Pressure P/T (K) (torr) (torr/K) 248 691.6 2.79 273 760.0 2.78 298 828.4 2.78 373 1,041.2 2.79 P P =k T T C. Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P P =k T T D. Combined Gas Law P V PV PV = k T P1V1 P2V2 = T1 T2 P1V1T2 = P2V2T1  E. Gas Law Problems A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: T V WORK: V1 = 473 cm3 P1V1T2 = P2V2T1 T1 = 36°C = 309K (473 cm3)(367 K)=V2(309 K) V2 = ? T2 = 94°C = 367K V2 = 562 cm3  E. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: P V WORK: V1 = 100. mL P1V1T2 = P2V2T1 P1 = 150. kPa (150.kPa)(100.mL)=(200.kPa)V2 V2 = ? V2 = 75.0 mL P2 = 200. kPa  E. 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 cm 3)(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  E. 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 WORK: P1 = 765 torr P1V1T2 = P2V2T1 T1 = 23°C = 296K (765 torr)T2 = (560. torr)(296K) P2 = 560. torr T2 = 217 K = -56.3°C T2 = ? A. Avogadro’s Principle Volume Mass V/n Gas Moles, n (mL) (g) (L/mol) O2 100.0 0.122 3.81  10-3 26.2 N2 100.0 0.110 3.93  10-3 25.5 CO2 100.0 0.176 4.00  10-3 25.0 V V =k n n A. Avogadro’s Principle Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas V V =k n n B. Ideal Gas Law V PV =R k n nT T UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm kPa/molK 3 B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 Latm/molK R=8.315 dm kPa/molK 3  B. Ideal Gas Law Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: WORK: P = ? atm PV = nRT n = 0.412 mol P(3.25)=(0.412)(0.0821)(289) T = 16°C = 289 K L mol Latm/molK K V = 3.25 L P = 3.01 atm R = 0.0821Latm/molK  B. Ideal Gas Law Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW 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 dm3kPa/molK K R = 8.315 dm3kPa/molK V = 64 dm3 A. Gas Stoichiometry Moles  Liters of a Gas STP - use 22.4 L/mol Non-STP - use ideal gas law Non-STP Problems Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv. B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? CaCO3 → CaO + CO2 5.25 g ?L Looking for liters: Start with stoich and calculate moles of CO2. non-STP 5.25 g 1 mol 1 mol CaCO3 CaCO3 CO2 = 0.052 mol CO2 100.09g 1 mol Plug this into the Ideal CaCO3 CaCO3 Gas Law to find liters.  B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC? GIVEN: WORK: P = 103 kPa PV = nRT V=? (103 kPa)V =(.052mol)(8.315dm3kPa/molK)(298K) n =.052 mol T = 25°C = 298 K V = 1.26 dm3 CO2 R = 8.315 dm3kPa/molK  B. 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 → 2 Al2O3 15.0 L non-STP ?g GIVEN: WORK: Given liters: Start with Ideal Gas Law and P = 97.3 kPa PV = nRT calculate moles of O2. V = 15.0 L (97.3 kPa) (15.0 L) n=? = n (8.315dm3kPa/molK) (294K) NEXT → T = 21°C = 294 K n = 0.597 mol O2 R = 8.315 dm3kPa/molK  B. 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 → 2 Al2O3 Use stoich to convert moles 15.0L ?g of O to grams Al O. 2 2 3 non-STP 0.597 2 mol 101.96 g mol O2 Al2O3 Al2O3 = 40.6 g Al2O3 3 mol O2 1 mol Al2O3 A. 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 +... Patm = PH2 + PH2O  A. 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: WORK: PH2 = ? Ptotal = PH2 + PH2O Ptotal = 94.4 kPa 94.4 kPa = PH2 + 2.72 kPa PH2O = 2.72 kPa PH2 = 91.7 kPa Look up water-vapor pressure Sig Figs: Round to least number on p.859 for 22.5°C. of decimal places.  A. 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 DALTON’S LAW pressure and is a mixture of the “gas” and water vapor. GIVEN: WORK: Pgas = ? Ptotal = Pgas + PH2O Ptotal = 742.0 torr 742.0 torr = PH2 + 42.2 torr PH2O = 42.2 torr Pgas = 699.8 torr Look up water-vapor pressure Sig Figs: Round to least number on p.859 for 35.0°C. of decimal places. B. 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 B. 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 because… KE = ½mv 2 B. Graham’s Law Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. Ratio of gas vA mB A’s speed to = gas B’s speed vB mA B. 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 “v A/vB”. vA mB = vB mA v Kr m Br2 159.80 g/mol = = = 1.381 v Br2 m Kr 83.80 g/mol Kr diffuses 1.381 times faster than Br 2.  B. 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 mB vH 2 32.00 g/mol = = vB mA 12.3 m/s 2.02 g/mol vH 2 = 3.980 vH 2 mO2 Put the gas with 12.3 m/s = the unknown vO2 mH 2 speed as “Gas A”. vH 2 = 49.0 m/s  B. 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. 2 Square both  32.00 g/mol  vA mB sides to get rid  4.0 =  = of the square  m AA  vB mA root sign.   32.00 g/mol 16 = vA mO2 mA = v O2 mA 32.00 g/mol mA = = 2.0 g/mol 16