2 Real vs. Ideal Gases:Ideal gas = an imaginary gas that conforms perfectly to all the assumptions of the kinetic-molecular theoryWe will assume that the gases used for the gas law problems are ideal gases.
3 Real Gases vs. Ideal Gases: Real gas = a gas that does not behave completely according to the assumptions of the kinetic-molecular theory. All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.
4 Causes of non-ideal behavior: The kinetic-molecular theory is more likely to hold true for gases whose particles have NO attraction for each other.
5 Causes of non-ideal behavior: 1. High pressure (low volume):Space taken up by gas particles becomes significant Intermolecular forces are more significant between gas particles that are closer together
6 Causes of non-ideal behavior: 2. Low temperature:Gas particles move slower so intermolecular forces become more important
7 KMT (Kinetic Molecular Theory) Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences.theory developed in the late 19th century to account for the behavior of the atoms and molecules that make up matter
8 KMT (Kinetic Molecular Theory) based on the idea that particles in all forms of matter are always in motion and that this motion has consequencesSOLIDLIQUIDGAS
9 KMT (Kinetic Molecular Theory) can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them
10 B. Kinetic Molecular Theory – Ideal gas KMT describing particles in an IDEAL gas:have no volume.have elastic collisions.are in constant, random motion.don’t attract or repel each other.have an avg. KE directly related to Kelvin temperature.
11 C. Kinetic Molecular Theory - Real Gases KMT describing particles in a REAL gas:have their own volumeattract each otherGas behavior is most ideal…at low pressuresat high temperaturesin nonpolar atoms/molecules
12 D. Characteristics of Gases- using KMT Gases expand to fill any container.KMT - gas particles move rapidly in all directions without significant attraction or repulsion between particles
13 D. Characteristics of Gases- using KMT Gases are fluids (like liquids).KMT - No significant attraction or repulsion between gas particles; glide past each other
14 D. Characteristics of Gases- using KMT Gases have very low densities.KMT - particles are so much farther apart in the gas statesodium in the solid state:sodium in the liquid state:sodium in the gas state:
15 D. Characteristics of Gases- using KMT Gases can be compressed.KMT - gas particles are far apart from one anther with room to be “squished” together
16 D. Characteristics of Gases- using KMT Gases undergo diffusion & effusion.KMT – gas particles move in continuous, rapid, random motionEffusion
17 E. TemperatureAlways use Kelvin temperature when working with gases.ºFºCK-45932212-273100273373K = ºC + 273
18 F. PressureWhich shoes create the most pressure?
19 F. Pressure Why do gases exert pressure? Gas particles exert a pressure on any surface with which they collide!More collisions = increase in pressure!
20 F. Pressure Barometer = measures atmospheric pressure The height of the Hg in the tube depends on the pressureThe pressure of the atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure!Mercury Barometer
21 F. Pressure Pressure UNITS 101.3 kPa (kilopascal) 1 atm 760 mm Hg 760 torr
22 Standard Temperature & Pressure G. STPSTPStandard Temperature & Pressure0°C (exact) 1 atm (exact)273 K kPa760 mm Hg (exact)760 torr (exact)
26 A. Boyle’s LawThe pressure and volume of a gas are INVERSELY relatedat constant mass & tempP1V1 = P2V2PV
27 A. Boyle’s LawReal life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.
28 A. Boyle’s LawEx: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.
29 A. Boyle’s LawAs the volume increases, the pressure decreases!
42 Gas Law Problems CHARLES’ LAW V1_ = V2__ T V WORK: GIVEN: 1.) A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.CHARLES’ LAW V1_ = V2__T T2GIVEN:V1 = 473 cm3T1 = 36°C = 309KV2 = ?T2 = 94°C = 367KWORK:TVV2 = V1T2T1V2= (473 cm3)(367 K)(309 K)V2 = 562 cm3C. Johannesson
43 Gas Law Problems BOYLE’S LAW P1V1 = P2V2 P V WORK: GIVEN: V2 = P1V1 2.) A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.BOYLE’S LAW P1V1 = P2V2PGIVEN:V1 = 100. mLP1 = 150. kPaV2 = ?P2 = 200. kPaVWORK:V2 = P1V1P2V2 = (150.kPa)(100.mL)200.kPaV2 = 75.0 mLC. Johannesson
44 Gas Law Problems P T V V1 = 7.84 cm3 V2 = P1V1T2 P1 = 71.8 kPa P2T1 3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.COMBINED GAS LAW P1V1 = __P2V T T2GIVEN:V1 = 7.84 cm3P1 = 71.8 kPaT1 = 25°C = 298 KV2 = ?P2 = kPaT2 = 273 KP TVWORK:V2 = P1V1T2P2T1V2 = (71.8 kPa)(7.84 cm3)(273 K)(101.3 kPa) (298 K)V2 = 5.09 cm3C. Johannesson
45 Gas Law Problems GUY LUSSAC’S LAW P1_ = P2__ T P GIVEN: WORK: 4.) A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?GUY LUSSAC’S LAW P1_ = P2__T T2GIVEN:P1 = 765 torrT1 = 23°C = 296KP2 = 560. torrT2 = ?WORK:T2 = P2T1P1PTT2 = (560. torr)(296K)765 torrT2 = 217 KC. Johannesson
46 III. Standard Molar Volume, Gas Densities & Molar Mass GasesIII. Standard Molar Volume, Gas Densities & Molar Mass
47 A. Standard Molar Volume The volume occupied by one mole of a gas at STPL/mol or about 22.4 L/molIn other words, one mole of any ideal gas at STP will occupy 22.4 L
48 B. Example Problems-standard molar volume 1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?0.068 mol O L O21 mol O2=1.52 L O2
49 B. Example Problems-standard molar volume 1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?0.068 mol O L O21 mol O2=1.52 L O2
50 B. Example Problems-standard molar volume 2.) A chemical reaction produced 98.0 mL of SO2 at STP. What mass (in grams) of the gas was produced?g SO298.0 mL SO L SO mol O2=1000 mL SO222.4 L SO21 mol SO20.280 g SO2
51 C. Density The ratio of an object’s mass to its volume. D = M V The volume (and density) of a gas will change when pressure and temperature change.Densities are usually given in g/L at STP
52 C. Density For an ideal gas: Density (at STP) = molar mass standard molar volume
53 D. Example Problems- density 1.) What is the density of CO2 in g/L at STP?g CO mol CO21 mol CO L CO2=1.96 g/L CO2
54 D. Example Problems- density 2.) What is the molar mass of a gas whose density at STP is 2.08 g/L.2.08 g L1 L 1 mol=46.6 g/mol
55 IV. More Gas Laws: Ideal Gas Law & Avogadro's Law GasesIV. More Gas Laws: Ideal Gas Law & Avogadro's Law
56 A. Avogadro’s LawEqual volumes of gases contain equal numbers of molesat constant temp & pressureVn
57 UNIVERSAL GAS CONSTANT B. Ideal Gas LawBoyle’s, Charles, and Avogadro’s laws are contained within the ideal gas law!Merge the Combined Gas Law with Avogadro’s Law:VnPVTPVnT= R= kUNIVERSAL GAS CONSTANTR= Latm/molKR=8.315 dm3kPa/molKYou don’t need to memorize these values!
58 PV=nRT B. Ideal Gas Law P= pressure V = volume n = moles T = temperature (in K)R = the ideal gas constantGAS CONSTANTR= Latm/molKR=8.315 dm3kPa/molKYou don’t need to memorize these values!
59 C. Ideal Gas Law Problems 1.) Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.PV = nRTGIVEN:P = ? atmn = molT = 16°C = 289 KV = 3.25 LR = Latm/molKWORK:P = nRTVP =(0.412 mol)(0.0821Latm/molK) (289K)3.25LP = 3.01 atmC. Johannesson
60 C. Ideal Gas Law Problems 2.) Find the volume of 85 g of O2 at 25°C and kPa. PV = nRTGIVEN:V = ?n = 85 gT = 25°C = 298 KP = kPaR = dm3kPa/molKWORK:85 g O2 1 mol O2 = 2.7 molg O2 O2= 2.7 molV = nRTPV =(2.7 mol)(8.315 dm3kPa/molK) (298K)104.5 kPaV = 64 dm3C. Johannesson
61 C. Ideal Gas Law Problems 3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas has a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRTGIVEN:V = 1.00 Ln = ?T = 28°C = 301 KP = 98.7 kPaR = LkPa/molKWORK:n = PVRT(98.7 kPa) (1.00L)(8.314 LkPa/molK) (301 K)n = mol=m = gmol=131 g/molC. Johannesson
62 IV. Two More Laws: Dalton’s Law & Graham’s Law GasesIV. Two More Laws: Dalton’s Law & Graham’s Law
63 Ptotal = P1 + P2 + P3 ... 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 + P3 ...
65 A. Dalton’s LawWhen a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 AND water vapor.
66 A. Dalton’s Law1.) 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 kPaPH2O = 2.72 kPaWORK:Ptotal = PH2 + PH2O94.4 kPa = PH kPaPH2 = 91.7 kPaLook up water-vapor pressure on gas formula sheet for 22.5°C.Sig Figs: Round to least number of decimal places.C. Johannesson
67 A. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr 2.) A gas is collected over water at a temp of 35.0°C when the barometric pressure is 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 = torrPH2O = 42.2 torrWORK:Ptotal = Pgas + PH2O742.0 torr = PH torrPgas = torrLook up water-vapor pressure on gas formula sheet for 35.0°C.Sig Figs: Round to least number of decimal places.C. Johannesson
68 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
69 KE = ½mv2 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 vKE = ½mv2
70 KE = ½mv2 KE = kinetic energy m= molar mass v = velocity B. Graham’s LawKE = kinetic energym= molar massv = velocityKE = ½mv2
71 B. Graham’s LawGraham’s Law = Rate of effusion 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.C. Johannesson
72 C. Graham’s Law Problems 1.) Determine the relative rate of effusion 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”.Kr effuses times faster than Br2.
73 C. Graham’s Law Problems 2.) 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?Put the gas with the unknown speed as “Gas A”.C. Johannesson
74 C. Graham’s Law Problems 3.) 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.Square both sides to get rid of the square root sign.C. Johannesson
75 VI. Gas Stoichiometry at Non-STP Conditions GasesVI. Gas Stoichiometry at Non-STP Conditions
76 A. Gas Stoichiometry Moles Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas lawNon-STPGiven liters of gas?start with ideal gas lawLooking for liters of gas?start with stoichiometry conv.
77 B. Volume Mass Stoich. Thinking: Mole Ratio If gas A is at STP: Gas volume A moles A moles B mass BIf gas A is at STP:? L “A” 1 mol “A” ? mol “B” molar mass(g) “B”22.4 L “A” ? mol “A” mol “B”Mole Ratio
78 B. Volume Mass Stoich. If gas “A” is NOT at STP: Use PV = nRT to find moles of “A”? mol “A” ? mol “B” molar mass (g) “B”? mol “A” 1 mol “B”
79 Given liters: Start with Ideal Gas Law and calculate moles of O2. B. Volume Mass1.) How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?4 Al O2 2 Al2O315.0 Lnon-STP? gGIVEN:P = 97.3 kPaV = 15.0 Ln = ?T = 21°C = 294 KR = dm3kPa/molKWORK:n = PVRTn= (97.3 kPa) (15.0 L) (8.315dm3kPa/molK) (294K)n = mol O2Given liters: Start with Ideal Gas Law and calculate moles of O2.NEXT C. Johannesson
80 Use stoich to convert moles of O2 to grams Al2O3. B. Volume Mass1.) How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?4 Al O2 2 Al2O315.0Lnon-STP? gUse stoich to convert moles of O2 to grams Al2O3.0.597mol O22 mol Al2O33 mol O2g Al2O31 molAl2O3= 40.6 g Al2O3
81 B. Volume Mass ? g 12.6L 12.6 L SO2 1 mol SO2 22.4L SO2 1 mol S8 2.) What mass of sulfur is required to produce 12.6 L of sulfur dioxide at STP according to the equation:S8 (s) O2 (g) 8 SO2 (g)? g12.6L12.6L SO21 mol SO222.4L SO21 mol S88 molSO2g S81 molS8= 18.0 gS8
82 C. Mass Volume Thinking: If gas “B” is at STP: Mass A moles A moles B gas volume BIf gas “B” is at STP:? g “A” 1 mol “A” ? mol “B” L “B”molar mass (g) “A” ? mol “A” mol “B”
83 C. Mass Volume If gas “B” is NOT at STP: ? g “A” mol “A” ? mol “B”molar mass (g) “A” ? mol “A”Then use PV = nRT to find volume of “B”
84 C. Mass Volume 2 Na + Cl2 2 NaCl ? L 10.4 g = 5. 07 L Cl2 1.) What volume of chlorine gas at STP is needed to react completely with 10.4 g of sodium to form NaCl?2 Na + Cl2 2 NaCl10.4 g? L22.4 L Cl210.4 g Na1 mol Cl21 mol Nag Na2 mol Na1 mol Cl2= L Cl2
85 C. Mass Volume CaCO3 CaO + CO2 5.25 g ? L non-STP 5.25 g CaCO3 2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?CaCO3 CaO CO25.25 g? L non-STPLooking for liters: Start with stoich and calculate moles of CO2.5.25 gCaCO31 molCaCO3100.09g1 molCO2CaCO3= 1.26 mol CO2Plug this into the Ideal Gas Law to find liters.
86 C. Mass Volume2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?GIVEN:P = 103 kPaV = ?n = 1.26 molT = 25°C = 298 KR = dm3kPa/molKWORK:V = nRTPV = (1.26mol)(8.315dm3kPa/molK) (298K) (103 kPa)V = 30.3 dm3 CO2C. Johannesson