1. Tutorial on Ideal Gas Equation and RMM determination of volatile liquid and gas .Prepared by Lawrence Kok http://lawrencekok.blogspot.com

2. Ideal Gas Equation P = Pressure Unit Nm-2/Pa/kPa/atmn = number of molesPV = nRT V = Volume gas Unit dm3 or m3R = universal gas constant Unit - 8.314Jmol-1K-1 or 0.0821 atm L mol-1 K-14 different variables P, V, n, TPV = nRT Boyles LawT = Absolute Temperature in KCharless LawFix 2 variables change to different gas LawsPressure LawAvogadros Law PV = nRT (P, T fix) V = constant x n VnPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTPV = nRT (n, V fix) P = constant x T PTP1V1 = P2V2V1 = V2 T1 T2P1 = P2 T1 T2V1 = V2 n1 n2 3. PV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pP1V1 = P2V2Charless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTV1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 4. PV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pP1V1 = P2V2Boyles Law PV = nRT (n, T fix) PV = constant V = constant P V inversely proportional to P V 1 P P1V1 = P2V2Charless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTV1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 5. PV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pP1V1 = P2V2Charless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTV1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2 Boyles Law Lab SimulatorBoyles Law PV = nRT (n, T fix) PV = constant V = constant P V inversely proportional to P V 1 P P1V1 = P2V2Video on Boyles LawV1 = V2 n1 n2 6. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 7. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Charless Law PV = nRT (n, P fix) V = constant x T V directly proportional to T V T V1 = V2 T1 T2 Temp increase kinetic energy increase collision bet particles with container increase volume increase Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 8. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2Charless Law Lab SimulatorCharless Law PV = nRT (n, P fix) V = constant x T V directly proportional to T V T V1 = V2 T1 T2 Temp increase kinetic energy increase collision bet particles with container increase volume increase Video on Charless Law 9. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 10. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Pressure Law PV = nRT (n, V fix) P = constant x T P directly proportional to T P T P1 = P2 T1 T2 Temp increase kinetic energy increase collision bet particles with container increase pressure increase Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 11. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2Pressure Law Lab SimulatorPressure Law PV = nRT (n, V fix) P = constant x T P directly proportional to T P T P1 = P2 T1 T2 Temp increase kinetic energy increase collision bet particles with container increase pressure increase Video on Pressure Law 12. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 13. PV = nRT Boyles LawCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pPV = nRT (n ,P fix) V = constant x T V = constant T VTP1V1 = P2V2V1 = V2 T1 T2Avogadro Law PV = nRT (P, T fix) V = constant x n V directly proportional to n V n V1 = V2 n1 n2Fix 2 variables change to different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnP1 = P2 T1 T2V1 = V2 n1 n2 14. Standard Temp/PressureStandard Molar Volume 1 mole of any gas at fix STP (Std Temp/Pressure) occupies a volume of 22.4dm3/22400cm3/24LP - 1 atm = 760 mmHg = 101 325Pa (Nm-2) = 101.325kPa T 0C (273.15K) Unit conversion22.4L22.4L22.4L1 mole gas1 atm = 760 mmHg/Torr = 101 325Pa(Nm-2) =101.325kPa 1m3 = 103 dm3 = 106cm3 1dm3 = 1 litreAvogadros LawGasHeliumNitrogenOxygenMole/mol114.028.032.0Pressure/atm111Temp/K273273273Vol/L22.4L22.4L22.4LParticles6.02 x 10236.02 x 10236.02 x 1023 equal vol of gases at same temperature/pressure contain equal numbers of molecules1Mass/gmolar volume of all gases the same at given T and P 22.4Lhttp://leifchemistry.blogspot.kr/2011/01/molar-volume-at-stp.htmlVideo on Avogadros Law 15. PV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pCharless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTCharless LawFix 2 variables change different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnBoyles LawPressure Law2 different variablesAvogadros Law 16. PV = nRT Boyles LawCharless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTPV = nRT (n, T fix) PV = constant V = constant/P V 1/pCharless LawFix 2 variables change different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnBoyles LawPressure LawAvogadros Law2 different variablesBoyles LawCharless LawV 1 PV TCombined Boyle + Charles LawV T P Gas constant, RPV = constant T PV = R T Combined Boyle Law + Charles Law 2 different variablesCombined Gas LawP1V 1 = P2V2 T1 T2 Combined Gas Law 3 different variables3 different variables 17. PV = nRT Boyles LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pCharless LawPV = nRT (n ,P fix) V = constant x T V = constant T VTCharless LawFix 2 variables change different gas LawsPressure LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTPV = nRT (P, T fix) V = constant x n VnBoyles LawPressure Law2 different variablesAvogadros Law 18. PV = nRT Boyles LawPressure LawPV = nRT (n ,P fix) V = constant x T V = constant T VTCharless LawAvogadros LawPV = nRT (n, V fix) P = constant x T PTCharless LawPV = nRT (n, T fix) PV = constant V = constant/P V 1/pFix 2 variables change different gas LawsPV = nRT (P, T fix) V = constant x n VnBoyles LawPressure LawAvogadros Law2 different variablesAvogadros LawBoyles LawCharless LawV nV 1 PBoyle + Charles + Avogadro LawV TV nT P Ideal Gas EquationPV = n R T Proportionality constant Gas constant, RBoyle + Charles + Avogadro LawPV = nRT 4 different variablesIdeal Gas Equation 19. PV = nRT Boyles LawCharless LawV 1 PV TPressure LawAvogadros LawV nP T2 different variables+ +Combined Gas LawIdeal Gas EquationP1V 1 = P2V2 T1 T2PV = nRT3 different variables4 different variablesFor 1 mole PV = RT For n mole PV = nRTPV = nRT Pressure/P 101 325 Pa(Nm-2)n 1 mol Find R (Universal Gas Constant) at molar volume n = 1 mol T = 273K P = 101325Pa/Nm-2 V = 22.4 x 10-3m3 R=?Temp/T oC 273KR=P V n TVolume/V 22.4dm3 22.4 x 10-3 m3R = 101325 x 22.4 x 10-3 1 x 273R = 8.31 JK-1mol-1or NmK-1 When n = 1 mol Gas constant, R is 8.31 JK-1mol-1or NmK-1 20. Value of gas constant, R (Universal Gas Constant) at molar volumePV = nRTDifferent Units Usedn 1 molTemp/T oC 273KR=PV nTR = 101325 x 22.4 x 10-3 1 x 273Pressure/P 101 325 Pa(Nm-2)Volume/V 22.4dm3 22.4 x 10-3 m3R = 8.31 JK-1mol-1or NmK-1 21. Value of gas constant, R (Universal Gas Constant) at molar volumePV = nRT n 1 molDifferent Units UsedTemp/T oC 273KR=PV nTR = 101325 x 22.4 x 10-3 1 x 273n 1 molDifferent Units UsedPV = nRT R=PV nTTemp/T oC 273KR = 1 x 22.4 1 x 273 Unit conversion1 atm 760 mmHg/Torr 101 325Pa/Nm-2 101.325kPa 1m3 103 dm3 106cm3 1dm3 1000cm3 1000ml 1 litre x 103cm3 x 10-3x 103dm3 x 10-3m3Pressure/P 101 325 Pa(Nm-2)Volume/V 22.4dm3 22.4 x 10-3 m3R = 8.31 JK-1mol-1or NmK-1Pressure/P 1 atmVolume/V 22.4LR = 0.0821 atmLmol-1K-1 22. Determination of RMM using Ideal Gas Equation Volatile Liquid (Propanone)PV = nRTVolatile Gas (Butane)HeatedConverted to gasDirect WeighingSyringe MethodDirect Weighing 23. Determination of RMM using Ideal Gas Equation Volatile Liquid (Propanone)Volatile Gas (Butane)PV = nRTHeatedConverted to gasDirect WeighingSyringe MethodDirect WeighingRMM calculated if- m, T, P, V, are knownIdeal Gas EquationPV = n x R x T PV = mass x R x T M M =mxRxT PVn = mass MRMM = MorPV = n x R x T PV = mass x R x T M M=mxRxT V P M = xRxT Pn = mass MDensity, = m V 24. Determination of RMM (LIQUID) using Ideal Gas Equation ProcedureDirect Weighing1. Cover the top with aluminium foil. 2. Make a hole on aluminium foil 3. Record mass flask + foil 4. Pour 2 ml volatile liquid into flask 5. Place flask in water, heated to boiling Temp and record pressure 6. Vapour fill flask when heated 7. Cool flask in ice bath allow vapour to condense to liquid8. Take mass of flask + foil + condensed liquidVol gas = Vol water = Vol water in flask = Mass water Assume density water = 1g/ml 25. Determination of RMM (LIQUID) using Ideal Gas Equation Procedure 1. Cover the top with aluminium foil. 2. Make a hole on aluminium foil 3. Record mass flask + foil 4. Pour 2 ml volatile liquid into flaskDirect WeighingData CollectionMass flask Mass flask + foil + condensed vapour Mass condensed vapour Atmospheric pressure Temperature of boiling water Volume of flask5. Place flask in water, heated to boiling Temp and record pressure 6. Vapour fill flask when heated 7. Cool flask in ice bath allow vapour to condense to liquid8. Take mass of flask + foil + condensed liquidVol gas = Vol water = Vol water in flask = Mass water Assume density water = 1g/ml115.15 g 115.67 g 0.52 g 101325 Pa 100 0C 373K 284 cm3 2.84 x 10-4 m3 26. Determination of RMM (LIQUID) using Ideal Gas Equation ProcedureDirect Weighing1. Cover the top with aluminium foil.Data CollectionMass flask Mass flask + foil + condensed vapour Mass condensed vapour Atmospheric pressure Temperature of boiling water Volume of flask2. Make a hole on aluminium foil 3. Record mass flask + foil 4. Pour 2 ml volatile liquid into flask115.15 g 115.67 g 0.52 g 101325 Pa 100 0C 373K 284 cm3 2.84 x 10-4 m35. Place flask in water, heated to boiling Temp and record pressureData Processing6. Vapour fill flask when heatedPV = nRT7. Cool flask in ice bath allow vapour to condense to liquid8. Take mass of flask + foil + condensed liquidVol gas = Vol water = Vol water in flask = Mass water Assume density water = 1g/mlVideo on RMM determinationPV = n x R x T PV = mass x R x T M M =mxRxT PV = 0.52 x 8.314 x 373 101325 x 2.84 x 10-4 = 56.33 RMM calculated - m, T, P, V are knownClick here for lab procedure 27. Determination of RMM (LIQUID) using Ideal Gas EquationProcedure 1. Set temp furnace to 98C. 2. Draw 0.2ml liquid into a syringe Record mass syringe + liquid.4. Record vol of heated air. 5. Inject liquid into syringe6. Liquid will vaporise , Record vol of heated vapour + airSyringe Method 28. Determination of RMM (LIQUID) using Ideal Gas EquationProcedureSyringe MethodData Collection1. Set temp furnace to 98C.Mass syringe + liquid before injection15.39 g2. Draw 0.2ml liquid into a syringe Record mass syringe + liquid.Mass syringe + liquid after injection15.27 gMass of vapour Atmospheric Pressure0.12 g 100792PaTemp of vapour Volume heated air Volume heated air + vapour Volume of vapour371 K 7 cm3 79 cm34. Record vol of heated air. 5. Inject liquid into syringe6. Liquid will vaporise , Record vol of heated vapour + air72 7 = 72 cm3 72cm3 7.2 x 10-5 m3 29. Determination of RMM (LIQUID) using Ideal Gas EquationProcedureSyringe MethodData Collection1. Set temp furnace to 98C.Mass syringe + liquid before injection15.39 g2. Draw 0.2ml liquid into a syringe Record mass syringe + liquid.Mass syringe + liquid after injection15.27 gMass of vapour Atmospheric Pressure0.12 g 100792PaTemp of vapour Volume heated air Volume heated air + vapour Volume of vapour371 K 7 cm3 79 cm34. Record vol of heated air. 5. Inject liquid into syringe6. Liquid will vaporise , Record vol of heated vapour + air72 7 = 72 cm3 72cm3 7.2 x 10-5 m3Data ProcessingVideo on RMM determinationPV = nRT PV = n x R x T PV = mass x R x T M M =mxRxT PV = 0.12 x 8.314 x 371 100792 x 7.2 x 10-5 = 51.1 RMM calculated - m, T, P, V are knownClick here for lab procedure 30. Determination of RMM (GAS) using Ideal Gas Equation ProcedureDirect Weighing1. Fill a flask with water and invert it . 2. Record pressure + temp of water 3. Mass of butane + lighter (initial) 4. Release gas into flask5. Adjust water level in flask until the same as atm pressure 6. Measure the vol gas 7. Mass of butane + lighter (final)Daltons Law of Partial Pressures: Total pressure of mix of gases = sum of the partial pressures of all the individual gases 31. Determination of RMM (GAS) using Ideal Gas Equation Procedure 1. Fill a flask with water and invert it . 2. Record pressure + temp of water 3. Mass of butane + lighter (initial) 4. Release gas into flask5. Adjust water level in flask until the same as atm pressure 6. Measure the vol gasData CollectionDirect WeighingMass butane + lighter (initial) Mass butane + lighter (final) Mass butane Pressure gasTemperature water Volume of gas7. Mass of butane + lighter (final)Daltons Law of Partial Pressures: Total pressure of mix of gases = sum of the partial pressures of all the individual gases87.63 g 86.98 g 0.65 g 743.9 mmHg 760 mmHg 101325 Pa 743.9mmHg 99.17Pa 21.70C 294.75K 276cm3 2.76 x 10-4 m3 32. Determination of RMM (GAS) using Ideal Gas Equation ProcedureData CollectionDirect WeighingMass butane + lighter (initial) Mass butane + lighter (final) Mass butane Pressure gas1. Fill a flask with water and invert it . 2. Record pressure + temp of water 3. Mass of butane + lighter (initial) 4. Release gas into flask5. Adjust water level in flask until the same as atm pressureTemperature water Volume of gas6. Measure the vol gas 7. Mass of butane + lighter (final)86.98 g 0.65 g 743.9 mmHg 760 mmHg 101325 Pa 743.9mmHg 99.17Pa 21.70C 294.75K 276cm3 2.76 x 10-4 m3Data ProcessingDaltons Law of Partial Pressures: Total pressure of mix of gases = sum of the partial pressures of all the individual gases Total Pressure (atm) = partial P(butane) + partial P(H2O) P butane = P(atm) P(H2O) = (760 19.32) mmHg P butane = 743.911 mmHg 99.17Pa Video on RMM determination RMM butane87.63 gRMM butaneCollection gasPV = nRT PV = n x R x T PV = mass x R x T M M =mxRxT PV = 0.65 x 8.314 x 294.75 99.17 x 2.76 x 10-4 = 58.17 RMM calculated - m, T, P, V are knownClick here for lab procedure 33. IB Questions on Ideal Gas 1Density of gas at stp is 2.78gdm-3. Find RMM of gas23Density of gas is 2.6gdm-3 , T- 25C and P - 101kPa Calculate RMM of gas4Mass of 1 dm3 gas is 1.96g at stp. Find RMM of gasDensity gas is 1.25gdm-3 at T- 25C and P- 101kPa. Calculate RMM of gas 34. IB Questions on Ideal Gas 1Density of gas at stp is 2.78gdm-3. Find RMM of gas2Answer 1 dm3 gas at stp = 1.78g 22.4 dm3 at stp = 1.78 x 22.4g (Molar Volume) = 39.87g (Molar Mass) RMM gas = 39.873Density of gas is 2.6gdm-3 , T- 25C and P - 101kPa Calculate RMM of gas Answer PV = nRT PV = m x R x T M M =mxRxT Density = m (mass) V(volume) V P M = x RT P Density = 2.6 gdm-3 2.6 x 103 gm-3 P = 101kPa 101 x 103 Pa M = x RT P M = (2.6 x 103) x 8.31 x (298) 101 x 103 = 63.7Mass of 1 dm3 gas is 1.96g at stp. Find RMM of gas Answer Mass of 1dm3 at stp = 1.96g Mass of 22.4dm3 at stp = 22.4 x 1.96g = 43.9g RMM = 43.94Density gas is 1.25gdm-3 at T- 25C and P- 101kPa. Calculate RMM of gas Answer Using PV = nRT n=m M m = density, = 1.25gdm-3 PV = m x RT V M M = m x RT V P M = x RT = 1.25 x 103 x 8.31 x 298 P 1.01 x 103 = 30.6 35. IB Questions on Ideal Gas 5Find molar mass gas by direct weighing, at T-23C and P- 97.7kPa Mass empty flask = 183.257g Mass flask + gas = 187.942g Mass flask + water = 987.560g Mass gas = (187.942 183.257) = 4.685g Vol gas = Vol water = Mass water = (987.560 183.257) = 804.303cm3 Vol gas = 804.303cm3 804.303 x 10-6m3 Pressure = 97.7kPa 97700Pa Temp = 23C (273 + 23) = 296K Assume density water = 1g/cm368 73.376g gas occupies 2.368dm3 at T- 17.6C, P - 96.73kPa. Find molar massCalculate RMM of gas Mass empty flask 25.385g Mass flask filled gas = 26.017 Mass flask filled water = 231.985g Temp = 32C Pressure = 101kPa6.32 g gas occupy 2200cm3, T- 100C P -101kPa. Calculate RMM of gas 36. IB Questions on Ideal Gas 5Find molar mass gas by direct weighing, at T-23C and P- 97.7kPa Mass empty flask = 183.257g Mass flask + gas = 187.942g Mass flask + water = 987.560g Mass gas = (187.942 183.257) = 4.685g Vol gas = Vol water = Mass water = (987.560 183.257) = 804.303cm3 Vol gas = 804.303cm3 804.303 x 10-6m3 Pressure = 97.7kPa 97700Pa Temp = 23C (273 + 23) = 296K Assume density water = 1g/cm36Answer Mass gas = (26.017 25.385) = 0.632g Vol flask = (231.985 25.385) = 206.6 x 10-6 m3 P = 101kNm-2 101 x 103 Nm-2 M = m x RT PV = 0.632 x 8.314 x 305 101 x 103 x 206.6 x 10-6 = 76.8Answer PV = nRT PV = mass x R x T M M = mass x R x T PV = 4.685 x 8.314 x 296 97700 x 804.303 x 10-6 = 146.7g/mol 8 73.376g gas occupies 2.368dm3 at T- 17.6C, P - 96.73kPa. Find molar mass Unit conversion Answer PV = nRT T 17.6C (273 = 17.6 = 290.6) PV = mass x RT P 96.73kPa 96730Pa M M = mass x R x T PV = 3.376 x 8.314 x 290.6 96730 x 2.368 x 10-3 = 35.61g/molCalculate RMM of gas Mass empty flask 25.385g Mass flask filled gas = 26.017 Mass flask filled water = 231.985g Temp = 32C Pressure = 101kPa6.32 g gas occupy 2200cm3, T- 100C P -101kPa. Calculate RMM of gas Answer PV = nRT n = PV RT n = (101 x 103) (2200 x 10-6) 8.31 x (100+273) n = 7.17 x 10-2 mol n = mass M RMM = mass n RMM = 6.32 7.17 x 10-2 = 88.15 37. IB Questions on Ideal Gas 9Find empirical formula for composition by mass. S 23.7%, O 23.7%, CI 52.6% Density of its vapour at T- 70C and P- 98kNm-2 = 4.67g/dm3 What molecular formula? 38. IB Questions on Ideal Gas 9Find empirical formula for composition by mass. S 23.7%, O 23.7%, CI 52.6% Density of its vapour at T- 70C and P- 98kNm-2 = 4.67g/dm3 What molecular formula? AnswerAnswerElementSOCIComposition23.723.752.6Moles23.7 32.1 = 0.73823.7 16.0 = 1.4852.6 35.5 = 1.481.48 = 2 0.738 21.48 = 2 0.738 2Mole ratio0.738 = 1 0.738 1Empirical formula - SO2CI2PV = nRT PV = m x R x T M M =mxRxT V P Density = m (mass) V (volume) M = x RT P Density = 4.67gdm-3 4.67 x 103 gm-3 P = 98kN-2 9.8 x 104 Nm-2 M = (4.67 x 103) x 8.31 x (273 +70) 9.8 x 104 = 135.8 135.8 = n [ 32 + (2x16)+(2 x 35.5) ] 135.8 = n [ 135.8] n=1 MF = SO2CI2 39. IB Questions on Ideal Gas 10Find volume (dm3) of 2.00g CO at T 20C P 6250Nm-213A gas occupy at (constant P) V - 125cm3 T - 27C Calculate its volume at 35C1114Find volume (m3) of 1 mol of gas at T - 298K P - 101 325PaFind final vol, V2, gas at (constant T) compressed to P2 = 250kPa V1 - 100cm3 P1 - 100kPa V2 - ? P2 250kPa1512What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C3.0 dm3 of SO2 reacted with 2.0 dm3 of O2 2SO2(g) + O2(g) 2SO3(g) Find volume of SO2 in dm3 at stp 40. IB Questions on Ideal Gas 10Find volume (dm3) of 2.00g CO at T 20C P 6250Nm-211Answer: (Ideal Gas Eqn) PV = nRT T (20 + 273) = 293K V = nRT n 2.00/28 = 0.0714 mol P = 0.0714 x 8.314 x 293 6250 =0.0278m3 = 27.8dm3Find volume (m3) of 1 mol of gas at T - 298K P - 101 325Pa Answer: (Ideal gas eqn) PV = nRT V = nRT P V = 1 x 8.314 x 298 101325 = 0.0244m312What volume (dm3) of 1 mol gas at P - 101325Nm-2 T - 25C Answer: (Ideal gas eqn) pV = nRT V = nRT P V = 1 x 8.31 x (273 + 25) 101325 = 0.0244m3 = 24.4dm3Using PV = nRT (Ideal gas eqn) Need to convert to SI units 4 variables involved13A gas occupy at (constant P) V - 125cm3 T - 27C Calculate its volume at 35C Answer: (Charles Law) V1 = V2 (constant P) T1 T2 125 = V2 (27+273) (35 + 273) V2 = 128cm314Find final vol, V2, gas at (constant T) compressed to P2 = 250kPa V1 - 100cm3 P1 - 100kPa V2 - ? P2 250kPa Answer: (Boyle Law) p1V1 = p2V2 (constant T) 100 x 100 = 250 x V2 V2 = 40cm3153.0 dm3 of SO2 reacted with 2.0 dm3 of O2 2SO2(g) + O2(g) 2SO3(g) Find volume of SO2 in dm3 at stp Answer: (Avogadro Law) PV = nRT (at constant P,T) Vn 2SO2(g) + 1 O2(g) 2SO3(g) 2 mol 1 mol 2 mol 2 vol 1 vol 2 vol 3dm3 2dm3 ?SO2 is limiting 2dm3 SO2 2dm3 SO3 3dm3 SO2 3dm3 SO3 Boyle, Charles, Avogadro Law no need to convert to SI units cancel off at both sides 2 variables involved 41. IB Questions on Ideal Gas 16A syringe contains gas at V1 - 50cm3 P1 1atm T1 - 20C 293K What volume , V2, if gas heated to V2 - ? T2 - 100C 373K P2 - 5 atm18Which change in conditions would increase the volume by x4 of a fix mass of gas?A. B. C. D.Pressure /kPa Doubled Halved Doubled HalvedTemperature /K Doubled Halved Halved Doubled173 Find volume fixed mass gas when its pressure and temp are double ?19Fix mass ideal gas has a V1 = 800cm3 , P1, T1 Find vol, V2 when P and T doubled. V2 = ? P2 = 2P1 T2 = 2T1 42. IB Questions on Ideal Gas 16A syringe contains gas at V1 - 50cm3 P1 1atm T1 - 20C 293K What volume , V2, if gas heated to V2 - ? T2 - 100C 373K P2 - 5 atm173 Find volume fixed mass gas when its pressure and temp are double ? Answer: (Combine Gas Law) Initial P1 Final P2 = 2P1 Initial T1 Final T2 = 2T1 Initial V1 Final V2 = ? P1V1 = P2V2 T1 T2 P and T double P1V1 = 2P1V2 T1 2T1 V2 = V1 Volume no changeAnswer: (Combine Gas Law) P1V1 = P2V2 T1 T2 1 x 50 = 5 x V2 293 373 V2 = 13cm318A. B. C. D.19Which change in conditions would increase the volume by x4 of a fix mass of gas?Pressure /kPa Doubled Halved Doubled HalvedTemperature /K Doubled Halved Halved DoubledAnswer: (Combine Gas Law) Initial P1 Final P2 = 1/2P1 Initial T1 Final T2 = 2T1 Initial V1 Final V2 = ? P1V1 = P2V2 P halved T1 T2 T double P1V1 = P1V2 T1 2 x 2T1 V2 = 4V1 Volume increase by x4Fix mass ideal gas has a V1 = 800cm3 , P1, T1 Find vol, V2 when P and T doubled. V2 = ? P2 = 2P1 T2 = 2T1 A. B. C. D.Combined gas Law no need to convert to SI units cancel off at both sides 3 variables involved200 cm3 800 cm3 1600 cm3 3200 cm3Answer: (Combine Gas Law) Initial P1 Final P2 = 2P1 Initial T1 Final T2 = 2T1 Initial V1 800 Final V2 = ? P1V1 = P2V2 T1 T2 P1 x 800 = 2P1V2 T1 2T1 V2 = 800 43. Ideal Gas Law Lab simulationIdeal Gas Law Videos 44. Acknowledgements Thanks to source of pictures and video used in this presentation Thanks to Creative Commons for excellent contribution on licenses http://creativecommons.org/licenses/Prepared by Lawrence Kok Check out more video tutorials from my site and hope you enjoy this tutorial http://lawrencekok.blogspot.com