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GASESKinetic Molecular Theory of Gases-The volume of an individual gas molecule is negligible compared to the volume of thecontainer holding the gas. This means that individual gas molecules, with virtually novolume of their own, are extremely far apart and most of the containeris “empty” space-There are neither attractive nor repulsive forces between gas molecules-Gas molecules have high translational energy. They move randomly in all directions, instraight lines-When gas molecules collide with each other or with a container wall, the collisions areperfectly elastic. This means that when gas molecules collide, somewhat like billiardballs, there is no loss of kinetic energy-The average kinetic energy of gas molecules is directly related to the temperature. Thegreater the temperature, the greater the average kinetic energy.
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GASESGas molecules have high translational energy. They move randomly inall directions, in straight lines
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GASESPressureVolumeTemperatureALL related to oneanother.If you change one,you change theother.
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GASES: PressurePressure (P) is measured in kilopascals (kPa)Pressure = ForceAreaThe unit Pa is the same as N/m2For example:The standard atmospheric pressure at 0ºC is 101.3kPaHow to measure pressure in pascals (Pa)?Converting common units: 760 mm Hg = 760 Torr = 1 atm = 101.3 kPa
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GASES: VolumeVolume (V) is measured in Litres (L)
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GASES: TemperatureTemperature (T) is measured in Kelvin (K)These are the same degrees as ºC, but:0ºC = 273.15KTo convert Celcius to Kelvin,use the following formula:T (in K) = ºC + 273
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GASES: RelationshipsBoyle’s Law: Pressure and volume are inversely proportionalPiVi = PfVfCharles’ Law: Volume and temperature are directly proportionalVi = VfTi TfGay-Lussac’s Law: Pressure and temperature are directly proportionalPi = PfTi Tf(assuming constant temperature)(assuming constant pressure)(assuming constant volume)
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GASES: RelationshipsP1VIf a sample of gasat initial conditionshas an increase ofpressure appliedto it, its volumedecreasesproportionallyBoyle’s Law: Pressure and volume are inversely proportionalPiVi = PfVf
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GASES: RelationshipsBoyle’s Law: Pressure and volume are inversely proportionalPiVi = PfVf
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GASES: RelationshipsCharles’ Law: Volume and temperature are directly proportionalVi = VfTi Tf
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GASES: RelationshipsGay-Lussac’s Law: Pressure and temperature are directly proportionalPi = PfTi Tf
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GASES: RelationshipsCOMBINED GAS LAWPiVi = PfVfTi TfSince pressure, volume, and temperature are all related, theycan all be combined together:
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GASESSTANDARD TEMPERATURE and PRESSURE (STP)Pressure = 101.3 kPa Temperature = 273K (0°C)STANDARD AMBIENT TEMPERATURE and PRESSURE (SATP)Pressure = 100 kPa Temperature = 298K (25°C)One of two conditions will be used for gas calculations:
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GASES: CalculationsSandra is having a birthday party on a mild winter’s day. The weatherchanges and a higher pressure (103.0 kPa) cold front (-25°C) rushes intotown. The original air temperature was -2°C and the pressure was 100.8kPa. What will happen to the volume of the 4.2 L balloons that were tied tothe front of the house?Given:Pi = 100.8 kPa Pf = 103.0 kPaVi = 4.2 L Vf = ?Ti = -2°C = 271K Tf = -25°C = 248KPiVi = PfVfTi Tf(100.8 kPa) (4.2 L) = (103.0 kPa) Vf(271 K) (248K)Vf = 3.76 LTherefore the volume of the balloons will decrease to 3.8 L
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GASES: Dalton’s LawDalton’s Law of Partial Pressures:The total pressure of a mixture of gases is the sum ofthe pressures of each of the individual gasesPtotal = P1 + P2 + P3 + P4 + P5 +… + PnPractice: Page 460 #22-23
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GASES: Ideal Gas LawGay-Lussac:Mole ratios are the same as volume ratiosAvogadro’s Hypothesis:Equal volumes of all ideal gases at the sametemperature and pressure contain the same numberof molecules
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GASES: Ideal Gas LawFor example:2H2(g) + O2(g) 2H2O(g)2 mol + 1 mol 2 mol2 volumes + 1 volume 2 volumeSo if you react 2L of hydrogen gas with 1L of oxygengas, you will get…2L of water vapour!!!
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GASES: Ideal Gas LawIdeal Gas Law formula: (Most IMPORTANT formula)PV = nRTPressure (kPa)Volume (L)Number of moles (mol)Universal gas constant8.314 kPa∙Lmol∙KTemperature (K)
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GASES: CALCULATIONSSulfuric acid reacts with iron metal to produce gas and an ironcompound. What volume of gas is produced when excess sulfuricacid reacts with 40.0g of iron at 18°C and 100.3 kPa?Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPaSTEP 1: Write the balanced chemical equationFe(s) + H2SO4(aq) H2(g) + FeSO4(aq)n = m/M= (40.0g) / (55.85g/mol)= 0.716 mol FeSTEP 2: Use molar ratios to solve for amount of product made (stoichiometry!)1 mol H2 = x1 mol Fe 0.716 mol Fex = 0.716 mol H2STEP 3: Use the ideal gas law to solve for the volume
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GASES: CALCULATIONSSulfuric acid reacts with iron metal to produce gas and an ironcompound. What volume of gas is produced when excess sulfuricacid reacts with 40.0g of iron at 18°C and 100.3 kPa?Given: mFe= 40.0g T = 18.0°C = 291K P = 100.3 kPaSTEP 3: Use the ideal gas law to solve for the volumePV = nRTV = nRTP= (0.716 mol x 8.314 kPa∙L/mol∙K x 291 K)(100.3 kPa)= 17.3 LTherefore 17.3 L of hydrogen gas are producedPractice: Page 506 #30-34
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GASES: CALCULATIONSA student reacts magnesium with excess dilute hydrochloric acid toproduce hydrogen gas. She uses 0.15g of magnesium metal. Whatvolume of dry hydrogen does she collect over water at 28°C and 101.8kPA?Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPa?Pressure of water vapour at 28°C = 3.78 kPa(from page 507, Table 12.3)Ptotal = PH2O + PH2(101.8 kPa) = 3.78 kPa + PH2Dalton’s Law of Partial PressuresPH2= 98.0 kPa
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GASES: CALCULATIONSA student reacts magnesium with excess dilute hydrochloric acid toproduce hydrogen has. She uses 0.15g of magnesium metal. Whatvolume of dry hydrogen does she collect over water at 28°C and 101.8kPA?Given: mMg= 0.15g T = 28.0°C = 301K P = 98.0 kPaSTEP 1: Write the balanced chemical equationMg(s) + 2HCl(aq) MgCl2(ag) + H2(g)n = m/M= (0.15g) / (24.31g/mol)= 0.0062 molSTEP 2: Use molar ratios to solve for amount of product made (stoichiometry!)1 mol H2 = x1 mol Mg 0.0062 mol Mgx = 0.0062 mol H2STEP 3: Use the ideal gas law to solve for the volume
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GASES: CALCULATIONSSTEP 3: Use the ideal gas law to solve for the volumePV = nRTV = nRTP= (0.0062 mol x 8.314 kPa∙L/mol∙K x 301 K)(98.0 kPa)= 0.16 LTherefore the student collects 0.16 L of dry hydrogenPractice: Page 511 #37-39A student reacts magnesium with excess dilute hydrochloric acid toproduce hydrogen gas. She uses 0.15g of magnesium metal. Whatvolume of dry hydrogen does she collect over water at 28°C and 101.8kPA?Given: mMg= 0.15g T = 28.0°C = 301K P = 101.8 kPa
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