CHM1045 Lecture 14Outline of Last LectureI. Oxidation numberII. Types of Oxidation-Reduction ReactionsIII. The Activity Series for HalogensIV. Chemistry in Action: Breath AnalyzerV. Solution StoichiometryVI. Gravimetric AnalysisVII. TitrationsOutline of Current LectureI. Physical Characteristics of GasesII. Apparatus for Studying the Relationship Between Pressure and Volume of a GasIII. Variation in Gas Volume with Temperature at Constant PressureIV. Charles’s and Gay-Lussac’s LawV. Avogadro’s LawVI. Ideal Gas EquationVII. How Gas Laws WorkCurrent LectureChapter 5: Gases- Elements that exist as gases at 250 C and 1 atmosphereH,N,O,F,Cl, He, Ne , Ar, Kr, Xe, RnPhysical Characteristics of Gases- Gases assume the volume and shape of their container- Gases are the most compressible state of matter- Gases well mix evenly and completely when confined to the same container- Gases have much lower densities then liquids and solidsPressure= Force (force= mass * acceleration) AreaUnits of Pressure 1 pascal (pa) = 1N/M2 1 atm= 760 mmHg= 760 torr 1 atm=101,325 PaExample 1:The pressure outside a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore, the air inside the cabin must be pressurized to protectthe passengers. What is the pressure in atmospheres in the cabin if the barometer reading is 688 mmHg?688 mm Hg pres. Atm688 mm Hg * 1 atm = .905 atm 760 mm HgExample 2:The atmospheric pressure in San Francisco on a certain day was 732 mmHg. What was the pressure in kPa?732 mm Hg pres. K Pa732 mm Hg * 101,325 Pa * 1 K Pa = 97.6k Pa 760 mm Hg 1000 Pa- Manometers used to measure gas pressureClosed tube= measure below atmosphere pressureOpen tube= measure above atmosphere pressureApparatus for Studying the Relationship Between Pressure and Volume of a GasAs P (h) increases V decreasesBoyles Law: -P a 1/V Constant temperature-P x V = constant Constant amount of gas- P1 x V1 = P2 x V2Variation in Gas Volume with Temperature at Constant PressureAs T increases V increasesCharles’s & Gay-Lussac’s Law- V a T-V = constant x T Temperature must be in Kelvin-V1/T1 = V2 /T2T (K) = t (0C) + 273.15 Charles’s and Gay-Lussac’s Law• Charles’s law– V1/T1=V2/T2• At constant pressure• Constant amount of gas• Gay-Lussac’s law– P1/T1=P2/T2• At constant volume• Constant amount of gasAvogadro’s Law- V a number of moles (n) Constant temperature- V = constant x n Constant pressure- V1 / n1 = V2 / n2Ideal Gas Equation- Boyle’s law: P a 1/V (at constant n and T)- Charles’s law: V a T (at constant n and P)- Avogadro’s law: V a n (at constant P and T)- V a = nt/P-V = constant x nt/P = R nt/P R is the gas constant- PV = nRTThe conditions 0 0C and 1 atm are called standard temperature and pressure (STP).PV = nRTR= PV = (1 atm)(22.414L) nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)How Gas Laws Work• Ideal gas law works when nothing is changing.• What if we have a pressure change, a volume change, a temperature change, or achange in the number of moles?Example 3: Sulfur hexafluoride (SF6) is a colorless and odorless gas. Due to its lack of chemical reactivity, it is used as an insulator in electronic equipment. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5°C.N= 1.82 mol V=5.43LT= 342.65 K 69.5+ 273.15= 342.65 KPV=nRT p = (1.82m) (.082057 L*atm/Mol*K) ( 342.65K) 5.43 L = 9.42 atmExample 4:Calculate the volume (in L) occupied by 7.40 g of NH3 at STP.Volume= ? = 7.40 g of NH3 @ STP T: 0 °C =273.15K P= 1 atm7.40g NH3 * 1 mol NH3 = .435 mol NH3 17.024 g NH3PV=nRT = (.435 mol)* .(08257 L*atm/mol *k) * (273.15) 1 atm= 9.74 L Example 5:An inflated helium balloon with a volume of 0.55 L at sea level (1.0 atm) is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon?V1= .55 L V2= ?P1= 1 atm P2= .40 atmP1V1=P2V2( 1 atm) ( .55 L) = (.40 atm)(V2)V2= 1.4LExample 6:Argon is an inert gas used in lightbulbs to retard the vaporization of the tungsten filament. A certain lightbulb containing argon at 1.20 atm and 18°C is heated to 85°C at constant volume. Calculate its final pressure (in atm).P1= P2 P1= 1.20 atm T1= 18°C = 291.5 KT1 T2 P2= ? T2= 85°C= 358.15 KP2= P1*T2 (1.20)(358.15k)= 1.48 atmT1 291.5Example 7:A small bubble rises from the bottom of a lake, where the temperature and pressure are 8.0°C and 6.4 atm, to the water’s surface, where the temperature is 25°C and the pressureis 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL.T= 8.0 °C 281.5 K T2=25°C 298.15 KP1= 6.4 atm P2= 1.0 atmV2=? V1= 2.1 mLP1V1 = P2V2T1 T2 V2= ( 6.4 atm)(2.1ml) (298.15) = 4007.136 (281.5) ( 1.0 atm)
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