CHEM 150 1nd Edition Lecture 18Outline of Last Lecture • The gass Phase • The Kelvin scale (K)• Absolute Zero Outline of Current Lecture• Kinetic Molecular Theory (6.8)• Know the five assumptions of KMT.• Know how KMT defines pressure and temperature.• Know how KMT explains the gas laws.Current Lecture-KMT tells us that gas particles behave independently of each other. They occasionally collide and exchange energy, but that is it.-This is useful as it means that we can deal with each gas in a mixture separately.Q: Calculate the presure in a 1.000L container at 25.0°C if the Container holds:a- 0.364 moles argonP= nRT/V0.364moles x 0.0806 x 298.15K/ 1.000L =8.91atm b- 0.450 moles of chlorineP = nRT/V0.450moles x0.08206 x298.15K/ 1.000L =11.00 atmc- 0.364 mole of argon and 0.450 mole of chlorineP= 19.9 atm • The pressure contributed by one gas in a mixture of gases is called the partial pressure. The total pressure is the sum of the partial pressures.-mole fraction (x) of the gasPa =Xa . PtotXa = is the mole fraction of APa = %Va/100% PtotQ: What is the partial pressure of each gas in a mixture that is 42.0% nitrogen, 26.5% oxygen and 31.5% carbon dioxide by volume at STP?A: Ptot =1.00000atmPN2= 42.0%/100.0% x100000= 0.420atmPO2 = 0.265atmPCO2= 0.315atm-Rank these three gasses in order of increasing average kinetic energy:-Rank these gasses in order of increasing rms velocity (“average speed”).-Rank these three gasses in order of increasing time that it would take them to diffuse across a room.• From KMT : Urms=√3RT/m- where M stands for molar mass in kg/mol. R = 8.31447215 J/molAK• The average speed (rms velocity) of a gas molecule is inversely proportional to the square root of the molecular weight and directly proportional to the square root of the temperature.• Thus the rate of effusion (the process of moving through the pores in a solid) and of diffusion (the process of moving through another gas) is determined by the rms velocity. Q: What is the relative rate of effusion for hydrogen gas compared to nitrogen gas?A: rH2/rN2 =√mN2/mH2√28.02g/mol /2.016g/mol3.728g/molQ: 0.100 mole of CO2 diffuses across a room in 240.0 seconds at 35.0 °C . How long would it take 0.100 mole of bromine gas to diffuse across the room?A: TBr2/TCO2= √mBr2/mCO2tBr = √159.80g/mol /44.01g/mol x 240.0second=
View Full Document
Unlocking...