•Dalton’s Law of partial pressures states that the total pressure in a mixture is the sum of the partial pressures of the component gases.•The partial pressure of a gas is proportional to its mole fraction:PA= XAx PtotalXA=nAntotalMixtures of Gases
5-34Sample Problem 5.9Applying Dalton’s Law of Partial PressuresPROBLEM:In a study of O2uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mole % N2, 17 mole % 16O2, and 4.0 mole % 18O2. (The isotope 18O will be measured to determine the O2uptake.) The pressure of the mixture is 0.75 atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2in the mixture.PLAN:Find Xand Pfrom Ptotaland mol % 18O2. 18O218O2divide by 100multiply by Ptotalpartial pressure P 18O2mole % 18O2mole fraction, X18O2
5-35Sample Problem 5.9SOLUTION:= 0.030 atmP= Xx Ptotal= 0.040 x 0.75 atm 18O218O2= 0.040X18O2=4.0 mol % 18O2100
5-36Collecting a water-insoluble gaseous reaction product and determining its pressure.
5-37T(0C)T(0C)0510121416182022242628303540455055606570758085909510055.371.992.5118.0149.4187.5233.7289.1355.1433.6525.8633.9760.04.66.59.210.512.013.615.517.519.822.425.228.331.842.2Table 5.2 Vapor Pressure of Water (P) + at Different TH2OP (torr)H2OP (torr)H2O
5-38Sample Problem 5.10Calculating the Amount of Gas Collected over WaterPLAN:The difference in pressures will give Pfor the C2H2. The number of moles (n) is calculated from the ideal gas law and converted to mass using the molar mass.PROBLEM:Acetylene (C2H2) is produced in the laboratory when calcium carbide (CaC2) reacts with water:CaC2(s) + 2H2O(l) →C2H2(g) + Ca(OH)2(aq)A collected sample of acetylene has a total gas pressure of 738 torr and a volume of 523 mL. At the temperature of the gas (23oC), the vapor pressure of water is 21 torr. How many grams of acetylene are collected?
5-39Sample Problem 5.10multiply by MPtotalP of C2H2mass of C2H2use ideal gas lawnof C2H2subtract Pfor H2OPLAN:SOLUTION:PC2H2= (738 - 21) torr = 717 torr1 atm760 torr= 0.943 atmP = 717 torr x1 L103mL= 0.523 LV = 523 mL xT= 23°C + 273.15 K = 296 K
5-40= 0.0203 mol0.0203 mol x26.04 g C2H21 mol C2H2= 0.529 g C2H2SOLUTION:Sample Problem 5.100.943 atm0.523 Lxatm·Lmol·K0.0821x296 KnC2H2=RTPV=
5-41Kinetic-Molecular Theory - How It Explains the Gas LawsRationalize Five Key Questions at Molecular Level1.Origin of Pressure. How do individual gas particles create P?2.Boyle’s Law – V α1/P. What happens to particles when subjected to P?