PV=nRTP--- PaV .--m3n --- molT --- K (Kelvin)R--gas constantR=PVInT= (1atm)(22.4141L/(1mol)(273.15K)=0.082atm·L/mol·K=8.314 J/mol·K
6 PV = n RT P- Pa V -m3 n - mol T - K (Kelvin) R-gas constant R = PV/nT = (1atm)(22.4141L/(1mol)(273.15K) = 0.082 atm·L/mol·K = 8.314 J / mol • K
Calculatepand M:PV=nRT= mRT / Mp= m / V = PVM /RTV = PM /RT:. M=pRT/P
7 Calculateρand M： PV = nRT = m RT / M ρ= m / V = PVM / RTV = PM / RT M = ρRT/ P
Example:It's found that 0.896g of a pure gaseouscompound containing only N and O occupies 0.524L atapressure of 0.973 X 105Pa and a temperature of 28.0℃.What are the molecular weight and molecular formula ofthegas?Solution : n =m / M=PV / RT.M= mRT / PV= 0.896X 8.314 X301 / (0.973 X105X 0.524 X10-3)= 43.98g/mol.. The molecule is N,O
8 Example: It’s found that 0.896g of a pure gaseous compound containing only N and O occupies 0.524L at a pressure of 0.973×105Pa and a temperature of 28.0℃. What are the molecular weight and molecular formula of the gas? Solution ：n = m / M = PV / RT ∴M = mRT / PV = 0.896×8.314×301 / (0.973×105×0.524×10-3 ) = 43.98g/mol ∴ The molecule is N2O
1-2Dalton's Lawof Partial Pressurespartial pressure the pressure of each gas in a mixturePtotal =Pi + P2 +... = ZP;:PV= n,RTP,V = n,RT ...P,V = n,RTPtotal V=(P, + P, + ... + P) V= (ni + n2 + ... + n) RT = n total RT.. P / P total = n; / n total = XiP; = Ptotan XiXi:molarfraction（摩尔分数）
9 1-2 Dalton’s Law of Partial Pressures partial pressure :the pressure of each gas in a mixture . Ptotal = P1 + P2 + . = ∑Pi ∵P1V = n1RT P2V = n2RT . PiV = niRT Ptotal V = ( P1 + P2 + . + Pi ) V = (n1 + n2 + . + ni ) RT = n total RT ∴Pi / P total = ni / n total = Xi Pi = Ptotal·Xi Xi :molar fraction（摩尔分数）
Example:When a mixture of potassium chlorate and manganesedioxide was heated to decompose, its quantity decreased O0.480g andan oxygen of 0.377 dm3was obtained by draining water method (排水集气法)。Thetemperatureis294Kandpressureis9.96X104PaPlease calculate the molecular weight of oxygen.(And the saturatedvaporpressureis2.48X103Pa.)SolutionsMnO2,△2 KCI03(s.) → 2 KCI(s.)+ 3021Mo2 = mo2 / no2Ptotal =PH20 + Po2:: Po2= Ptotal -Ph20 = 9.96 X 104—2.48 X 103 = 9.71 X 104(Pa)no2 = Po02:Vtotal / RT= 9.71× 104X0.377X10-3/ (8.314X294)=0.0150 (m0l)Mo2 = mo2 / no2 = 0.480 / 0.0150 = 32.0 (g / mol)
10 Example : When a mixture of potassium chlorate and manganese dioxide was heated to decompose, its quantity decreased 0.480g and an oxygen of 0.377 dm3 was obtained by draining water method (排 水集气法)。The temperature is 294 K and pressure is 9.96×104 Pa. Please calculate the molecular weight of oxygen. (And the saturated vapor pressure is 2.48×103 Pa.) Solutions： MnO2 ,△ 2 KClO3 (s.) → 2 KCl(s.) + 3O2↑ MO2 = mO2 / nO2 Ptotal = PH2O + PO2 ∴PO2 = Ptotal－PH2O = 9.96×104－2.48×103 = 9.71×104 (Pa) nO2 = PO2·Vtotal / RT = 9.71×104×0.377×10－3 / (8.314×294) = 0.0150 (mol) MO2 = mO2 / nO2 = 0.480 / 0.0150 = 32.0 (g / mol)