Example :A sample of PCl(g.) weighing 2.69g was placed in a1.0Lflask and completelyvaporized at temperatureof 250C.Thetotal pressure observed at this temperature was 1.oOatm.Thepossibility exists that some of the PCl, may have dissociatedaccordingtotheequationPCl(g.)=PCl(g.)+Cl(g.)Whatmustbethefinal partialpressures ofPCls,PCl,and Cl,under suchexperimentalconditions?Solutions:npCI5 = 2.69 / Mpcis = 2.69 / 208 = 0.0129 (mol)P = n RT / V = 0.0129 × 8.314X 523 / 10-3 = 5.609 X 104(Pa)Since P<1.0Oatm , some dissociation ofPCl, must have occurred
11 Example :A sample of PCl5 (g.) weighing 2.69g was placed in a 1.0L flask and completely vaporized at temperature of 250℃.The total pressure observed at this temperature was 1.00atm.The possibility exists that some of the PCl5 may have dissociated according to the equation PCl5 (g.) = PCl3 (g.) + Cl2 (g.) What must be the final partial pressures of PCl5 ,PCl3 and Cl2 under such experimental conditions? Solutions： nPCl5 = 2.69 / MPCl5 = 2.69 / 208 = 0.0129 (mol) P = n RT / V = 0.0129×8.314×523 / 10－3 = 5.609×104 (Pa) Since P < 1.00atm ，some dissociation of PCl5 must have occurred
PCl, (g.)) = PCl,(g) + Clz (g.)00t=00.0129te0.0129—nnnPV = n total RT: ntotal = PV /RT = 1.013X 105 X1X 10-3 / (8.314X 523)= 0.0233= 0.0129 + n.. n = npc3 = ncn = 0.0104 molPPCI3 = PtotalX npCB / ntotal= 1.013X 105 × 0.0104 / 0.0233 = 0.4520X 105 (Pa)Pc2 = PpcI3Ppcis = 1.013 × 105 × (0.0129 -0.0104) / 0.0233= 0.1087 X 105 (Pa)
12 PCl5 (g.) = PCl3 (g.) + Cl2 (g.) t = 0 0.0129 0 0 te 0.0129－n n n PV = n total RT ∴ntotal = PV / RT = 1.013×105×1×10-3 / (8.314×523) = 0.0233 = 0.0129 + n ∴n = nPCl3 = nCl2 = 0.0104 mol PPCl3 = Ptotal×nPCl3 / ntotal = 1.013×105×0.0104 / 0.0233 = 0.4520×105 (Pa) PCl2 = PPCl3 PPCl5 = 1.013× 105×(0.0129－0.0104)/ 0.0233 = 0.1087×105 (Pa)
When the temperature and pressure is constant, the totalvolume of a mixture of gases is equal to the sum of the partialvolumes of thegases inthemixture.Vtotal = Vi + V, + .X; = V; / V totalProof:PV,=n,RTPV, = n, RTPV,= n;RTP(Vi + V, + ... + V) = (ni + n + ... + n) RTPVtotal = Nntotal RT:. V; / V total = n; / ntotal = X;店
13 When the temperature and pressure is constant, the total volume of a mixture of gases is equal to the sum of the partial volumes of the gases in the mixture. Vtotal = V1 + V2 + . Xi = Vi / Vtotal Proof： PV1 = n1 RT PV2 = n2 RT PVi = ni RT P (V1 + V2 + . + Vi ) = (n1 + n2 + . + ni ) RT P Vtotal = ntotal RT ∴ Vi / Vtotal = ni / ntotal = Xi
Several questions concerning gas moleculesmovement:How to move?guandP、V、n、TThe relations amongThe distribution of uIf u>100m/s, why we cannot smell the odor ofH,S instantly?
14 Several questions concerning gas molecules movement: ▪ How to move? ▪ The relations among u and P、V、n、T ▪ The distribution of u ▪ If u>100m/s, why we cannot smell the odor of H2S instantly?
1-3 The Kinetic-Molecular Theory--A ModelforGas Behavior(1)Suppose-Gas is composed of molecules-Molecules are in constant motion, in a random way.0~lightspeed3X108m/sAlldirectionsSpeed changes rapidly-Molecules don't affect each other except elastic collision
15 1-3 The Kinetic-Molecular Theory -A Model for Gas Behavior (1)Suppose ▪Gas is composed of molecules. ▪Molecules are in constant motion, in a random way. 0 ~ light speed 3×108 m / s All directions Speed changes rapidly ▪Molecules don’t affect each other except elastic collision