Question

13. Calculate the effusion rate of the following gases. (4.0 pts, Gas Molar mass (g/mol) AAAAAAAAAAAA the following gases. (4.0 pts, show working out) Effusion rate (a) a MM SO2 NO CO2 CHA yyyyyyyyyyyyyyyy

Answer 1

Solution Effusions not only do gaseous harticles move with high kinetic energy but their small size enables them to move through small openings as well. this process is knows as effusion for effusion to occur the holes diameter must be smaller than the molecules mean free hath (the aucrage distance that a gas particle travels b/w succesine cellaisions with other gas hasticles). The opening of the hole must be smaller than the mean free hath because otherwise the gas could move back and forth through hole. Rate of effusion is in wisly proportional to {quare rout of mass of its particles effusion sate () DO OS JM.M. 'molar mass of

given since x = 1 SMM eyjusien sati @) = 1 SMM SO2 Molar mass of so2 = mass of S+ 2 mass of - 32+ 2x16 - Guglmal 1 I = 0-125 (g/mall 564 8. (2) NO Malas mais of No= (N+ omass 14+ 16 = 30 g/mal x= = 0. 1825741 (g/mil) -1/2 Co2 Molar mass of CO₂ = mass of (c +20) . 12 + 2x16 2 12 + 32- 44. = 0.1507556] 23 (otmal)

CHY Melao mass of Chy (gimae) = 12 +4 (1) = 16 g/mol 2 to 0.25 (91ml) dur 1 Molao mass (almal effusion raty 2) & = 1 0.125- 2 Nol 30 0 18 25741 0.150755 0.25 CHy will effuse fastly where as 502 will affuse slowly

given since x = 1 SMM eyjusien sati @) = 1 SMM SO2 Molar mass of so2 = mass of S+ 2 mass of - 32+ 2x16 - Guglmal 1 I = 0-125 (g/mall 564 8. (2) NO Malas mais of No= (N+ omass 14+ 16 = 30 g/mal x= = 0. 1825741 (g/mil) -1/2 Co2 Molar mass of CO₂ = mass of (c +20) . 12 + 2x16 2 12 + 32- 44. = 0.1507556] 23 (otmal)