Department of Chemistry of Texas at Austin University Name Gas UNIT READINESS ASSESSMENT QUIZ RAQ THIS QUIZ WILL BE PACED WITH CLICKER QUESTIONS 1 A 5 0 mol sample of Ne is confined in a 3 14 L vessel at a pressure of 2 5 atm What is the number density of the gas What is the mass density of the gas Number Density mol L Number Density 5 0 mol 3 14L 1 6 mol L Mass Density mass L We do not have the mass of the Ne gas sample yet So we will have to do a conversion using the periodic table 5 0 mol Ne 20 g Ne 1 mol Ne 100 g Ne Mass Density 100 g 3 14 L 32 g L 2 What is the total pressure of the gas mixture that contains 0 267 atm He 0 317 atm Ar 0 277 atm Ne Ptotal PHe PAr PNe Dalton s Law of Partial Pressures Ptotal 0 267 atm 0 317 atm 0 277 atm Ptotal 0 861 atm 3 What is the mole fraction of Ar in a mixture that contains 0 267 atm He 0 317 atm Ar 0 277 atm Ne PAr XArPtotal Xar PAr Ptotal Xar 0 317atm 0 861atm 0 368 Remember mole fractions have no units 4 Think about marshmallow in syringe Using macroscopic ideas of P T and V and using the ideas from KMT explain the result of pushing in the plunger In the description include a picture of gas inside and outside marshmallow before and after plunger pushed in Use the concept of velocity pressure collisions and molar mass of particles to explain the observed change The temperature remains the same therefore the velocity of the particles remains the same The number and molar mass of the particles remain the same As the plunger is depressed the volume decreases As the volume decreases the space between the particles decreases inside the syringe so the number of collisions per square inch increases therefore the pressure increases Revised SH 9 17 13 LaBrake Vanden Bout 2013 Department of Chemistry of Texas at Austin University Name Pictures below Before and After the plunger of the syringe is depressed 5 Think about the balloon in liquid N2 Using macroscopic ideas of P T and V and the ideas from KMT explain the result of putting the balloon in the liquid N2 In the description include a picture of gas inside the balloon before and after it is dipped Use the concept of velocity pressure collisions and molar mass of particles to explain the observed change The pressure inside the balloon remains the same the balloon is always at constant pressure It adjusts it volume to maintain a constant pressure because it is flexible The temperature decreases The volume decreases as temperature decreases The number of gas particles and molecular weight of the gas particles remain the same Comparing before and after the major change is the velocity of the gas particles Because the temperature is lower the velocity is lower This means their collision have a smaller impact Therefore to maintain a constant pressure there must be more collisions This can be achieved if the volume decreases Revised SH 9 17 13 LaBrake Vanden Bout 2013 Department of Chemistry of Texas at Austin University Name 6 Think about the balloon in the vacuum chamber Using macroscopic ideas of P T and V the ideas from KMT explain the result of putting the balloon in the chamber and turning it on In the description include a picture of gas inside the balloon and outside the balloon before and after the pump is turned on Use the concept of velocity pressure collisions and molar mass of particles to explain the observed change The number of particles inside the balloon does not change The number of particles outside the balloon but still inside the vacuum chamber decreases The molecular weight of the particles does not change The temperature and therefore the velocity of the particles remain the same The volume of the balloon increases when the chamber is turned on When the vacuum chamber is turned on the number of particles outside the balloon decreases therefore the number of collisions decreases therefore pressure outside the balloon decreases The collisions inside the balloon are greater than the collisions outside so the volume of the balloon increases against the low external pressure to create a matching low pressure inside P and V vary inversely Pictures below Before and After the vacuum chamber is turned on Revised SH 9 17 13 LaBrake Vanden Bout 2013 Department of Chemistry of Texas at Austin University Name 7 You have two gases under identical conditions One gas has a density that is double that of the other gas What is the ratio of the rate of diffusion of the lowdensity gas compared higher density gas Our goal is to solve for the ratio of diffusion rates of a low density gas compared to a high density gas v v low v high which is the same ratio as low v high Key Idea We assume that gases take up the same volume so the differences in density must be due to a difference in MASS So the high density gas has twice the mass as a low density gas mhigh 2mlow Now we can use the equation for ratio of diffusion to compare the two rates 3RT mlow v low 3RT 2mlow 2mlow x 2 v high mlow 3RT mlow 3RT 2mlow We substitute the mass of the high density gas to be twice the mass of the lowdensity gas The gas constant R and the temperature T cancel out the gases were at identical conditions So the final ratio between the rates of diffusion is the square root of 2 8 Given the following reaction 2CO O2 2CO2 Initially you have a container with 2 moles of CO gas and 3 moles of O2 gas at a constant temperature of 25 C and a constant pressure of 1 atm What is the final volume after the reaction is complete This is a LIMITING REACTANT problem There are several strategies for determining which reactant is the limiting reactant Conceptual strategy For every 2 moles of CO that react 1 mole of O2 reacts and 2 moles of CO2 are created If we use up the 2 moles of CO that we are given in the problem we would ONLY use up 1 mole of the O2 according to the moles ratios we identified in the previous bullet point Revised SH 9 17 13 LaBrake Vanden Bout 2013 Department of Chemistry of Texas at Austin University Name Therefore 2 moles of O2 gas would be leftover and we would make 2 moles of CO2 Mathematical strategy Divide the number of moles given in the problem by the molar coefficient from the reaction equation 2 moles CO 1 2 3 moles O2 3 1 Compare the two results from above The smaller number corresponds to the limiting reagent Because 1 is less than 3 CO is the limiting reagent We will use up ALL of the limiting reagent and …
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