Answer Key, Problem Set 7 Full

Transcription

1 Chemistry Mies, Fall 07 Aswer Key, Problem Set 7 Full. NT;. 5.43; ; ; ; ; 7. NT; ; ; & 5.90;. NT3;. 5.70; & 5.9; 4. NT Pressure uits ad measuremet with maometer [I Masterig oly] Simple Gas Laws [Not Dalto s Law]. NT. A cylider with a moveable pisto cotais some argo gas at a certai T, P, ad. If the umber of moles of argo is quadrupled (i.e., made to be four times greater tha the origial value) ad the Kelvi temperature is doubled, how must the volume be chaged to make the ew pressure ed up beig three times the origial pressure? (Assume ideal gas behavior.) 8 Aswer: must become or.66 times the origial value 3 Reasoig : P P RT T R (a costat) T T P T If is quadrupled, the 4; if T is doubled, the is ; ad T if P is to be three times P, the P P P 3 P Thus, 8 4, meaig must be adjusted to eight-thirds of its origial value. 3 3 Note: This makes sese i that if the T aloe were doubled, P would double. Ad if aloe were quadrupled, P would quadruple. So those two factors would make P become 8 times what it was. Thus, i order for the fial P to ed up 3 times (ad ot 8 times) the origial value, the must chaged to make the P become 3/8 (if that were the oly chage). Sice P ad are iversely proportioal, should become 8/3 its value to achieve this desired effect. Reasoig : P 3P T T P RT RT RT 4RT 8 RT 8 P P 3P 3 P A automobile tire has a maimum ratig of 38.0 psi (gauge pressure). The tire is iflated (while cold) to a volume of.8 L ad a gauge pressure of 36.0 psi at a temperature of.0 C. Drivig o a hot day, the tire warms to 65.0 C ad its volume epads to. L. Does the pressure i the tire eceed its maimum ratig? (Note: The gauge pressure is the differece betwee the total pressure ad atmospheric pressure. I this case, assume that atmospheric pressure is 4.7 psi.) Aswer: Yes, it will. Fial gauge pressure will become 43.5 psi Importat Commets: I either approach below, you must recogize that the actual iitial pressure i the tire is ot 36.0 psi, but rather psi (because of the defiitio of gauge pressure : Pgauge = Pgas Pbar Pgas = Pgauge + Pbar; see Note at the ed of the problem tet above). Likewise, oce a fial (actual) pressure is obtaied (here, 58. psi), the gauge pressure is that value mius 4.7 psi ( 43.5 psi). You must also covert the T s ito Kelvi: T.0 C K T 65.0 C K P T T PS7-

2 Approach (as i the prior two problems!): A ew P is asked for after a chage i T ad, with costat. Use: P P P P P P RT P T costat R (a costat) P T T T T Substitutig i (see Importat Commets above): P 50.7 psi.8 L. L K 85.5 K psi (actual pressure) T T gauge pressure = psi > 38.0 psi (maimum ratig) NOTE: This approach has advatages over usig the ideal gas law to fid first, ad the usig the ideal gas law a secod time ( traditioal approach). ) You do ot eed to covert the pressures ito atm (ad back) or the volumes ito L, because i this approach, R cacels out (or rather, it is the ratios of P ad that matter, so their uits cacel out.) ) There eds up beig fewer overall calculatios, sice oe mai setup is all that is eeded (rather tha two, ot coutig the coversios). Approach : Use the ideal gas law to fid, the use it agai to fid P. Setups are ot show i key at this time, but turs out to be.739 moles, ad P eds up beig atm (actual pressure). This leads to the same result as above, as it must.] Desity ad Molar Mass Problems A sample of N O gas has a desity of.85 g/l at 98 K. What is the pressure of the gas (i mmhg)? Aswer: mmhg Strategy (Mies): ) Recogize that sice o amout of gas is give, you ca pick a coveiet amout. Sice desity is give, a coveiet amout is just L. Not oly does this give you a, it also gives you a amout of gas:.85 g. ) Use grams ad molar mass to calculate moles (this gives you ) 3) Now, use,, ad T, to calculate P (usig P RT) 4) Covert from atm to mmhg by multiplyig by 760 mmhg/atm Eecutio of Strategy: MM(NO) = (4.0) = 44.0 g/mol =.85 g RT P RT P Latm ( mol) (98 K) molk L mmhg mo NO mol NO 44.0 g 760 mmhg atm 03...mmHg NOTE: You could also use Equatio 5.6 i your tet to solve this problem, but that just meas oe more equatio to memorize (or lear how to derive), whe there is really absolutely o eed to do so! Sice you will ot be allowed a equatio sheet o the eam, ad I will ot provide you with ay equatios o the eam, does t it make sese to lear the most basic equatios ad the lear how to use them to solve problems (rather tha memorizig a buch of [related] equatios so you ca try to plug ad chug everythig)? A quatity of CO gas occupies a volume of 0.48 L at.0 atm ad 75 K. The pressure of the gas is lowered ad its temperature is raised util its volume is.3 L. Fid the desity of the CO uder the ew coditios. Aswer: 0.46 g/l Strategy (otice the parallel to the prior problem. This is a very similar problem i my opiio): PS7-

3 ) Recogize that to fid desity, you eed two thigs: grams ad Liters (mass ad ) ) They give you the ew volume (.3 L), so the oly thig left to do is fid the mass of the gas! 3) The iitial coditios give you, P, ad T, so you ca easily get (moles) usig the ideal gas equatio. 4) Sice the gas is kow (it is CO), you ca use its molar mass ad moles to calculate grams. NOTE: If you focus o what you eed, this problem is actually quite straightforward. The oly reaso I ca thik of that this is i the Challege Problems sectio is because you might thik you have to do somethig with the ew pressure ad ew temperature (sort of like teasers i this problem). Also, you have to realize that the mass of a gas does ot chage with T,, or P if remais the same. Perhaps this problem might seem harder because you ca t use Equatio 5.6 (sice you do t kow the ew P?! Eecutio: P P RT RT Stoichiometry atm0.48 L mol CO Latm (75 K) molk 0.0..mol CO ( ) g/mol g CO m g d g/l.3 L Cosider the chemical reactio (represeted by): H O(l) H (g) + O (g) What mass of H O is required to form.4 L of O at a temperature of 35 K ad a pressure of atm? Aswer:.9 g HO eeded Strategy: ) Recogize that the T, P, ad ifo provided are for O, ad so ideal gas equatio. O ca be calculated from the ) Recogize that the questio asks about the mass of HO (ot O!), so this really is a stoichiometry problem ot just a ideal gas law problem. Use the balaced equatio to get the mole ratio betwee O ad HO, the calculate moles of HO. 3) Use the molar mass of HO ((.008) = 8.0 g/mol) ad moles (#) to calculate mass. Eecutio: atm.4 L P P RT O RT Latm mol O (35 K) molk mol O mol HO 8.0 g g HO eeded mol O mol H O Kietic Molecular Theory (icludig Speed Distributio Curve Cocept) What are the basic postulates of kietic molecular theory (KMT)? How does the cocept of pressure follow from KMT? NOTE: Your tet combies a couple of postulates to ed up with oly three, but I like to list them as four (eve though there is some overlap this way). I ay case, you should ot have copied, word-for-word the postulates from the tet o your paper (uless you have actually memorized the postulates word-for-word). If you do t make the material your ow, you will likely ot lear much. Aswers: PS7-3

4 ) Gas particles are i costat motio, movig i straight lies util they collide with each other or a wall of the cotaier. The collisios with the walls are the cause of the pressure eerted by the gas. ) The volume of a gas particle is egligibly small compared to the size of the cotaier (because the distace betwee particles is so large). (For a ideal gas, the particles are cosidered to have o volume at all. Obviously real gas particles have a fiite volume.) 3) Collisios are completely elastic, ad betwee collisios, particles eert o force (attractive or repulsive) o each other. 4) The average kietic eergy of a gas particle (i a sample with lots of gas particles) is depedet oly o the Kelvi temperature, to which it is directly proportioal. The cocept of pressure directly follows from theory because each collisio imparts a force o the wall it hits, ad so whe billios of collisios each secod hit a uit area of a wall, the collective force per area is just the sum force of all the idividual collisios. Sice P = force / area, the collective force per area associated with the collisios is the pressure of the gas. 7. NT. A small balloo with some air iside is placed iside of a syrige, ad the syrige (which also has air i it) is placed iside of a large bo with the barrel (pluger) of the syrige pulled about halfway out (see sketch to the right). If the air i the bo were pumped out, describe what would iitially happe to the (i) barrel (Would it move iward, outward, or ot move) ad (ii) the balloo (Would it epad, cotract, or stay the same size?) IF: (a) The ed of the syrige is capped. Give reasoig. (b) The ed of the syrige is left ope. Give reasoig. (a) Pressure outward moves barrel because there s gas i syrige; the balloo moves. (b) Pressure outward oly o ski of the balloo because gas i syrige goes out. ANSWERS. (a) the barrel would move outward, ad the balloo would epad. (b) the barrel would ot iitially move, but the balloo would epad. EXPLANATIONS. (a) Whe gas is pumped out of the chamber, the pressure o the barrel from the outside decreases, ad thus a et force from the iside (i the outward directio) develops. Thus the barrel moves outward. Oce the barrel moves outward, the situatio is idetical (from the balloo s perspective) to the situatio i my demo (where I pull the barrel out) the pressure agaist the outside of the balloo decreases (because air iside syrige epads ito a bigger volume, fewer collisios agaist the outside of the balloo), while the pressure agaist the iside of the balloo remais the same (temporarily). Thus there is a et force outward o the ski of the balloo ad the balloo epads. (b) If the syrige is NOT capped, the the NET FORCE o the barrel does ot chage, sice air leaves the syrige whe air is pumped out of the chamber. I other words, there will be (equally) less pressure o both eds of the barrel because the cocetratio of gas will decrease ot oly o the outside, but also o the iside. However, NO GAS CAN ESCAPE FROM THE BALLOON, so it will push out with the same force as always ad as the pressure ON IT from the outside decreases, there will be a et force i the outward directio (same as (a)). PS7-4

5 Eplai how Boyle s Law, Charles s Law, Avogadro s Law, ad Dalto s Law all follow from KMT. Aswers: (Note: your authors directly aswer these questios o p.07. I hope you did ot just copy the tet aswers o your paper. See ote i problem #6 (5.9) above.) ) Boyles Law: P is iversely proportioal to at costat T ad. Thus, eplai why a decrease i leads to a icrease i P (at costat T & ). Shortest aswer: A decrease i leads to a icreased particle cocetratio ad thus a icrease i the # of collisios per sec with the walls. More collisios greater pressure (at costat T). More detailed aswer: Sice T is fied, force per collisio is fied. Thus, the pressure depeds o the umber of collisios per sec with a uit area of a wall. If is decreased while is costat, the cocetratio of particles goes up, raisig the collisioal frequecy ad thus the pressure. Or you could say that the walls are, i effect, beig moved closer to the particles, so more of them will hit the walls each secod after the decrease. ) Charles s Law: is proportioal to Kelvi T at costat P ad. Thus, eplai why a icrease i T leads to a icrease i (at costat P & ). Shortest aswer: A icrease T leads to a icrease i average KE, makig speed greater ad thus collisioal force ad frequecy greater (iitially). The oly way to keep P from icreasig uder such coditios is to icrease, which will decrease collisioal frequecy despite the icreased speed. More detailed aswer: A icrease i T leads to a icrease i average kietic eergy of particles. Thus, they move faster ad hit harder (i.e., with greater force per collisio ). Thus, Pgas must iitially icrease. But i this case, P is costat, which meas the vessel ca chage its volume (if mechaical equilibrium is disturbed). Oce Pgas becomes greater tha Peteral, the walls of the cotaier will feel a et force outward, so they will move outward icreasig the volume. This will decrease Pgas (see Boyles Law eplaatio above) util it becomes equal to Peteral. Alteratively, you could say (as the book authors do) that i order to keep Pgas from icreasig whe T icreases, must icrease. 3) Avogadro s Law: is proportioal to at costat T ad P. Thus, eplai why a icrease i leads to a icrease i (at costat T & P). Shortest aswer: A icrease i leads to a icreased particle cocetratio (iitially) ad thus a icrease i the # of collisios per sec with the walls, ad thus pressure. The oly way to keep P from icreasig uder these coditios is to icrease the volume proportioately, which will keep the cocetratio costat. More detailed aswer: A icrease i leads to a icreased particle cocetratio (iitially) ad thus a icrease i the # of collisios per sec with the walls, ad thus pressure. But if the gas is i a costat pressure cotaier like a syrige, the pressure ca t icrease! Rather, the icreased Pgas will make Pgas temporarily greater tha Peteral, makig the walls move outward, icreasig the volume. This offsets the iitial pressure icrease (if icreases proportioally to, / will remai costat, so the collisioal frequecy will remai costat). Alteratively, you could say that i order to keep P from icreasig uder these coditios, must icrease." Sice T is costat, force per collisio remais costat. This oe is all about collisioal frequecy (ad cocetratio). 4) Dalto s Law: Ptotal is the sum of all partial pressures. I a sese, this is sayig that pressure does ot deped o the idetity of the particles at the same T ad, P depeds oly o, regardless of the type of particle. Thus, eplai why o matter what the particle type is, P depeds oly o (at costat T & ). NOTE: I ll tell you right ow that this oe is, by far, the most difficult oe to eplai properly. Eve the tetbook does ot get it quite right. I will aswer it correctly here, but I will ot hold you to this level of eplaatio o a eam. What I do wat you to kow how to eplai is why, for a give gas, if you icrease at costat T ad, P icreases (see ote i #9 (5.49) PS7-5

6 below [et problem]). That is more straightforward icreased cocetratio meas greater collisioal frequecy, ad if T remais the same, the force per collisio remais costat. Correct aswer: Pressure i KMT ca be thought of as due to two compoets: force per collisio ad # collisios per sec (collisioal frequecy): force # collisios P collisio sec Force per collisio is proportioal to (average) mv (mometum). The # collisios per sec is proportioal to both cocetratio ad velocity. Thus, To justify this, ote that if cocetratio is the same, the average distace to a wall is the same, ad so a icreased velocity will result i a greater umber of collisios per sec. Ad if average velocity were the same, a icrease i cocetratio would mea a smaller average distace to a wall, ad thus result i a greater umber of collisios per sec. P mv v Mathematically, this is the simplest way to show you why it is oly T that matters. Note that if you move the oe v from the secod epressio to the first, you get: mv P Sice the first epressio (mv ) is proportioal to average KE (½ mv ), ad average KE depeds oly o T, this shows that eve if mparticle chages, P will ot chage uless / chages. Ad thus at costat T ad, P is proportioal oly to. QED (sort of) Coceptually is where it gets subtle. If m particle is larger for oe gas (let s say it s 4 larger, for simplicity) but T is the same, the force / collisio actually will become greater for that gas (twice as great i this case!) eve though the T is the same. This is because v will oly become half as great (because the average KE s are the same--½ mv ½ mv ) But the collisios / sec will become smaller eve though the cocetratio is the same (half as great i this case because the speed is half as great), so the two effects will offset oe aother, leavig the P the same. The bottom lie is this, for two gases at the same T, the oe whose m particle is larger will have a smaller collisioal frequecy (because v is smaller), but a proportioately greater force / collisio (because m is bigger by more tha v decreases). Whew! Which gas sample represetatio has the greatest pressure? Assume that all the samples are at the same temperature. Eplai. Aswer: (b) [the oe with 0 gas particles i it] NOTE: I put this questio here i the PS sequece because I wated you to be able to use KMT to eplai your aswer. However, you ca aalyze it with either law or theory as I ll show: Reasoig: Gas Laws P RT P RT R T T costat This shows that whe T is costat, P is determied oly by particle cocetratio (/). Sice is the same i this set of three pictures, the bo that has the most particles i it has the greatest cocetratio ad therefore the greatest pressure. Thus, the aswer is (b). Reasoig: Kietic Molecular Theory I KMT, pressure is caused by the collisios of the gas particles with the walls of the cotaier. You ca imagie that P is proportioal to both the force per collisio ad the # collisios per sec (see prior problem). For a give gas, if T is costat, the force per collisio is costat. P PS7-6

7 But if the cocetratio of particles is greater, the frequecy of collisios with the walls will also be greater, ad thus so will be the pressure. So at costat T, P is proportioal to cocetratio, ad bo (b) has the greatest pressure. Although techically thigs are a bit more complicated whe you compare differet gases (see techical ote below as well as prior problem), this idea is sufficiet for my class (i.e., o a eam). Techical ote: The above argumet is ot techically correct because, of course, differet gases move with differet velocity at the same T (this you are resposible for kowig o a eam! See et questio, amog others!), ad so collisioal frequecies will ot all be idetical at the same T. But as oted i the prior problem (subtle part of eplaatio), the chages i mass, velocity, mometum, ad collisioal frequecy (at costat T) offset themselves such that at costat T, oly cocetratio affects pressure & Substace A has the greater molar mass. The average speed of A particles is less tha that of B (peak for A is at a further left velocity about 500 m/s vs. about 00 m/s for B). Thus A must have the greater mass per particle (ad thus molar mass) T is greater. The average speed of the particles is greater at T (peak is farther to the right ~00 m/s vs. ~800 m/s at T). Sice average KE ( ½ mv ) is greater at higher T (KMT), ad m here is fied (same substace), average v must be higher at the higher T.. NT3. Cosider two differet cotaiers, each filled with moles of eo gas. Oe of the cotaiers is rigid ad has costat volume. The other cotaier is fleible (like a balloo) ad is capable of chagig its volume to keep the eteral pressure ad iteral pressure equal to each other. If you raise the temperature i both cotaiers, what happes to the pressure ad desity of the gas iside each cotaier? Assume a costat eteral pressure. Aswers: I the costat- cotaier, whe T is icreased, P icreases ad desity stays the same I the costat P cotaier (Pgas Peteral), whe T is icreased, P remais the same ad desity decreases. Reasoig, Gas Laws: I both cotaiers: ) is costat sice the cotaiers are sealed ad o chemical reactio is occurrig iside. Thus, P RT reduces to P kt i both cases (for ow). ) It also follows that the mass of gas i each cotaier is also costat ( 0.8 g/mol = g). 3) Lastly, desity (d) is defied as mass/ = mgas/cotaier. st Cotaier: If (i additio to ) is also held costat, the P kt reduces to P k T, ad P is directly proportioal to T. Thus whe T icreases P icreases. Also, sice cotaier is costat ad mgas is costat, dgas does ot chage. d Cotaier: If P (i additio to ) is also held costat, the P kt reduces to k T, ad is directly proportioal to T. Thus whe T icreases, icreases ad thus dgas decreases. Sice, by defiitio i this cotaier, Pgas always equals Peteral, Pgas must remai the same. Reasoig, Kietic Molecular Theory: I both cotaiers, whe T icreases, the average kietic eergy of the gas particles icreases, which meas that their average velocity icreases (sice the gas particles masses do ot chage). I cotaier, sice the is kept the same, the result is that there are a greater umber of collisios per secod with a give area of the cotaier walls, as well as a greater force per collisio, resultig i a greater P. Sice the same umber of gas particles are movig aroud i the same volume after the T icrease, the desity does ot chage (i.e., desity is idepedet of T as log as is held costat). I cotaier, iitially after the T icrease, the P temporarily does icrease as i cotaier. The differece is that i cotaier, the walls are ot rigid, so that oce the Piteral is greater tha Peteral, there will be a et force outward o the walls of the cotaier pushig them outward, icreasig the volume. As icreases, the pressure will decrease because the umber of collisios per secod will PS7-7

8 decrease (o average, particles are farther from a wall whe is larger [cocetratio is smaller]). So evetually, the Pgas will agai become equal to Peteral eve though the particles are movig faster. Sice the same umber of particles are movig aroud i a greater cotaier, dgas decreases. Gas Mitures, Partial P, Dalto s Law A helio deep-sea divig miture cotais.0 g of oyge to every 98.0 g of helium. What is the partial pressure of oyge whe this miture is delivered at a total pressure of 8.5 atm? Aswer: 0.0 atm Strategy: Note: This would result i a dead diver! The problem authors seemigly made a mistake here! I thik they thought they were makig a miture that would result i a partial pressure of 0. atm (which would be close to ormal ). Oops!! ) Note that partial pressure ca be obtaied i multiple ways. I this problem, sice Ptotal is give, A recall that the partial pressure of a gas A is related to Ptotal by mole fractio: PA Ptotal total ) Sice masses of each gas are give, moles of each ca be calculated from molar masses. 3) Sum the moles to get total, ad use the equatio i () to get PO Eecutio: mol MM(O) (6.00) 3.00 g/mol O.0 g mol O 3.00 g mol MM(He) g/mol He 98.0 g g total mol P O O Ptotal total mol mol mol He 8.5 atm atm Real Gases [Deviatios from Ideal Gas Behavior ad qualitative eplaatio of] & 5.9. Which postulate of the KMT breaks dow uder coditios of (a) high pressure? (b) low temperature? Eplai. (a) At high pressure, the postulate that states that the volume of a gas particle is egligible with respect to the volume of the cotaier is o loger a good assumptio. Thik of high pressure as low volume here that is, i very compressed states, where cocetratio (/) is very high. Whe the particles are so close together, their actual volume becomes evidet, ad the behavior is o loger ideal. The fact that the gas particles take up space makes the actual volume of the gas larger tha what it would be if it were behavig like a ideal gas. Thik of it this way: O page 4, your tetbook authors state that at STP, the sum of the volumes of the idividual atoms i a sample of Ar is oly about 0.0% of the volume of the gas. That meas that i L (000 ml) of Ar at STP, the sum of the volume of all of the idividual atoms is oly 0. ml! This is a egligibly small volume (about the volume of - drops of water)! However, assumig ideal behavior, if the pressure were made to be 000 atm, the volume of the gas would become /000 th of L, which equals ml. Now the volume of the idividual atoms (still 0. ml) would be 0% of the volume of the gas! Clearly this amout of volume is ot egligible. I fact, the volume would ot ed up beig oly ml it would be somewhat larger tha that (see Figure 5.4, which is eve better tha my eample sice it is at elevated T which miimizes the effects of itermolecular forces, which I coveietly igored.) PS7-8

9 (b) At low temperature, the postulate that states that there are o forces betwee the particles betwee collisios is o loger a good assumptio. The fact of the matter is that there are attractive forces betwee particles (called itermolecular forces ). This should be obvious to everyoe whe you cosider that all gases will tur ito liquids or solids if the temperature is made low eough. Why would molecules of liquid water stick together i a liquid if there were o forces of attractio betwee them? They would t! But those forces of attractio have decreasig effect o the motio of particles whe the particles are movig with high kietic eergy (high T). So at very high temperatures for all substaces (ad lesser temperatures for those substaces with the weakest forces, like typical gases at room temperature), the forces become egligible relative to the average kietic eergy (which icreases with T), ad that eplais why gases behave ideally at high T. But thik about the reverse as T decreases, the forces (which remai the same ) become less ad less egligible sice the average kietic eergy is steadily decreasig. At some poit, the forces become apparet, ad the assumptio that they are egligible is ot a good oe. That s whe gas behavior begis to deviate from ideal behavior. Cumulative Problems The fact that gas particles do attract oe aother makes the pressure with which they strike the walls of their cotaier less tha what it would be if the forces were egligible (See Fig. 5.5). I a very real sese, the particles are beig attracted toward the ceter of the sample by their mutual forces of attractio, so they do t hit the walls as hard. 4. NT4. Which graph below would best represet the distributio of molecular speeds for the gases acetylee (C H ) ad N give the followig coditios? Both gases are i the same flask (ad thus at the same T) with a total pressure of 750 mm Hg. The partial pressure of N is 500 mm Hg. Give your reasoig. # particles with a give speed N CH (a) CH (b) (c) N N CH speed speed speed Aswer: (a) Reasoig: Note: There are several aspects to a (speed) distributio curve: a) the shape, b) the highest poit o the curve, whose associated speed is called the most probable speed, ad c) the overall height of the curve (really the area udereath the curve), which is proportioal to how may total particles are i the sample (because the y-ais is umber of particles which have a particular speed ). **Tro has chose to use relative umber of particles (rather tha actual umber) o its y-aes for distributio curves. This differs from the web simulatios that I showed you (ad what is i the preset problem. So be aware of this possible differece. Always check aes!** SPEED ARGUMENT. Same T same average kietic eergy, ad thus the oe whose molecules are less massive will move more quickly. I this case, the molecular mass of CH is () + () = 6 amu/molecule ad the molecular mass of N is (4) = 8 amu/molecule. So they are actually ERY CLOSE i mass ad should be very close i average speed; but the molecules i the sample of CH should be movig slightly faster (6 amu < 8 amu). That would be idicated o a distributio curve plot by havig the peak of the curve for the sample of CH lie just to the right (higher speed) of the peak for the sample of N. The oly picture that shows this is (a). (b) has the N sample havig the greater average speed, ad (c) idicates that CH is the faster oe, but by quite a substatial amout rather tha a small amout. PS7-9

10 PARTIAL PRESSURE/AMOUNT ARGUMENT. If you were t sure about whether or ot it was (a) or (c), you could use the partial pressure iformatio to make it crystal clear. Sice the partial pressure of O is 500 mm Hg ad the total pressure is 750 mm Hg, the partial pressure of CH must be 50 mm Hg. That meas that there are TWICE AS MANY O MOLECULES i the flask as CH. That meas that the curve for O should have about twice as much area uder it (or be about twice as high at every poit). The curves i Figure (a) look like this, but ot i figure (c). ==================== END OF SET ==================== Problems below are from a prior key. Use these for practice! 5.8. Covert 65.5 mmhg (lowest pressure ever recorded at sea level iside Typhoo Tip) to: Aswers: (a) 65.5 torr; (b) atm; (c) 5.69 i Hg; (d).6 psi Note: See Table 5. for equivaleces (a) torr 65.5 torr (because torr = mmhg) atm (b) atm 65.5 mmhg atm (recall that 760 here is eact) 760 mmhg Aswer is reasoable. 65 is somewhat less tha 760, so P should be somewhat less tha atm 9.9 i Hg (c) i Hg 65.5 mmhg i Hg (or you could multiply by 760 mmhg i cm.54 cm 0 mm ) Aswer is reasoable. 65 is somewhat less tha 760, so P should be somewhat less tha 9.9 i Hg 4.7 psi (d) psi 65.5 mmhg psi (your tet oly gives psi to 3 SF) 760 mmhg Aswer is reasoable. 65 is somewhat less tha 760, so P should be somewhat less tha 4.7 psi Give a barometric pressure of 75.5 mmhg, calculate the pressure of each gas sample as idicated by the maometer. Aswers: (a) 775 mmhg; (b) 89 mmhg Strategy: ) Reaso out the proper relatioship betwee Pgas ad Pbar. How? As show i class, I like to look at it this way: The side that has more Hg i it eeds the help ( Petra, which equals h, i uits of mmhg if the maometer is filled with Hg) to equal the P o the other side. So i (a), where the height is higher o the gas side, Pgas + h Pbar Pgas Pbar h But i (b), where the height is higher o the atmosphere side, Pgas Pbar + h ) Calculate h ( hhigher hlower) from the maometer pictures. Note the uits! 3) Covert the uits (cm) ito mm to be cosistet with those of Pbar. The use the equatio i (). Eecutio: PS7-0

11 0 mm (a) h =. (-.) cm =.3 cmhg 3 mm Pgas = 78.5 = 79 mmhg cm 0 mm (b) h = 3.3 (-3.4) cm = 6.7 cmhg 67 mm Pgas = 88.5 = 89 mmhg cm 5.3. A sample of gas has a iitial volume of 3.9 L at a pressure of. atm. If the sample is compressed to a volume of 0.3 L, what will its pressure be? Note: You eed to assume costat T ad here, as well as ideal behavior! Aswer:.65 atm Reasoig : If ad T are costat, makig gas smaller makes P bigger (iversely proportioal, Boyle s Law). So if becomes (0.3 / 3.9) ths of what it was, the P must get (3.9 / 0.3) times what it was. Reasoig : If you do ot ituitively see how the iverse proportioal relatioship betwee P ad works, you ca either memorize P P, or derive it from the ideal gas law as show i class ad below: P P P,T costat P RT R (a costat) P P P P T T T Substitutig i with values (make sure you carefully assig the values for State ad State!) yields: 3.9 L. atm atm 0.3 L A syrige cotaiig.55 ml of oyge gas is cooled from 95.3 C to 0.0 C. What is the fial volume of oyge gas? Aswer:.5 ml Reasoig : If ad P are costat (which must be assumed here!), makig Tgas smaller makes smaller. The relatioship is proportioal (Charles Law) oly if T is i Kelvi). So first covert the T s to Kelvi by addig 73.5 to each: T = K; T = 73.5 K. So if T becomes (73.5 K / K) ( ) times what it was, the must become times what it was also (directly proportioal) ml = =.5 ml Reasoig : If you do ot ituitively see how the direct proportioal relatioship betwee T ad works, you ca either memorize /T /T, or derive it from the ideal gas law as show i class ad below: P P P T P RT T T Substitutig i with values yields:,p costat R (a costat) T T T T ( ) K.55 ml.49.. ( ) K NOTE: Make sure to always carefully assig values for State ad State. I particular, here you must realize that mol is ot, because it is the umber of moles added. Thus = ml A cylider with a moveable pisto cotais mol of gas ad has a volume of 53 ml. What will its volume be if a additioal mol of gas is added to the cylider? (Assume costat temperature ad pressure.) Agai, you must assume ideal behavior here. Aswer: 40. ml Reasoig : If P ad T are costat, makig gas bigger makes bigger (proportioal, Avogadro s Law). So if becomes (( ) / 0.553) (.66) times what it was, the must get.66 times what it was also ml = 40. ml PS7-

12 Reasoig : If you do ot ituitively see how the direct proportioal relatioship betwee ad works, you ca either memorize / /, or derive it from the ideal gas law as show i class ad below: P P P P RT T Substitutig i with values yields: P,T costat R (a costat) T T ( ) mol 53 ml mol 40. ml NOTE: Make sure to always carefully assig values for State ad State. I particular, here you must realize that mol is ot, because it is the umber of moles added. Thus = A sample of gas has a mass of g. Its volume is 7 ml at a temperature of 85 C ad a pressure of 753 mmhg. Fid the molar mass of the gas. Aswer: 4 g/mol Strategy (Mies): ) Recogize that to fid molar mass (g/mol), you eed two thigs: grams ad moles (just divide them to get MM) ) The mass i grams of the sample is already i the problem! So all you eed to do is fid, the moles of gas usig the ideal gas equatio (sice, T, ad P are give)! Just remember to get quatities ito proper uits. Eecutio: P P RT RT MM g mol 753 mmhg g/mol atm 760 mmhg Latm molk 7 ml ( K) L 000 ml mol = K (ucertaity i the uits place b/c of additio rule 3 SF) NOTE: The authors of the Solutios Maual do this problem usig Equatio 5.6! I my opiio, that is just plai silly (they actually calculate the desity first. Why? Because that formula has desity ad molar mass it i, so it would seem that you eed d to get MM. How slavish to formulas this seems! My approach is much more direct. Please try to lear to solve the problem. Do t try to fid a equatio from the book that might work A flask at room temperature cotais eactly equal amouts (i moles) of itroge ad eo. (a) Which of the two gases eerts the greater partial pressure? A NEITHER. Partial pressure depeds oly o of the gas P A Ptotal (Okay, so I just total realized that I covered partial pressure last i lecture ad put the partial pressure problems later i the sequece. Sorry this oe slipped i.). Sice the moles of gases i this problem are eactly equal, their partial pressure must also be equal. (b) The molecules or atoms of which gas have the greater average velocity? NITROGEN. At the same T, sice average kietic eergy is the same, more massive particles move more slowly. Xeo atoms have a average mass of about 3 amu, while itroge molecules oly have a mass of about 8 amu. (c) The molecules or atoms of which gas have the greater average kietic eergy? NEITHER. Same T same avg KE! PS7-

13 Do t mi up KE with velocity! The itroge molecules move more quickly at the same T because they are less massive, ot because they have more kietic eergy. I fact, we use the assumptio that average KE s are the same i order to reaso out that they must move more quickly! (d) If a small hole were opeed i the flask, which gas would effuse more quickly? NITROGEN. Effusio ad diffusio rates are both directly related to the speed of the particles. If you are movig more quickly, you will collide more frequetly. Thus, i the case of effusio, there will be more collisios with the hole each secod, which meas more particles per secod will escape (i.e., effuse ) A gas miture with a total pressure of 745 mmhg cotais each of the followig gases at the idicated partial pressures: CO, 5 mmhg; Ar, 4 mmhg; ad O, 87 mmhg. The miture also cotais helium gas. (a) What is the partial pressure of the helium gas? (b) What mass of helium gas is preset i a.0-l sample of this miture at 73 K? (a) Aswer: 9 mmhg Apply Dalto s Law: Ptotal sum of all Pgas s: PHe PHe 745 ( ) 9 mmhg (b) Aswer: 0.68 g He Strategy: ) Recogize that the T ad of the gas miture are also the T ad for each of its compoets (e.g., He). Thus, He ca be calculated from the ideal gas law with T,, ad PHe (after covertig mmhg to atm). ) Use moles ad molar mass of He (4.003 g/mol) to calculate mhe. He (Alteratively, you could use Ptotal,, ad T to calculate total. The use PHe Ptotal to total calculate He. That is slightly loger, but equally valid.) Eecutio: P RT He PHe RT 9 mmhg atm 760 mmhg Latm molk (73 K).0 L mol He mol He g mol He g He Coceptual Coectio 5.5 (o p. 3) Nitroge ad hydroge react to form ammoia accordig to the followig equatio: N + 3 H NH 3 Cosider the followig represetatios of the iitial miture of reactats ad the resultig miture after the reactio has bee allowed to react for some time (pictures ot show): If the volume is kept costat, ad othig is added to the reactio miture, what happes to the total pressure durig the course of the reactio? Aswer: The pressure decreases Reasoig: Do t be fooled here. It may iitially soud like everythig is costat (i.e., T,, ad [particularly because they say that othig was added!]), but if you actually look at the pictures ad cout up the molecules (remember PS ad PS? Make sure you kow the differece betwee a molecule ad a atom!), you ca see that there are fewer afterward. At the same T ad, P is proportioal to! So pressure decreases here. NOTE #: There is o law of coservatio of molecules i chemical reactios oly coservatio of atoms! PS7-3

14 NOTE #: You ca see from the balaced chemical equatio that as forward reactio occurs, the umber of moles of gas must go dow, because for every four molecules that get used up (dismatled), oly two molecules get made (assembled upo reparterig of the atoms) A 75 ml flask cotais pure helium at a pressure of 75 torr. A secod flask with a volume of 475 ml cotais pure argo at a pressure of 7 torr. If the two flasks are coected through a stopcock ad the stopcock is opeed, what are the partial pressures of each gas ad the total pressure? NOTE: You must assume that the temperature of the two gases is the same to solve the problem. I thik it is helpful to see a sketch of this sceario (usually roudbottom flasks are used for these): 475 ml Ar 7. torr stopcock 75 ml He 75 torr Aswers: PHe = 76 torr; PAr = 457 torr; Ptotal = 733 torr Strategy: ) The key to this problem (i my opiio) is recogizig that EACH GAS will spread out ito the FULL OLUME (of both flasks) oce the stopcock is opeed, so that for BOTH gases will equal = 750 ml after the stopcock is opeed. ) Thus, you ca use Boyle s Law (twice) to calculate PAr ad PHe after the stopcock is opeed. These are the partial pressures asked for! 3) Sum the partial pressures to obtai Ptotal (Dalto s Law) NOTE: You could solve this problem by assumig ay temperature you wat (just make it the same o both sides), ad the solvig the ideal gas equatio (twice) to get the moles of each gas before you ope the stopcock. The you could use total, total, ad T total to calculate P total usig the ideal gas equatio (a third time). The you could get the partial pressures usig mole fractios. But that would be a lot more work! There is aother simple way to do this problem, but it ivolves derivig a formula that I do ot believe is obvious, so I wo t provide it here (ask me if you wish). Eecutio of Strategy: For He: P For Ar: P P P P 75. torr P P P P P 7. torr Ptotal torr 75 ml torr 76 torr 750 ml 475 ml torr 457 torr 750 ml PS7-4