Uit 5. Gases (Aswers) Upo successful completio of this uit, the studets should be able to: 5. Describe what is meat by gas pressure.. The ca had a small amout of water o the bottom to begi with. Upo heatig the water vaporized ad displaced the air i the ca, leavig mostly steam iside the ca. Whe the ca was capped ad the immersed i the cool water bath, much of the steam iside the ca codesed. This left a partial vacuum iside the ca (i.e. very low pressure). The atmospheric pressure the crushed the ca because it was ow much greater tha the pressure iside the ca. 5. Idetify, memorize coversio factors, ad kow how to itercovert betwee the followig commo uits of pressure: atm, mm Hg, ad torr.. atm? atm 645 mm Hg 0.49 atm 5. Idetify ad, if give the coversio factors, kow how to itercovert betwee the followig commo uits of pressure: a, i Hg, ad psi.. mm Hg 0,5 a? a.05 m Hg 0 m Hg -.7 0 5 a.? mm Hg 5.0 psi 776 mm Hg 4.7 psi 5.4 State Boyle s, Charles s, ad Avogadro s Laws.. At costat temperature ad moles, the volume of a gas is iversely proportioal to its pressure. V k (where k is a costat) 5.5 Solve qualitative ad quatitative problems that ivolve the applicatio of Boyle s, Charles s, ad Avogadro s Laws (Avogadro s law should be uderstood as stated i Sectio 5. for a gas at costat temperature ad pressure, the volume is directly proportioal to the umber moles of gas ad as stated i Sectio. at the same temperature ad pressure, equal volumes of differet gases cotai the same umber of molecules ).
. decrease..9 atm.50 atm V.90 cm V? T o C + 7 K T 5 o C +7 K V V VT (.9 atm)(.90 cm )( K) T T V.9 cm T (.50 atm)( K). No. Boyle s law says that the pressure ad volume of a gas are iversely proportioal at costat moles ad temperature. Whe you put air i your tire it is likely the temperature is stayig relatively costat, but the moles of gas is icreasig cosiderably. 4. V V.00 mol? T 5 o C + 7 9 K T 5 o C +7 50 K V T V T (.00 mol)(9 K) 4.69 mol T T (50 K).00 4.69. mol of O eeds to be released 5. b 6. Oe versio of Avogadro s Law (see page 4 i tet) states that equal volumes of differet gases uder the same coditios cotais equal umbers of moles of gas. Therefore, if we ca calculate the umber of moles i oe gram of helium, we will kow how may moles of the diatomic gas we have. mol He? mol He.0 g He 0.5 mol He 4.00 g He Therefore there must also be 0.5 mol of the diatomic gas. 7 g 0.5 mol g mol So the diatomic gas must be N (4.0 )
7. Avogadro s Law. (mm Hg) V (L) Other uits of pressure ad volume are also acceptable 9. 75 torr 500 torr V V T o C + 7 94 K T? V T V T (500 torr)(94 K) T 969 K T (75 torr) 5.6 Eplai the differece betwee a ideal gas ad a real gas.. Ideal gases are those which have o iterparticle iteractios ad zero volume for the idividual gas molecules. No real gases are ideal gases. Some gases ca approach ideal behavior at high temperatures ad low pressures (these are the coditios which favor miimal particle iteractio) but ay real gas is ever truly a ideal gas. Real gases do have at least some iterparticle iteractios ad certaily each particle (i.e. molecule) has to occupy some volume.. Real gases have particles (i.e. the molecules) which occupy volume ad have iterparticle iteractios. Ideal gases are assumed to have either of these. Real gases behave most ideally at high temperatures ad low pressures. 5.7 State the values for stadard temperature ad pressure ad the molar volume of a gas at ST...4 L
5. Solve problems ivolvig the ideal gas law (V ).. Assume eactly g of gas. mol UF6 g UF6 0.004 mol UF 5.0 g UF6 0.90 atm V? 0.004 mol T 60.0 o C + 7 K (0.004 mol)(0.006 )( K) V V 0.079 L 0.90 atm 6 m g D V 0.079 L g.6 L. a) 56mm Hg - 0 L V 5 ml 0.5 L ml o T.0 C + 7 96 K? atm 0.7 atm V V (0.7 atm)(0.5 L) (0.006 )(96 K) 0.000 mol Molar Mass 0.05 g 0.000 mol 7.6 g mol b) m 0.05 g D 0.40 V 0.5 L g L
. Assume eactly g of fluorie gas. mol F g F.00 g F 0.06 mol F atm V? 0.06 mol T 0 o C + 7 7 K (0.06 mol)(0.006 )(7 K) V V 0.59 L.00 atm m g D V 0.59 L g.70 L OR Aother way of doig this problem is to realize that the coditios are ST. Therefore we kow right away that there are.4 L/mol of F. mol.00 g.4 L mol.70 g L 4. 50.0 g He? V 50 L T 5 o.5 mol mol He 4.00 g He C + 7 9 K.5 mol He V (.5 mol)(0.006 )(9 K) 0.7 atm V 50 L
5. ) Fid Molar Mass of S F y. mm Hg 0 L V 9 ml 0.09 L ml? T 45 o - C + 7 atm o C 0.0 atm V (0.0 atm)(0.09 L) V.75 0 (0.006 )( K) 0.0955 g g Molar Mass 54-4.75 0 mol mol ) Fid the Empirical Formula of S F y Assume 00 g total. 00 5. 74.77 g F 5. g S 74.77 g F.95 mol F 0.767 mol S mol S.07 g S mol F 9.00 g F 5 mol F mol S 0.767 mol S.95 mol F Therefore empirical formula is SF5 Fid formula mass of empirical formula, the divide that ito molecular mass (which is the same umber as the molar mass) S (.07).07 F 5(9.00) 95.00 7.97-4 mol 54 7.97 (SF 5) SF0
5.9 erform stoichiometric calculatios that ivolve gaseous reactats ad/or products.. a. Ca(s) + H O(l) H (g) + Ca(OH) (aq) b..05 atm V 7.00 L T 0 o C + 7 0 K? V (.05 atm)(7.00 L) V 0.96 mol (0.006 )(0 K) H mol H O.0 g H O cm H O ml H O? ml H O 0.96 mol H 0.7 ml H O mol H mol H O.00 g H O cm H O. mol S mol H? mol H.7 g S mol S.0 g S 0.06 mol H 70 torr V? T 7.0 0.06 mol o atm 760 torr C + 7 00 K 0.94 atm (0.06 mol)(0.006 )(00 K) V V 0.6 L 0.94 atm. C H 6 + 7O 4CO + 6H O For C H 6 : V (.4 atm)(0.56 L) V 0.0 mol C H 6 (0.006 )(00 K)
For O : V (.05 atm)(.65 L) V 0.0704 mol (0.006 )(00 K) Therefore we HAVE 0.0 mol C H 6 ad 0.0704 mol O. O 7 mol O 0.0 mol C H 6 0.099 mol O mol CH 6 eed mol C H 0.0704 mol O 6 0.00 mol CH 6 7 mol O Have Need C H 6 0.0 0.00 O 0.0704 0.099 Have < Need therefore O is limitig 4 mol CO? mol CO 0.0704 mol O 0.040 mol CO 7 mol O (0.040 mol)(0.006 )( K) V V 0.694 L.56 atm ml 0.694 L 0 L - 694 ml 5.0 State Dalto s Law of artial ressures, state the meaig of mole fractio, ad solve related problems.. - 0 m Hg mm Hg 44.0 cm Hg - cm Hg 0 m Hg 00 mm Hg atm mm Hg. m Hg - 0 m Hg.05 atm atm atm.9 atm.7 atm tot.9 +.05 +.7 4.65 atm
. You ca solve this by either addig the umber of moles together ad the usig the total moles i V (ad solvig for tot ) or you ca solve V (for pressure) for each gas ad the add the idividual partial pressures. The former method is used here. 0.0 mol Ar + 0.06 mol N 0.0494 mol total V (0.0494 mol)(0.006 )( K) tot V 0.00 L tot tot tot.5 atm. a. tot O + H O O tot H O 755 mm Hg 6.7 mm Hg 7 mm Hg b. atm 7 mm Hg 0.957 atm V (0.957 atm)(0.55 L) V 0.0 mol O (0.006 )(00 K).00 g O 0.0 mol O 0.44 g O mol O 4. O XOtot 0.0(747 mm Hg) 57 mm Hg 57 torr 5. First, fid how may moles of Ar would be eeded to fill the cylider. 50 atm V 5.4 L? T 0 + 7 9 V X V (50 atm)(5.4 L) mol Ar (0.006 )(9 K) Ar mol Ar 4.7 0 X mol Ar Ar 0.0094 mol total Ar Ar tot tot mol total
V V (.7 0 4 mol)(0.006 )(7 K) 5. 0 atm 5 L 5. Describe the differece betwee effusio ad diffusio ad solve qualitative problems related to these cocepts.. Effusio is the term used to describe the passage of a gas through a tiy orifice ito a EVACUATED chamber (see Fig. 5. i tet). Diffusio, o the other had, is the term used to describe the rate of MIXING of DIFFERENT gases.. The precipitate of NH 4 Cl would form closer to the HCl ed. This is because the velocity of the NH molecules would be faster tha the HCl molecules due to their smaller molecular mass. See Fig. 5.4 i tet. 5. Use the kietic-molecular theory to eplai the various gas laws. As the temperature is raised, the speeds of the particles icreases, thereby hittig the walls of the cotaier with greater force ad greater frequecy. Assumig the volume was kept costat, this would result i greater pressure.. (ay three below) ) The volume of the idividual particles ca be assumed to be egligible. ) The particles are i costat motio ad the collisios of the particles with the walls of the cotaier are the cause of the pressure eerted by the gas. ) The particles are assumed to ot iteract with each other i ay way. 4) The average kietic eergy of a collectio of gas particles is assumed to be directly proportioal to the Kelvi Temperature of the gas. Additioal Uit 5 Sample Questios:. The absolute easiest way to solve this problem is to remember oe of the versios of Avogadro s Law (Equal volumes of differet gases cotai equal umbers of moles uder the same coditios). Therefore, it follows that the volume ratio which reacts will be the same as the mole ratio which reacts. So the fact that we are ot give the coditios (temperature ad pressure) does t matter, because the coditios for all gases are the same. We will eed to write a balaced equatio first ad the determie the limitig reactat.
C H (g) + 5O ( g) CO (g) + 4H O(g) Have.00 L C.00 L O H 5 L O L C H L CH 5 L O 0.0 L O 0.600 L C H Need Sice have < eed for O, O is the limitig reactat..00 L O.00 L O L CO 5 L O.0 L CO 4 L HO.40 L H 5 L O O To prove that the above result is valid, you ca also solve this problem by assumig certai coditios. It does ot matter what coditios you assume as log as you carry the same values throughout the etire problem. To make thigs easiest, we will assume we have ST coditios (T 7 K, atm), this way we kow that the molar volume of every gas ivolved is.4 L/mol. We could also assume other coditios but the we would eed to use the ideal gas law, V. Have: mol CH.00 L CH 0.09 mol CH.4 L mol O.00 L O 0.4 mol O.4 L Need : 0.4 mol O 0.09 mol C H mol C H 5 mol O 5 mol O mol C H 0.06 mol C H 0.447 mol O For O, have < eed, therefore O is limitig. 0.4 mol O 0.4 mol O mol CO 5 mol O.4 L CO mol CO.0 L CO 4 mol H O.4 L H O.40 L H 5 mol O mol H O O
. a. Yes. Avogadro s law tells us that equal volumes of differet gases cotai equal umbers of moles (i.e. equal umbers of molecules) uder the same coditios. Sice every tire has the same volume ad the same temperature ad pressure, each tire has a equal umber of molecules. b. 0 times as heavy. Sice both gases cotai equal umbers of molecules, we ca simply divide 60 g ukow/6.0 g He. c. All tires would have the same average kietic eergy because they are all at the same temperature. d. The molecules i the helium tire would have the greatest average speed because they are the lightest molecules.. We have 0.50 mol N ad 0.50 mol H. Need: 0.50 mol N 0.50 mol H mol H mol N mol N mol H.5 mol H 0.7 mol N Sice have < eed for H, H is the limitig reactat. mol NH 0.50 mol H 0. mol NH mol H (0. mol)(0.006 )(4 K) V V L NH 0.0 atm