Chapter 13 Properties of Gases Many chemical insights are gained by studying the simplest chemical phase, gases

Transcription

1 Chapter 13 Prperties f Gases Many chemical insights are gained by studying the simplest chemical phase, gases 13-1 Gases Slid - fixed vlume and shape Liquid - fixed vlume but shape can change Gas - bth vlume and shape can change Will explre the ther phases in chapter 15 A picture is wrth a 100 wrds. Figure 13.1 vlume f slids and liquids fixed because atms/mlecules in cntact with each ther (cndensed state) Slids further fixed int array s cannt mve, that is why shape fixed Liquids cndensed, but nt fixed s atms/mlecules can mve, that is why vlume nt fixed Gases mstly empty space, mlecules nt tuching s free t mve as well as vlume can change 13- Measurement f Pressure & 13-3 Atmspheric Pressure mixed tgether Gases are in cnstant mtin and the mlecules are clliding with each ther as well as the wall f the cntainer they are in. The cllective frce f these mlecules hitting a surface is respnsible fr the pressure f a gas. Strictly speaking pressure is defined ad P = F/A where P is pressure, F is frce, A is area In SI units frce is measured in Newtns (1N 1 kg m/s ) Rughly the frce exerted by gravity n an apple at earth s surface Area wuld be in m The SI unit f pressure is a Pascal (Pa) = 1 Newtn/m This is a very small frce spread ut ver a very large area, s it is a very small unit. It is s small that we usually dn t use it, and instead use the unit bar where 1 bar = 10,000 Pa The air arund us is a gas, and that gas exerts a pressure n every chemical experiment we d in the lab, s it is an imprtant pressure t knw. First, hw d we measure it? This frce can be easily measured with a barmeter. At its simplest, a barmeter is a tube filled with mercury that is turned upside dwn. When the

2 tube is turned upside dwn mst peple think intuitively that it is the vacuum in the clsed end that hlds the mercury in the tube. If fact yu have it bassackwards. It is the pressure f the air n the bttm end f the tube that is pushing t keep the mercury in the tube. The height f the clumn f mercury supprted by the pressure f a gas is directly prprtinal t the pressure f the gas pushing n the bttm f the tube. Thus ne f the mst cmmn units used t express pressure is related t hw high a clumn f mercury it supprts. S yu will see pressure expressed as mm f Hg, referring t the height f the mercury clumn. Ntice this is nt a measure f frce/area, instead it is an easily measured number that is prprtinal t pressure. Anther unit yu will see is Trr, named after Trricelli, the man wh invented the barmeter. 1 mm f Hg = 1 Trr Key cncept: S what is the pressure f the atmsphere arund us? At sea level, n a clear day it is 760 mm Hg r 760 Trr r Pa r bar r 14.7 psi r 1 Atm. Example Prblem 1. I have a vacuum system that measures pressures dwn t.05 mbar. Express these pressures in bar, mm Hg, Trr, atm and psi -3-3,05 mbar x 1x10 bar =.05x10 bar 1 mbar -3,05 mbar x 1x10 bar x 1 Atm x 760 mm Hg =.0188 mm Hg 1 mbar bar 1 Atm -3,05 mbar x 1x10 bar x 1 Atm x 760 Trr =.0188 Trr 1 mbar bar 1 Atm -3-5,05 mbar x 1x10 bar x 1 Atm =.47x10 Atm 1 mbar bar,05 mbar x 1x10 bar x 1 Atm x 14.7 psi = 3.63x10 psi mbar bar 1 Atm

3 Prblem The pressure at the center f the earth is estimated t be 350 Gpa. Express this pressure in bar, mm Hg, Trr, atm and psi Gpa x 1x10 Pa x 1 bar = 3.50x10 bar 1 Gpa 10,000 Pa Gpa x 1x10 Pa x 1 bar x 1 Atm x 760 mm Hg =.63x10 mmhg 1 Gpa 10,000 Pa bar 1 Atm Gpa x 1x10 Pa x 1 bar x 1 Atm x 760 Trr =.63x10 Trr 1 Gpa 10,000 Pa bar 1 Atm Gpa x 1x10 Pa x 1 bar x 1 Atm = 3.4x10 atm 1 Gpa 10,000 Pa bar Gpa x 1x10 Pa x 1 bar x 1 Atm X 14.7 psi = 5.08x10 psi 1 Gpa 10,000 Pa bar 1 atm 5 =.54x10 tns/sq in 13-4 Byle s and Charles s Laws The first prperty f gases was discvered by Byle in the 1660's What he fund was that when the temperature is cnstant, when V, P That is, pressure and vlume are inversely prprtinal Draw figure like 13.8 n bard Mathematically this can be expressed as V 1/P r V =c/p where c is a cnstant Ging t ur picture f a gas cmpsed f particles buncing arund, and P cming frm them buncing ff a surface Draw figure like 13.9 n bard Makes sense, as vlume increases the # f times a mlecule hits the wall t P Rughly 100 years later Jacques Charles uncvered the relatinship that as T,P r that T and P were directly prprtinal r that V T Sketch a direct relatinship n bard Nt nly that, but when he wrked at different pressures r with different gases all the line came t the same spt Figure 13.11

4 4 S with this series f experiment he discvered that all gases have zer vlume at C. Why? Because at that temperature the gas mlecules stp mving s they can exert pressure. He had uncvered abslute zer 13-5 Avgadr s Law The final key t understanding gases, and indeed t understanding chemical and mlecules came in the early 1800 When Gay-Lussac discvered that there was a direct relatinship between vlumes f gas reactants and prducts This, in turn was interpreted by Avgadr s say that any gas at a specific vlume and temperature wuld cntain the same number f mlecules, and that V n 13-6 Ideal-Gas Equatin S at this pint we have V 1/P V T V n Cmbining this tgether we have V nt/p Putting in a prprtinality cnstant V = K nt/p and multiplying bth sides by P we have PV = K nt and giving K a special name and symbl: the mlar gas cnstant R, we have the ideal gas law PV =nrt Key Equatin: PV=nRT and R =.0806 L atm/k ml ( R has ther values when P is measured in ther units) Example prblems A. Full blwn Ideal gas prblem... given any 3 f P,V,n,r T, calculate the missing variable. One f thse big gas cylinders cntains abut 50L f gas. IF it was filled t

5 ,500 psi with N gas at 5 C, hw much gas (in mle and grams and lbs) is in the tank 5 First ntice that this prblem has P, V and T defined s we must be lking fr n, the number f mles f gas PV=nRT Let s rearrange the questin t get n alne n ne side n = PV/RT Befre we fill in the blanks, what are the units f R.0806 L atm/k ml s V must be in L P in atm T in K S we have sme cnversins t d befre we fill in the blanks P = 500 psi x 1atm = 170 atm 14.7psi T = 5 C = 98.15K n = (170 atm 50L)/ (.0806 L atm/k ml 98K) = 347 mles 347 mle x (14 g/ml) = 973g 973 g x.046 lbs/1000g = 1 lbs f N gas B. Prblems where 1 variable changes (and sme f the ther variables may nt be given!) B.1 I fill a balln with 1 L f helium fr my daughter s birthday n a day that is raining because with lw pressure frnt just mved thrugh s the air pressure is mm f mercury. What is the vlume f the balln tw days later when a high pressure frnt has mved thrugh and its nice and bright and sunny, and the air pressure is mm f Hg. Ntice the set up, nly tw f the variables are given P and V. What d we d nw? We assume the ther variables are cnstant and dn t change. (S T is the same bth days and n the # f mles f He is the same - nthing leaked ut f the balln) IN this case put all the variables n ne side f the equatin, and all the cnstants n the ther

6 6 Variables Cnstants PV = nrt Cnstants are cnstants are the same bth days s P*V = a cnstant s PV(day 1)=PV(day) The starting vlume was 1 l 1l x 736.6mm = V x 774.7mm V = V 1 X 736.6/774.7 =.951 l, s my ballns are 95% f their riginal size - they have shrunk 5%! Ntice I did NOT change pressure t atm. Why nt? Didn t use R t slve the prblem Had same units n bth sides f equatin s canceled ut! S it is easier t slve this way than a full blwn PV=nRT B. Let s cntinue with my daughter s ballns. If a balln cntains 1 liter f helium when it is filled in ur huse at 65 F (91.5K). I put it in the car, in the sun, and the temperature gets up t 100 F (310.9 K). What is its vlume nw? PV=nRT Variables : V,T Cnstants: n,p Rearranging Variables V/T Cnstants = nr/p V 1/T 1 = V /T Whenever yu use T in a gas prblem, yu ALWAYS have t cnvert t K V 1=1 Liter, T 1 = 65 F =18.33 C = 91.48K), the ther temp is V = X, T = 100 F = C = K 1 1 V T = b = V /T and 1/91.48 = X/310.93, X=1.07 L my balln is anther 7% larger.

7 B.3 Suppse we have 1. L f xygen cntaining 0.5 ml f O at 11 atm and 5 C. IF we cnverted this t zne (O 3) what is the vlume f the zne (at same P and T)? st 1 what is the balanced reactin? 3O O3 Is this a redx reactin? (N, elemental O xidatin state 0) Variables: n,v Cnstants T,P PV=nRT Variables Cnstants V/n = RT/P V 1/n 1 = V /n We have V and n What is n? 1 1 Mlar cnversin using sthicimetric cefficeints.5 mle O x ml O =.333 ml O 3 = n 3 ml O 3 1. L /.5 mle = X/.333 ml; X = 8.13L Des this make sense? We have a smaller number f mles, s the vlume shuld have decreased and it did. I think 4 examples are enugh Clicker questins Set up a wrd prblem, identify the equatin they need t slve it r set up sme 3 f 4 prblems with an errr 13-7 STP and Determinatin f Mlar Masses (I am deviating a little bit frm the text) STP If we have 1 mle f an ideal gas at RT and 1 atm what is its vlume? PV=nRT V = nrt/p =1(.0806)(73.)/1.00 =.4 L.4L is a vlume abut 8x8x8 cm 7

8 The vlume.4 L is called the mlar vlume Nte the cnditins f 1 atm and 73.K are called standard temperature and pressure (STP), and this is a cmmn reference fr gases. Let s check if sme f ur cmmn gases are ideal gases at STP Gas Mlar Vlume (L) O.397 N.40 H.433 He.434 All abve ideal NH3.079 Nt ideal, abut % lw 8 This cnfirms Avgdr s hypthesis that a mle f each gas shuld ccupy the same vlume. Key Cncept: 1 mle f an ideal gas has a vlume STP Example calculatin: What is the vlume f 6 mles f H gas at STP The lng way - use PV=nRT STP = 73.15K & 1 atm n=6 PV=nRT V = nrt/p = /1 = 134 L Quick and Dirty use 1 STP, 1 mle.4 L 6 ml x.4l = 134L 1 ml Density and Mlecular mass One very imprtant feature f the ideal gas law is that it helps yu t see te clse relatinship between a gas's density and it mlecular weight. First, d yu remember what density is? It is the mass per vlume r: density=mass/vlume I just shwed yu that 1 mle f all gases, n matter what their mlecular weight

9 9 will have the same vlume Mlar vlume =.4 L If yu think abut it, the mlecular mass f 1 mle f a gas depends n its chemical cmpsitin, thus H hydrgen gas has a mlecular weight f.016 grams s its density is g/.4 L, while CO, a gas used t put ut fires, has a mlecular mass f r 44.01, and its density is 44.01/.4 L. Let's see if we can use PV=nRT t make an equatin abut density. Let s start with the term n, number f mles n=mass/mlar mass Therefre PV=(mass/mlar mass) RT Our equatin fr density is: density = mass/vl S hw d we rearrange the equatin t lk like this? We already have mass n the right-hand side f the equatin, s we need t divide it by vlume and then mve mlar mass, R and T t the left-hand side Thus we get P x Mlar mass/rt = mass/vl = density -r- Mlar mass = density (RT/P) Key Equatins: Mlar mass = density (RT/P) density f a gas = mlar mass x (P/RT) Let's try ne example using this equatin Yu may have heard (Or d as a dem?) that when yu have yur lungs full f helium yu talk like Dnald Duck. This is because He has a different density than the N /O mix f ur atmsphere, s yur vcal crds vibrate differently and the sund waves are transmitted differently when the gas has a different density. Let s calculate the density f He and N and cmpare them. Atmic mass f He= Mlar Mass = density (RT/P); Mlar Mass X (P/RT) = density at STP P=1, T=73. d=mlar mass*1/.0806*73.=0.18 g/l at STP (gram/mle) *(atm) * Kml/Latm)* 1/K

10 10 What is the density f N? 8.01/(.0806*73.) = 1.5 g/l Nw can yu tell be why they use He t fill ballns? And CO t put ut fires? 13-8 Partial Pressures and Mle Fractin Partial Pressure Nw what happens if we have a mixture f gases? If yu calculate the vlume f an ideal gas, and cmpare it t the vlume f the mlecules in the gas, we will find that the gas mlecules ccupy a very small part (<.%) f the vlume Fr myself, class des nt have t d: Vlume f N mlecule 65 A, Vlume f 1 mle f N = 65 x 6.0x10 = 3.9x10 A A = 1x10 cm s 1A = 1x10 cm = 1x10 ml x10 x 1x10 = 39 ml Vlume f 1 mle f N Gas =.4 L = 40mL S 39/40 =.0017 r.17% f gas vlume is ccupied by actual gas mlecule! Bttm line. When yu mix gases there is s much space, the gas mlecules ignre each ther s yu can treat each gas in the mixture independently! S, if yu have a mixture f tw gases Key Equatins: P gas 1 = n1rt/v P gas = nrt/v and the pressure f the mixture = P + P 1 We call the pressure f each individual gas the partial pressure. Mle Fractin Mle fractin is anther way t measure the cncentratin f a material in a unifrm mixture like a mixture f gases. The Definitin f mle fractin is: Key Equatin: Mle fractin ( ) = n /(n + n + n...) = n /n ttal

11 11 Ntice that as a fractin the sum f the fractins in a whle =1 s in a mixture =1. 1 I talk abut mle fractin right nw because there is an very useful relatinship between mle fractin and partial pressures f gases Ging back t ur equatin n partial pressures P ttal = P 1 + P +... = n1rt/v + nrt/v...= (n 1 + n +...) RT/V =nttalrt/v Let s lk at a single gas, P 1. What is P 1/ P ttal? =( n1rt/v ) / (n1rt/v + nrt/v...) But all the RT/V s drp ut s we have P 1/P ttal = n 1/n ttal = mle fractin f gas 1! Key equatin: P /P = n /n = mle fractin 1 ttal 1 ttal Practice prblems invlving partial pressures and mle fractins: A typical prblem starts with figure ml 950 Trr O 500 ml 700 Trr N Part 1. What is the ttal pressure inside the system when yu pen the stpcck? Treating each gas separately If yu have 400 ml f O at 950 Trr and yu let it expand int 500 ml f additinal space PV=nRT Variables P,V, cnstant n,r,t PV = nrt =PV P1V 1 = PV P1V 1/V = P V = the new TOTAL vlume = = 900 ml (950 Trr x 400 ml)/(900ml) P = 4. Trr O If yu have 500 ml f N at 700 Trr and yu let it expand int 400 ml f additinal space PV=nRT Variables P,V, cnstant n,r,t

12 1 PV = nrt =PV P1V 1 = PV P1V 1/V = P V = the new TOTAL vlume = = 900 ml (700 Trr x 500 ml)/(900ml) P = 388.9Trr N P ttal = P O + N N = = Trr Part. What is the mle fractin f each gas in the mixture? = P /P = 4./811.1 =.5 O O ttal N = P N/P ttal = 388.9/811.1 =.48 Quick check des = 1? Yes! S n bvius errr Gases prduced in the lab ver water The bk ges n t talk abut the fact that if yu make gas in the lab, yu usually t it ver water, s the gas yu create has sme water vapr in it that yu have t accunt fr in yur calculatins. Since we wn t d a lab like that, let s skip it! 13-9 Maxwell-Bltzmann Distributin Thrughut this chapter we have talked a gas as being a bunch f particles wizzing arund in a large space. Befre we develp this int a cmplete thery we need t think a little bit abut the speed f the gas mlecules. We can measure hw fast a gas particle mves. We can actually measure the speed f gas mlecules using a set up like the ne shwn in figure 13. Just like all the cars n the freeway dn t all g the same speed, all the mlecules in a gas dn t all g the same speed. Instead there is a distributin f speeds. This distributin, called a Maxwell-Bltzmann distributin, is shwn in figure 13.3 Ntice tw things. First, the distributin changes with T, and secnd it is nt a nice symmetric distributin (Sme f yu may have had statistics, and wrked with a Gaussian distributin that is nice and symmetric) In the next sectin we will talk abut the Kinetic Energy f the gas mlecules. Hw d yu calculate kinetic energy? ½ mv

13 I just shwed yu that gas mlecules have a range f v s. Which V shuld yu use? What d I d when I have a range f grades n a test? I use a mean r an average. S that is what we will d in the next sectin, talk abut the mean (average) KE and velcity f a distributin f gas mlecules Kinetic Thery and Rt-Mean-Square Speed S let s put all the pieces tgether and make a cmprehensive thery that describes gases. This thery is called the kinetic thery f gases Key Thery: Kinetic thery f Gases 1. The mlecules f a gas are in cnstant mtin. They randmly cllide with each ther and with the walls f the cntainer. (The randm cllisins with the cntainer is what we measure as pressure). All cllisins are elastic, that is n energy is lst due t heat r frictin 3. The average distance between the mlecules is much larger than the actual size f the mlecules 4. Any interactins between particles are negligible. That is there is n net attractin r repulsin between particles 5. The mean (average) kinetic energy f the mlecules is directly prprtinal t the K temperature f the gas The last statement is very imprtant. It says that K.E. T Remembering that we have a range f kinetic energies we have: average kinetic energy = 3/ RT nw KE = ½ mv s if we knw an average KE, we can als figure ut an average velcity! Cutting ut the math intermediates Key Equatin: 13

14 14 Where v rms is the rt mean square average velcity. What des rt mean square actually mean? Well we have a distributin f velcities right? Nw we square each velcity and get a new distributin f v Next we get the mean (average) f the v distributin Finally we take the square rt f (the average f the v distributin) R is ur gas cnstant, but here we are dealing with energy s we use the gas cnstant f J/K ml M kg is the mlar mass f the mlecule IN KG (needs t match units in R) Sample calculatin u rms =(3RT/M kg) 1/ M N =8.0 g =.0801 kg T=abslute = =98. Plug and chug sqrt[3(8.3145kg (m/s) /K ml)98. K/.081 kg/ml] = m/s m/s = abut 560 yd/sec =.3 mile/sec = 19 mi/min=1,145 mph!!!!! (Yu might cmpare this t the sped f a helium atm at 5 C. Helium is much lighter (mm=4) s shuld it g faster r slwer?) Anther questin - If a gas mlecule is ging 1000 mph. Hw cme if smebdy breaks a bttle f a smelly substance in the lab, it may take a minute r tw befre everybdy is gassed ut? The answer is t remember that there are lts f gas mlecules in the air all buncing int each ther. Remember, What is mlar vlume at STP?.4 L 3 and hw many mlecules f air are there? 6.0x10 s that is 6.0/.4 r.7 x10 mlecules/liter. That s a lt f chances fr cllisins. In -7 fact in a typical gas at STP an atm will nly g 1x10 M befre a cllisin (0.1 ìm) S the mlecules are clliding with each ther frequently and in each cllisin the atms change bth directin and speed sp it takes a lng time t get acrss the rm Graham s law f Effusin & 13-1 Mean Free Path Fun and interesting calculatins, we can figure ut that ur N STP actually ges abut 63 nm befre its first cllisin -see example fr details 9 and that it cllides abut 8.1 x10 times /secnd! But nthing I want yu t memrize, s let s skip these sectins

15 13-1 Van der Waals Equatin An ideal gas is ne that fllws the ideal gas law, PV=nRT Mst gases are ideal lw pressures, say belw 10 atm but what happens then? Figure First, what is the Y axis? PV/RT Frm the ideal gas law PV=nRT PV/RT =n in this plt n=1, s if yu are an ideal gas yu fllw the dashed line Yu can see that all the gases, even He are NOT ideal at high pressure. The deviatins frm the ideal gas can be traced back t a cuple f pstulates in ur kinetic thery f gases Pstulate 3. The average distance between the mlecules is much larger than the actual size f the mlecules At high pressures, we have s many gas mlecules crammed int a small space that this pstulate is n lnger true. T crrect fr this we have t subtract the vlume f the gas mlecules frm V. S the V crrected = V-nb where V = ur riginal V n= # f mles b = is the vlume f a mle f gas If ur ideal gas law is PV=nRT we can thrw in the abve crrectin fr V and get P(V-nb) =nrt th The next prblem with ur kinetic thery is in the 4 pstulate; Any interactins between particles are negligible. That is there is n net attractin r repulsin between particles There can be attractins between particles. If there is an attractin, this makes the mlecules stick tgether a little bit, and makes the pressure lwer s P crrected = P - a(n/v) Where a is a empirical attractin factr. Why (n/v)? n/v wuld be the cncentratin f ne particle. Since this is an interactin between tw mlecules we have t deal with the cncentratin f bth mlecules s we have an (n/v) term 15

16 16 These crrectins were frmulated by Jhannes van der Waals and put int a single equatin called the vander Waals equatin Key equatin: T use this equatin yu need the tw crrectin factrs, a & b This is nt an equatin I expect yu t memrize. Typical test questins I will ask n this sectin are: Given the abve equatin, and cnstants a & b, calculate a crrected pressure -r- Given the abve equatin explain what all the symbls mean -r- Explain what is wrng with the kinetic mlecular gas thery as riginally written and explain hw the abve equatin crrects these issues