GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW

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1 GASES Pressure & Byle s Law Temperature & Charles s Law Avgadr s Law IDEAL GAS LAW PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2 Earth s atmsphere: 78% N 2 21% O 2 sme Ar, CO 2

2 Sme Cmmn Gasses Frmula Name Characteristics N 2 Nitrgen Inert O 2 Oxygen Explsin Hazard Life-sustaining HCN Hydrgen Very Txic Cyanide H 2 S Hydrgen Very Txic Sulfide Rtten Eggs CO Carbn Txic Mnxide Odrless CO 2 Carbn Plant-Fd Dixide Odrless CH 4 Methane Flamable Odrless N 2 O Nitrus Laughing Gas Oxide Sweet Odr

3 GAS MOLECULES CHARACTERISTICS mlecules are mving they are far apart (10 times as far apart as they are big) mve in straight lines cllisins with each ther cllisins with walls: pressure higher temp yields faster mtin lwer temp yield slwer mtin and eventually cndensatin

4 PRESSURE Pressure = frce per unit area = F / A 760 mm at sea level BAROMETER

5 UNDER CONSTANT PRESSURE 101,325 N/m lb/in 2 yur bdy has ~2 m 2 surface area ~200,000 N r ~45,000 lb pressing n yu right nw!

6 ORIGIN OF PRESSURE Kinetic Thery f Gasses N N N N N N N N O=O

7 UNDERSTANDING GASSES 3 fundatinal relatinships Byle s Law (P and V) Charles Law (V and T) Avgadr s Law (V and mles)

8 PRESSURE / BOYLE S S LAW PV = cnstant (fixed T,n) V 1 P 1 atm 2 atm 4 atm x L x 2 L x 4 L Vlume and pressure are inversely related (fr fixed T and n)

9 PRESSURE / BOYLE S S LAW PV = cnstant (fixed T,n) V 1 P Vlume and pressure are inversely related (fr fixed T and n)

10 TEMPERATURE CHARLES S S LAW V T = cnstant (fixed P, n) V T abslute temperature K = C

11 V T TEMPERATURE CHARLES S S LAW = cnstant (fixed P, n) V T abslute temperature K = C

12 AVOGADRO S S LAW V n = cnstant (fixed P, T) V n 2 mle 1 mle Fixed P and T

13 AVOGADRO S S LAW mlar vlumes nearly identical fr all gasses If n = 1, then V = L fr all gases at STP STP: 1 atm 0 C

14 IDEAL GAS LAW PV = nrt abslute temp K gas cnstant Units f R are imprtant R = L atm ml K R = J ml K

15 Simple prblems: Given 3 quantities, slve fr the 4th Prblem: What is the V f 2.35 mles f H 2 at 22.0 C and 700 trr? V = n R T P (2.35 ml)( L atm/ml K)(295 K) = atm 700 trr = 61.7 L 760 trr Mst prblems invlve a change in cnditins (P, V, T) f a gas

16 CHANGES IN P, V, T Given: Initial cnditins P, V, T Final cnditins: any tw Find: Value fr the third final value P i V i = n R T i P i V i = P f V f P f V f = n R T i T f T f

17 The gas in a 750 ml vessel at 105 atm and 27 C is expanded int a vessel f 54.5 L and 10 C. What is the final P? P 1 V 1 = P 2 V 2 T 1 T 2 V P 2 = P 1 T 2 1 V 2 T 1 = (105 atm) (0.750 L) ( K) (54.5 L) ( K) P 2 = 1.3 atm

18 GAS DENSITY AND MOLAR MASS Start with PV = nrt m but n = M m s PV = RT M m density is d = M rearrange t get m PM = RT V P d = M RT m = mass (g) M = mlar mass (g/ml) density mlar mass

19 Weigh a 1.00 L bulb f air t find air s average mlecular weight. V = 1.00 L T = 25 C = 298 K P = 760 trr = 1.00 atm Weight f air = 1.20 g RT 1.20 g M = d d = = 1.20 g/l P 1.00 L M = (1.20 g/l) ( L atm/ml K)(298 K) 1.00 atm M = 29.3 g/ml air is ~80% N 2 + ~20% O 2 (0.80)(28) + (0.20)(32) = 28.8 g/ml

20 PARTIAL PRESSURE PV = nrt P = RT V n what is this? Number f mles f gas What gas? Hw abut a mixture? n ttal = n 1 + n 2 + Daltn s Law: ttal pressure is the sum f partial pressures

21 PARTIAL PRESSURE If 5 ml CO 2, 2 ml N 2, 1 ml Cl 2 are mixed in a 40 L vessel at 0 C, what is P? nrt P = V P T = (n c2 + n N2 + n Cl2 ) RT V n RT CO2 V = P CO 2 Partial Pressure f CO 2 Daltn s Law: Ttal pressure is the sum f partial pressures P T = ( ) (0.0821)(273) 40 = 4.5 atm

22 EXAMPLE Cllect N 2 ver water Barmetric pressure = 742 trr Vlume = 55.7 ml Temp = 23 C Hw much N 2 is cllected? n = PV RT R = L atm/ml K T = ( ) = 296 K V = 55.7 ml = L P tw gases P H2 O at 23 C is 21 trr (frm Tables) P N2 = = 721 trr r 0.95 atm n = PV RT = (0.95)(0.0557) (0.0821)(296) = ml N ml x 28 g/ml = g r 62 mg N 2

23 KINETIC MOLECULAR THEORY PV = nrt explains hw gases behave Need mre t explain why Lk at gases n the mlecular level Five key pstulates f KMT: straight-line mtin, randm directin mlecules are small n intermlecular frces elastic cllisins mean kinetic energy T (in K) E k = ½ mv 2

24 KINETIC MOLECULAR THEORY S. Kinetic mlecular thery prvides understanding f why gases behave as they d Increase T at cnst. V P increase T increase, E increase, v increase, mre cllisins per unit time and harder cllisins, s P increases Increase V at cnst. T P decrease Cnst. T means cnst. E and cnst. v, lnger distances between cllisins, fewer cllisins per unit time with walls, s P decrease

25 Temperature and Mlecular Speeds N 2 at 0 C N 2 at 100 C v v Sme mlecules mve slwly Sme mlecules mve fast Mean speed in middle Entire curve (and mean) shift upwards fr higher temperature

26 Temperature and Kinetic Energy Average kinetic energy f a mlecule E = ½ m v 2 rt mean square speed mass E = ½ m v 2 = 3 RT 2 N N Avgadr s number At a given T, all gases have the same average kinetic energy, E v = 3RT M ½ M = mlar mass

27 Frm KMT: rms speed EFFUSION & DIFFUSION v = 3RT M mlar mass lighter gases have higher rms mlecular speed Graham s Law f effusin r 1 r 2 = M 2 M 1 r is rate M is mlar mass effusin = escape f gas thrugh pinhle diffusin = spread f ne gas thrugh anther Even thugh diffusin is much mre cmplicated (due t gas mlecular cllisins) it still beys Graham s Law well

28 EFFUSION RATES time N 2 : M = 28 v is smaller He: M = 4 v is larger r v 1 M r N 2 r He smaller larger

29 IDEAL GASES PV = nrt Ideal gas law Wrks best at lw P & high T Assumptins: n intermlecular interactins n mlecular vlume REAL GASES Deviatins frm ideal gas law Reasns: (1) Mlecules have finite size, they ccupy space (2) Mlecules have attractive frces that becme strnger when they are clse tgether

30 INTERMOLECULAR POTENTIAL repulsive ENERGY attractive d distance High pressure pushes gas mlecules clse tgether, s the attract each ther Very high pressure pushes gas mlecules even clser s they repel each ther

31 High Pressure at high P attractive frces lead t the appearance f a smaller n Example: CO 2 at 200 atm

32 Very High Pressure At very high P, finite mlecular vlume leads t repulsin and the appearance f a larger n Example: CO 2 at 800 atm

33 NON-IDEAL BEHAVIOR

34 Real Gas Fr a gas, measure P, V, T Yu find, at higher P that P is t small (due t attractin between mlecules) The amt f P missing is prprtinal t (1) size f attractive interactins (a) (2) freq f cllisins (n/v) 2 T cmpensate use: P + n 2 V 2 a a = cnstant At high P, V is t large due t excluded vlume The actual V = V measured V excluded S, use: V nb b = cnstant

35 Real Gas van der Waals Equatin adjustments relative t measurement ( P + n 2 V 2 a ) ( V nb) = nrt Crrectin fr attractive frces between gas mlecules Pressure crrectin Adjusts P Crrectin fr excluded vlume f gas mlecules Vlume crrectin Adjusts V

36 van der Waals Cnstants mlecular shape and interactins Substance a(l 2 -atm/ml 2 ) b(l/ml) He Xe H Cl CH CCl dispersin interactins increase with mlar mass CCl 4 is largest mlecule, has largest b value

37 Real Gas Example What is the P f 1.0 ml Cl 2 in 2.0 L at 273 K? Ideal Gas Law P = nrt V = (1)(0.0821)(273) 2 = 11.2 atm van der Waals P = ( P + n2 V 2 nrt V nb n2 a V 2 a ) ( V nb) = nrt a = 6.49 L2 atm ml 2 b = L/ml P = (1)(0.0821)(273) 2 (1)(0.0562) (1)(6.49) 2 2 P = = 9.9 atm The term cntaining a is mre imprtant % difference: x 100% = 12%

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