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
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
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
PRESSURE Pressure = frce per unit area = F / A 760 mm at sea level BAROMETER
UNDER CONSTANT PRESSURE 101,325 N/m 2 14.7 lb/in 2 yur bdy has ~2 m 2 surface area ~200,000 N r ~45,000 lb pressing n yu right nw!
ORIGIN OF PRESSURE Kinetic Thery f Gasses N N N N N N N N O=O
UNDERSTANDING GASSES 3 fundatinal relatinships Byle s Law (P and V) Charles Law (V and T) Avgadr s Law (V and mles)
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)
PRESSURE / BOYLE S S LAW PV = cnstant (fixed T,n) V 1 P Vlume and pressure are inversely related (fr fixed T and n)
TEMPERATURE CHARLES S S LAW V T = cnstant (fixed P, n) V T abslute temperature K = C + 273.15
V T TEMPERATURE CHARLES S S LAW = cnstant (fixed P, n) V T abslute temperature K = C + 273.15
AVOGADRO S S LAW V n = cnstant (fixed P, T) V n 2 mle 1 mle Fixed P and T
AVOGADRO S S LAW mlar vlumes nearly identical fr all gasses If n = 1, then V = 22.41 L fr all gases at STP STP: 1 atm 0 C
IDEAL GAS LAW PV = nrt abslute temp K gas cnstant Units f R are imprtant R = 0.08206 L atm ml K R = 8.314 J ml K
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)(0.0820 L atm/ml K)(295 K) = 0.921 atm 700 trr = 61.7 L 760 trr Mst prblems invlve a change in cnditins (P, V, T) f a gas
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
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) (273 10 K) (54.5 L) (273 + 27 K) P 2 = 1.3 atm
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
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) (0.0821 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
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
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 = (5 + 2 + 1) (0.0821)(273) 40 = 4.5 atm
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 = 0.0821 L atm/ml K T = (273 + 23) = 296 K V = 55.7 ml = 0.0557 L P tw gases P H2 O at 23 C is 21 trr (frm Tables) P N2 = 742 21 = 721 trr r 0.95 atm n = PV RT = (0.95)(0.0557) (0.0821)(296) = 0.0022 ml N 2 0.0022 ml x 28 g/ml = 0.0616 g r 62 mg N 2
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
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
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
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
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
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
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
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
High Pressure at high P attractive frces lead t the appearance f a smaller n Example: CO 2 at 200 atm
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
NON-IDEAL BEHAVIOR
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
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
van der Waals Cnstants mlecular shape and interactins Substance a(l 2 -atm/ml 2 ) b(l/ml) He 0.0341 0.02370 Xe 4.19 0.0510 H 2 0.244 0.0266 Cl 2 6.49 0.0562 CH 4 2.25 0.0428 CCl 4 20.4 0.1383 dispersin interactins increase with mlar mass CCl 4 is largest mlecule, has largest b value
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 = 0.0562 L/ml P = (1)(0.0821)(273) 2 (1)(0.0562) (1)(6.49) 2 2 P = 11.5 1.6 = 9.9 atm The term cntaining a is mre imprtant % difference: 11.2 9.9 11.2 x 100% = 12%