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Chapter 5 Gases 3 5.1

2 NaN 3 (s) 2Na(s) + 3N 2 (g) Air bag 45.5L sodium azide?,,? 3 5.2

? 1.,,, 2. P, V, n, T ( ) 3. 3 5.3

5.1,, = 1L = 10 3 cm 3 = 10-3 m 3 m=m n ( ) T k = t c + 273.15 : psi, mmhg, atm( ) 1.013 bar = 1atm =760mmHg SI Unit 1Pa = 1 newton/m 2 (= 101.3 kpa) 3 5.4

Barometric Pressure Standard Atmospheric Pressure 1.00 atm 760 mm Hg, 760 torr 101.325 kpa 1.01325 bar 1013.25 mbar 3 5.5

5.2 V = volume (liters, cm 3, m 3 ) n = amount in moles T = temperature (in K) P = pressure (atmospheres) 3 5.6

Boyle s Law Boyle 1662 P 1 V PV = constant 3 5.7

Charles s Law Charles 1787 Gay-Lussac 1802 V T V = b T 3 5.8

STP Gas properties depend on conditions. Define standard conditions of temperature and pressure (STP). P = 1 atm = 760 mm Hg T = 0 C = 273.15 K 3 5.9

Avogadro s Law Gay-Lussac 1808 Small volumes of gases react in the ratio of small whole numbers. Avogadro 1811 Equal volumes of gases have equal numbers of molecules (V n )and Gas molecules may break up when they react. 3 5.10

Formation of Water 3 5.11

3 5.12

P T n V Boyle V 1/P Charles V T Avogadro V n P nt/v : PV nt: PV/nT = const. Const. = (1atm. 22.414L)/(1 mol 273.13K) = R PV = nrt : Ideal Gas Law R, 0.0821 L atm/(mol K)= 8.31J/(mol K), No attraction between particle! 3 5.13

: 255 lb/in 2 75%?,. V and T are constant, so P 2 /P 1 = n 2 /n 1 ; n 2 = 0.25 n 1 P 2 = 0.25 P 1 = 0.25 255 lb/in 2 = 64 lb/in 2 3 5.14

P, V, n, T 25 C 50.0 L 15.0 mol O 2? nrt P = = V (15.0 mol)(0.0821 L atm/mol K)(298 K) 50.0 L = 7.34 atm R = 0.0821 L atm/(mol K) V liter, P, T K, and n 3 5.15

27 C, 735 mm Hg O 2 (g)?, (1L) O 2 n= PV RT = (735/760 mm Hg)(1.00 L) (0.00821 L atm /mol K)(300 K) mass = 0.0393 mol 37.00 g/mol = 1.26 g d = 1.26 g/l (M ) n M P D = = = M ( P/RT = n/v) V RT = 0.0393 mol 3 5.16

( ) 52.693 g 100 C 752 mm Hg 53.117 g. 226.2 ml? n, g, n = PV RT = (752 / 760 atm)(0.2262 L) 0.0821 L atm / mol K)(373 K) = 0.00731 mol 0.424 g = M 0.00731 mol M = 58.0 g/mol 3 5.17

5.4 20 C, 735 mm Hg 16.0 L H 2 (g) Zn? Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Path to follow: V H 2 n H 2 n Zn m Zn n H2 PV (735/760 atm)(16.0 L) = = = 0.643 mol H RT 2 m Zn (0.0821 L atm/mol K)(293 K) 1 mol Zn 65.38 g Zn = 0.643 mol H 2 = 42.0 g Zn 1 mol H 2 1 mol Zn 3 5.18

2 NaN 3 (s) 2Na(s) + 3N 2 (g) 45.5L Air bag sodium azide?, dkq 828mmHg, 22 C PV n N2 = = =2.05 mol, n NaN3 = RT (828/760)atm 45.5L (0.0821 L atm/mol K) (295.2K) 2 mol NaN 2 3 mol N 2 = 1.37 mol NaN 2 65.0 g/mol = 88.8 g,,? 3 5.19

5.5 Dalton : ) 2NH 4 NO 3 (s) 2N 2 (g) + 4H 2 O(g) + O 2 (g) 2.0 mol NH 4 NO 3, 10.0 L, 300K:? P tot = P A + P B + P C +... P A A, P B B,... : P H2 O P gas = P tot P H2 O 3 5.20

Partial Pressure P tot = P a + P b + V a = n a RT/P tot and V tot = V a + V b + V a V tot n a RT/P = tot = n a n tot RT/P tot n tot Recall n a n tot = χ a P a P tot n a RT/V = tot = n a n tot RT/V tot n tot 3 5.21

Dalton s Law P A = X A P tot 734 mm Hg (X = 0.2095)? P O2 = 0.2095 (734 mm Hg) = 154 mm Hg 3 5.22

5.6 (p128) Particles are point masses in constant, random, straight line motion. Particles are separated by great distances. Collisions are rapid and elastic. No force between particles. Total energy remains constant. 3 5.23

Pressure Assessing Collision Forces Translational kinetic energy, E t e = k 1 2 mu 2 Frequency of collisions, v = u N V Impulse or momentum transfer, I = mu Pressure proportional to impulse times frequency P N V 2 mu 3 5.24

Pressure and Molecular Speed P = m u Three dimensional systems lead to: 2 1 3 N V u m is the modal speed u av is the simple average u rms = 2 u 3 5.25

E t T E t = ½ (mass)(avg. speed) 2 = ½ mu 2 ½mu 2 T E t = 3RT 2N A - ( mu 2 = 3PV/N) - T - T E t 3 5.26

Assume one mole: PV = 1 3 N A m u 2 PV=RT so: 3RT = N A m u 2 N A m= M: 3RT = M u 2 Rearrange: u rms = 3RT M 3 5.27

(Distribution of Molecular Speeds) u rms = 3RT M 3 5.28

R = 8.31 10 3 (in the proper units.) 0 C H 2? u = ([3 8.31 10 3 ] 273/2.02) 1/2 = 1.84 10 3 m/s 3 5.29

Graham s Law : T m u. m 2 u 2 2 = m 1 u 2 1 (rate2) / (rate1) = (M 1 / M 2 ) 1/2 T, P 2.42.? (M / 32.0 g/mol) 1/2 = 2.42 M = (2.42) 2 32.0 g/mol = 187 g/mol 3 5.30

5.7 : V m -V m V m < 0 (finite volume) V m -V m V m > 0 (high P, low T). 3 5.31

3 5.32

7, 23, 29, 33, 43, 45, 55, 61, 79, 81 3 5.33