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1 Chapter 5 Gases 3 5.1

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

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

4 5.1,, = 1L = 10 3 cm 3 = 10-3 m 3 m=m n ( ) T k = t c : psi, mmhg, atm( ) bar = 1atm =760mmHg SI Unit 1Pa = 1 newton/m 2 (= kpa) 3 5.4

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

6 5.2 V = volume (liters, cm 3, m 3 ) n = amount in moles T = temperature (in K) P = pressure (atmospheres) 3 5.6

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

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

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

10 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

11 Formation of Water

12 3 5.12

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

14 : 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 = lb/in 2 = 64 lb/in

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

16 27 C, 735 mm Hg O 2 (g)?, (1L) O 2 n= PV RT = (735/760 mm Hg)(1.00 L) ( L atm /mol K)(300 K) mass = mol g/mol = 1.26 g d = 1.26 g/l (M ) n M P D = = = M ( P/RT = n/v) V RT = mol

17 ( ) g 100 C 752 mm Hg g ml? n, g, n = PV RT = (752 / 760 atm)( L) L atm / mol K)(373 K) = mol g = M mol M = 58.0 g/mol

18 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) = = = mol H RT 2 m Zn ( L atm/mol K)(293 K) 1 mol Zn g Zn = mol H 2 = 42.0 g Zn 1 mol H 2 1 mol Zn

19 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 ( L atm/mol K) (295.2K) 2 mol NaN 2 3 mol N 2 = 1.37 mol NaN g/mol = 88.8 g,,?

20 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

21 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

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

23 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

24 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

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

26 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

27 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

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

29 R = (in the proper units.) 0 C H 2? u = ([ ] 273/2.02) 1/2 = m/s

30 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) g/mol = 187 g/mol

31 5.7 : V m -V m V m < 0 (finite volume) V m -V m V m > 0 (high P, low T)

32 3 5.32

33 7, 23, 29, 33, 43, 45, 55, 61, 79,

슬라이드 제목 없음

슬라이드 제목 없음 물리화학 1 문제풀이 130403 김대형교수님 Chapter 1 Exercise (#1) A sample of 255 mg of neon occupies 3.00 dm 3 at 122K. Use the perfect gas law to calculate the pressure of the gas. Solution 1) The perfect gas law p