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Chapter 5 The Gaseous State HST Mr. Watson. HST Mr. Watson Properties of Gases can be compressed exert pressure on whatever surrounds them expand into.

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Presentation on theme: "Chapter 5 The Gaseous State HST Mr. Watson. HST Mr. Watson Properties of Gases can be compressed exert pressure on whatever surrounds them expand into."— Presentation transcript:

1 Chapter 5 The Gaseous State HST Mr. Watson

2 HST Mr. Watson Properties of Gases can be compressed exert pressure on whatever surrounds them expand into whatever volume is available easily diffuse into one another can be described in terms of their temperatures and pressure,the volume occupied, and the amount (number of molecules or moles) present

3 HST Mr. Watson Mercury Barometer

4 HST Mr. Watson Composition of Air at Sea Level

5 HST Mr. Watson Important Units of Pressure Conversion factor to learn

6 HST Mr. Watson Boyle’s Law At constant temperature and mass of gas: V  1/P V = a * 1/P where a is a proportionality constant thus VP = a V 1 P 1 = a = V 2 P 2 V 1 P 1 = V 2 P 2

7 HST Mr. Watson Boyle’s Law

8 HST Mr. Watson Boyle’s Law

9 HST Mr. Watson Charles’ Law At constant pressure and mass of gas: V  T V = b * T where b is a proportionality constant V/T = b V 1 /T 1 = b = V 2 /T 2 V 1 /T 1 = V 2 /T 2

10 HST Mr. Watson Charles’ Law

11 HST Mr. Watson Combined Gas Law At constant mass of gas V  T/P V = d * (T/P) where d is a proportionality constant (VP)/T = d V 1 P 1 = d = V 2 P 2 T 1 T 2 V 1 P 1 = V 2 P 2 T 1 T 2

12 HST Mr. Watson Avogadro’s Law At constant pressure and temperature V  n V = c * n where c is a proportionality constant V/n = c V 1 /n 1 = c = V 2 /n 2 V 1 /n 1 = V 2 /n 2

13 HST Mr. Watson Ideal Gas Law V  (n * T)/P V = R * (n * T)/P where R is proportionality constant P * V = n * R * T (P*V)/(n*T) =R Thus, (P 1 *V 1 )/(n 1 *T 1 ) = (P 2 *V 2 )/(n 2 *T 2 )

14 HST Mr. Watson What will be volume of an ideal gas at absolute zero? - 10 mL/mole 0 mL/mole 10 mL/mole

15 HST Mr. Watson Ideal Gas Constant R = 0.08205 L*atm/mol*K R has other values for other sets of units. R = 82.05 mL*atm/mol*K = 8.314 J/mol*K = 1.987 cal/mol*K

16 HST Mr. Watson Molar Mass from Gas Density gas density = #g/V = d PV = nRT where n = #g/MM PV = (#g/MM)*RT MM = (#g*R*T)/(P*V) MM = (#g/V)*((R*T)/P) = (d*R*T)/P

17 HST Mr. Watson Dalton’s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of the individual gases (partial pressures). P T = P 1 + P 2 + P 3 + P 4 +.... whereP T => total pressure P 1 => partial pressure of gas 1 P 2 => partial pressure of gas 2 P 3 => partial pressure of gas 3 P 4 => partial pressure of gas 4

18 HST Mr. Watson Dalton’s Law of Partial Pressure

19 HST Mr. Watson Collecting Gases over Water

20 HST Mr. Watson Example: What volume will 25.0 g O 2 occupy at 20 o C and a pressure of 0.880 atm? (25.0 g)(1 mol) n = ---------------------- = 0.781 mol (32.0 g) V =?; P = 0.880 atm; T = (20 + 273)K = 293K R = 0.08205 L*atm/mol*K V = nRT/P = (0.781 mol)(0.08205L*atm/mol*K)(293K) 0.880atm = 21.3 L

21 HST Mr. Watson Example: A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? (P water = 23.8 torr at 25 o C) P O2 = P bar - P water = (758 - 23.8) torr = 734 torr

22 HST Mr. Watson Example A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758 torr at 25 o C, what is the volume of the “dry” oxygen gas at STP? P water = 23.8 torr at 25 o C; P O2 = P bar - P water = (758 - 23.8) torr = 734 torr P 1 = P O2 = 734 torr; P 2 = SP = 760. torr V 1 = 245mL; T 1 = 298K; T 2 = 273K; V 2 = ? (V 1 P 1 /T 1 ) = (V 2 P 2 /T 2 ) V 2 = (V 1 P 1 T 2 )/(T 1 P 2 ) = (245mL)(734torr)(273K) (298K)(760.torr) = 217mL

23 HST Mr. Watson Kinetic Molecular Theory Matter consists of particles (atoms or molecules) in continuous, random motion.

24 HST Mr. Watson Kinetic Molecular Theory: Gases particles in continuous, random, rapid motion collisions between particles are elastic volume occupied by the particles has a negligibly small effect on their behavior attractive forces between particles have a negligible effect on their behavior gases have no fixed volume or shape, take the volume and shape of the container

25 HST Mr. Watson Maxwell’s Distribution of Speeds

26 HST Mr. Watson Real Gases have a finite volume at absolute zero have attractive forces between gas particles

27 HST Mr. Watson Van der Waals Equation (P + a/V 2 )(V - b) = nRT wherea => attractive forces b => residual volume

28 HST Mr. Watson

29 HST Mr. Watson

30 HST Mr. Watson Carbon Dioxide and Greenhouse Effect

31 HST Mr. Watson Some Oxides of Nitrogen N 2 O NO NO 2 N 2 O 4 2 NO 2 = N 2 O 4 brown colorless NO x

32 HST Mr. Watson Air Pollution in Los Angles


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