I. Physical Properties Ch. 12 - Gases. A. Kinetic Molecular Theory b Particles in an ideal gas… have no volume. have elastic collisions. are in constant,

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Presentation transcript:

I. Physical Properties Ch Gases

A. Kinetic Molecular Theory b Particles in an ideal gas… have no volume. have elastic collisions. are in constant, random, straight- line motion. don’t attract or repel each other. have an avg. KE directly related to Kelvin temperature.

B. Real Gases b Particles in a REAL gas… have their own volume attract each other b Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

C. Characteristics of Gases b Gases expand to fill any container. random motion, no attraction b Gases are fluids (like liquids). no attraction b Gases have very low densities. no volume = lots of empty space

C. Characteristics of Gases b Gases can be compressed. no volume = lots of empty space b Gases undergo diffusion & effusion. random motion

D. Temperature ºF ºC K K = ºC b Always use absolute temperature (Kelvin) when working with gases.

E. Pressure Which shoes create the most pressure?

E. Pressure b Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

E. Pressure b Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

E. Pressure b KEY UNITS AT SEA LEVEL kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

F. STP Standard Temperature & Pressure 0°C 273 K 1 atm kPa -OR- STP

II. The Gas Laws BOYLES CHARLES GAY- LUSSAC Ch Gases

A. Boyle’s Law P V PV = k

A. Boyle’s Law b The pressure and volume of a gas are inversely related at constant mass & temp P V PV = k

A. Boyle’s Law

V T B. Charles’ Law

V T b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

B. Charles’ Law

P T C. Gay-Lussac’s Law

P T b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

= k PV T D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =( kPa) V 2 (298 K) V 2 = 5.09 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

III. Ideal Gas Law (p , ) Ch. 10 & 11 - Gases

V n A. Avogadro’s Principle

V n b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R

B. Ideal Gas Law UNIVERSAL GAS CONSTANT R= L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

GIVEN: P = ? atm n = mol T = 16°C = 289 K V = 3.25 L R = L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm B. Ideal Gas Law b Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = kPa R = dm 3  kPa/mol  K B. Ideal Gas Law b Find the volume of 85 g of O 2 at 25°C and kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3 IDEAL GAS LAW

b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses times faster than Br 2. B. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

b A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? B. Graham’s Law Put the gas with the unknown speed as “Gas A”.

b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. B. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.