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

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Presentation transcript:

I. Physical Properties

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.

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

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

V T B. Charles’ Law

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

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

= kPV PTPT VTVT T D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

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