The Gas Laws

Characteristics of Gases highly compressible. occupy the full volume of their containers. exert a uniform pressure on all inner surfaces of a container diffuse (mix) easily and quickly have very low densities.

Kinetic Molecular Theory Gases consist of a large number of molecules in constant random motion. Volume of individual molecules negligible compared to volume of container. Intermolecular forces (forces between gas molecules) negligible. Collision of gas particles are elastic so no kinetic energy is lost As temperature increases the gas particles move faster, hence increased kinetic energy. Gases only occupy about 0.1 % of the volume of their containers.

Four Physical Quantities for Gases Phys. Qty. Symbol SI unit Other common units pressure P Pascal (Pa) atm, mm Hg, torr, psi volume V m3 dm3, L, mL, cm3 temp. T K °C, °F moles n mol

Temperature Always use absolute temperature (Kelvin) when working with gases. ºF ºC K -459 32 212 -273 100 273 373 K = ºC + 273

Kelvin Practice What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0C = _______ K 100C = _______ K 100 K = _______ C – 30C = _______ K 300 K = _______ C 403 K = _______ C 25C = _______ K 0 K = _______ C Absolute zero is – 273C or 0 K 273 373 – 173 243 27 130 298 – 273

Pressure Which shoes create the most pressure? Pressure (P ) is defined as the force exerted per unit area. The atmospheric pressure is measured using a barometer. Which shoes create the most pressure?

Pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi 1 atm = 760 mmHg = 760 torr = 101325 Pa.

Pressure Barometer measures atmospheric pressure Aneroid Barometer Mercury Barometer Aneroid Barometer

Standard Temperature & Pressure STP STP Standard Temperature & Pressure 273 K 101.325 kPa

Standard Laboratory Conditions STP SLC Standard Laboratory Conditions 25°C or 298 K 101.325 kPa

-BOYLES -CHARLE -GAY-LUSSAC The Gas Laws P V T -BOYLES -CHARLE -GAY-LUSSAC

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

A. Boyle’s Law

Practice A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure V T

Charles’ Law

Practice A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

For more lessons, visit www.chalkbored.com A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com

V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K

Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T

Gay-Lussac’s Law

Combined Gas Law P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1

E. Gas Law Problems A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

E. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

Practice A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

E. Gas Law Problems COMBINED GAS LAW P T V V1 = 7.84 cm3 A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K P T V WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa) V2 (298 K) V2 = 5.09 cm3

E. Gas Law Problems A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C

Avogadro’s Principle Equal volumes of all gases contain equal numbers of moles at constant temp & pressure. V n

The Ideal Gas Equation 11.5 The gas laws can be combined into a general equation that describes the physical behavior of all gases. Boyle’s law Charles’s law Avogadro’s law PV = nRT rearrangement R is the proportionality constant, called the gas constant.

PV=nRT UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·K R=0.0821 Latm/molK B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·K R=0.0821 Latm/molK

R = 0. 0821 liter·atm/mol·K R = 8. 3145 J/mol·K R = 8 R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·K R = 8.2057 m3·atm/mol·K R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

B. Ideal Gas Law IDEAL GAS LAW P = ? atm n = 0.412 mol Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821L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 IDEAL GAS LAW V = ? n = 85 g T = 25°C = 298 K Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK WORK: 85 g 1 mol = 2.7 mol 32.00 g = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3