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gas_laws.ppt

The document discusses the key gas laws and concepts: 1) Gases are highly compressible, occupy their container fully, exert uniform pressure, diffuse easily, and have low densities according to kinetic molecular theory. 2) The gas laws describe the relationships between pressure, volume, temperature, and moles of gas including Boyle's law, Charles' law, Gay-Lussac's law, Avogadro's principle, and the combined ideal gas law. 3) The ideal gas law combines all gas laws as PV=nRT, relating pressure, volume, moles of gas, temperature, and the universal gas constant R. It can be used to calculate gas properties.

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The Gas Laws
• 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. Characteristics of Gases
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.
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 ºF ºC K -459 32 212 -273 0 100 0 273 373   32 F C 9 5     K = ºC + 273 Always use absolute temperature (Kelvin) when working with gases.
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 Kelvin Practice 273 373 –173 243 27 130 298 –273 Absolute zero is –273C or 0 K
Pressure area force 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.
Pressure 2 m N kPa  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 Mercury Barometer Aneroid Barometer
STP Standard Temperature & Pressure 273 K 101.325 kPa STP
STP Standard Laboratory Conditions 25°C or 298 K 101.325 kPa SLC
-BOYLES -CHARLE -GAY-LUSSAC The Gas Laws
Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp. P V PV = k 2 2 1 1 V P V P   
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?
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 V1 = 946 mL P2 = ? V2 = 154 mL P2 = P1 x V1 V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg
k T V  V T Charles’ Law The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure 2 2 1 1 T V T V 
Charles’ Law
Practice 2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 3. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?
2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 3. If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? 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 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
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 = 3.20 L T1 = 398.15 K V2 = 1.54 L T2 = ? T2 = V2 x T1 V1 1.54 L x 398.15 K 3.20 L = = 192 K V1/T1 = V2/T2
k T P  P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume
Gay-Lussac’s Law
Combined Gas Law P1V1 T1 = P2V2 T2 P1V1T2 = P2V2T1
GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K WORK: P1V1T2 = P2V2T1 E. Gas Law Problems A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW T V (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3
GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa WORK: P1V1T2 = P2V2T1 E. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW P V (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.
GIVEN: V1 = 7.84 cm3 P1 = 71.8 kPa T1 = 25°C = 298 K V2 = ? P2 = 101.325 kPa T2 = 273 K 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 occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. P T V COMBINED GAS LAW
GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? WORK: P1V1T2 = P2V2T1 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 P T (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C
k n V  V n Avogadro’s Principle  Equal volumes of all gases contain equal numbers of moles at constant temp & pressure. 2 2 1 1 n V n V 
The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. 11.5 nT V P  V P  1 Boyle’s law V n  Avogadro’s law V T  Charles’s law nT V R P  PV = nRT rearrangement R is the proportionality constant, called the gas constant.
B. Ideal Gas Law UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·K R=0.0821 Latm/molK PV=nRT
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
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 Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW
GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm3kPa/molK B. Ideal Gas Law Find the volume of 85 g of O2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K V = 64 dm3 IDEAL GAS LAW