# II. The Gas Laws - Midland High School

The Gas Laws A. Boyles Law PV = k P V A. Boyles Law The pressure and volume of a gas are inversely related oat constant mass & temp

PV = k P V Example Problems pg 335 # 10 &11 1. The pressure on 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temp remains constant?

P1 = 105 kPa P2 = 40.5 kPa V1 = 2.5 L V2 = ? P1 V1 = P2 V2 (105) (2.5) = (40.5)(V2) 262.5 = 40.5 (V2) 6.48 L = V2 Example Problems pg 335 # 10 &11

2. A gas with a volume of 4.00L at a pressure of 205 kPa is allowed to expand to a volume of 12.0L. What is the pressure in the container if the temp remains constant? P1 = 205 kPa P2 = ? V1 = 4.0 L V2 = 12.0 L P1 V1 = P2 V2 (205) (4.0) = (P2)(12) 820 = (P2) 12 68.3 kPa = P2

B. Charles Law V T V k T B. Charles Law The volume and absolute temperature (K) of a gas are directly related o at constant mass & pressure

V T V k T Example Problems pg. 337 # 12 & 13 3. If a sample of gas occupies 6.80 L at 325C, what will be its volume at 25C if the pressure does not change? V1= 6.8L

V2 = ? T1 = 325C = 598 K T2 = 25C = 298 K 6.8 = V2 598 298 6.8 X 298 = V2 X 598 2026.4 = V2 X 598 3.39 L = V2 Example Problems pg. 337 # 12 & 13 4. Exactly 5.00 L of air at -50.0C is warmed to 100.0C. What is the new volume if the pressure remains

constant? V1= 5.0L T1 = -50C = 223 K 5 = V2 223 373 5 X 373 = V2 X 223 1865 = V2 X 223 8.36L = V2 V2 = ? T2 = 100C = 373 K

C. Gay-Lussacs Law P k T P T C. Gay-Lussacs Law The pressure and absolute temperature (K) of a gas are directly related o at constant mass & volume

P k T P T Example Problems 5. The gas left in a used aerosol can is at a pressure of 103 kPa at 25C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928C? P1= 103 kPa

T1 = 25C = 298 K 103 = P2 298 1201 298 P2 = 123,703 P2 = ? T2 = 928C = 1201 K Example Problem pg. 338 # 14 6. A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does not change? P1= 6.58 kPa

T1 = 539 K 6.58 = P2 539 211 6.58 X 211 = P2 X 539 1388.38 = P2 X 539 2.58 kPa = P P2 = ? T2 = 211 K D. Avogadros Law Under the same condition of

temperature and pressure, equal volumes of all gases contain the same number of molecules. E. Combined Gas Law P V PV PV = k T P1V1 P2V2 = T1 T2

P1V1T2 = P2V2T1 F. Gas Law Problems A gas occupies 473 cm3 at 36C. Find its volume at 94C. CHARLES LAW GIVEN: T V V1 = 473 cm3 WORK: P1V1T2 = P2V2T1 T1 = 36C = 309K

(473 cm3)(367 K)=V2(309 K) V2 = ? T2 = 94C = 367K V2 = 562 cm3 F. AGas Law Problems gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLES LAW

GIVEN: P V WORK: V1 = 100. mL P1V1T2 = P2V2T1 P1 = 150. kPa (150.kPa)(100.mL)=(200.kPa)V2 V2 = ? V2 = 75.0 mL P2 = 200. kPa F. Law7.84

Problems A Gas gas occupies cm at 71.8 kPa 3 & 25C. Find its volume at STP. COMBINED GAS LAW GIVEN: P T V WORK: V1 = 7.84 cm3 P1V1T2 = P2V2T1 P1 = 71.8 kPa 3 T1 = 25C = 298 K

V2 = ? (71.8 kPa)(7.84 cm )(273 K) =(101.325 kPa) V2 (298 K) 3 V = 5.09 cm 2 P2 = 101.325 kPa F. Gas Law Problems A gas pressure is 765 torr at 23C. At what

temperature will the pressure be 560. torr? GAY-LUSSACS LAW GIVEN: P T WORK: P1 = 765 torr P1V1T2 = P2V2T1 T1 = 23C = 296K (765 torr)T2 = (560. torr)(309K) P2 = 560. torr T2 = ? T2 = 226 K = -47C

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