II. The Gas Laws

A. Boyle’s Law P V PV = k

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

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? P 1 = 105 kPa P 2 = 40.5 kPa V 1 = 2.5 L V 2 = ? P 1 × V 1 = P 2 × V 2 (105) (2.5) = (40.5)(V 2 ) = 40.5 (V 2 ) 6.48 L = V 2 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? P 1 = 205 kPa P 2 = ? V 1 = 4.0 LV 2 = 12.0 L P 1 × V 1 = P 2 × V 2 (205) (4.0) = (P 2 )(12) 820 = (P 2 ) kPa = P 2 Example Problems pg 335 # 10 &11

B. Charles’ Law V T

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

Example Problems pg. 337 # 12 & If a sample of gas occupies 6.80 L at 325°C, what will be its volume at 25°C if the pressure does not change? V 1 = 6.8LV 2 = ? T 1 = 325°C = 598 K T 2 = 25°C = 298 K 6.8 = V X 298 = V 2 X = V 2 X L = V 2

4. Exactly 5.00 L of air at -50.0°C is warmed to 100.0°C. What is the new volume if the pressure remains constant? V 1 = 5.0L V 2 = ? T 1 = -50°C = 223 K T 2 = 100°C = 373 K 5 = V X 373 = V 2 X = V 2 X L = V 2 Example Problems pg. 337 # 12 & 13

C. Gay-Lussac’s Law P T

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

Example Problems 5. The gas left in a used aerosol can is at a pressure of 103 kPa at 25°C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928°C? P 1 = 103 kPaP 2 = ? T 1 = 25°C = 298 K T 2 = 928°C = 1201 K 103 = P × P 2 = 123,703 P 2 = 415 kPa

Example Problem pg. 338 # 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? P 1 = 6.58 kPaP 2 = ? T 1 = 539 K T 2 = 211 K 6.58 = P X 211 = P 2 X = P 2 X kPa = P 2

D. Avogadro’s Law Under the same condition of temperature and pressure, equal volumes of all gases contain the same number of molecules.

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

F. Gas Law Problems A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. 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 CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

F. Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. 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 BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

F. Gas Law Problems A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. 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 P  T  VV COMBINED GAS LAW

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