Boyle and Charles' Law Problem Solving PPT

Solving problems regarding Boyle's Law and Charle's Law

1. P R E S S U R E

2. T E M P E R T U R E A

3. V O L U M E

4. H E A T

5. Gas Laws: Boyle’s Law Charles Law

6. Opening thoughts… Have you ever: Seen a hot air balloon? Had a soda bottle spray all over you? Baked (or eaten) a nice, fluffy cake? These are all examples of gases at work!

7. Properties of Gases You can predict the behavior of gases based on the following properties: Pressure Volume Amount (moles) Temperature

8. STP Standard Temperature & Pressure 0°C 273.15 K 1 atm 101.325 kPa -OR- STP

9. Equivalent Air Pressure Units

10. A. Boyle’s Law- Robert Boyle P V

11. A. Boyle’s Law The pressure and volume of a gas are inversely related at constant mass & temp P V P V = P V 1 1 2 2

12. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 = P 2 V 2 Boyle’s Law Sample Problem 1 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)V 2 V 2 = 75.0 mL

13. Boyle’s Law Sample Problem 2 G as occupies a volume of 1 L and exerts a pressure of 400 kPa on the walls of its container. What would be the pressure exerted by the gas if it is completely transferred into a new container having a volume of 3 litres (assuming that the temperature and amount of the gas remain the same.)? GIVEN: V 1 = 1 L P 1 = 400 kPa V 2 = 3 L P 2 = ? WORK: P 1 V 1 = P 2 V 2 ( 400 kPa )( 1 L) = P  2  X 3 L P  2  = 400 / 3 P  2  = 133.33 kPa

14. Practice A helium balloon has a volume of 735 mL at ground level. The balloon is transported to an elevation of 5 km, where the pressure is 0.8 atm. At this altitude, the gas occupies a volume of 1286 mL. Assuming that the temperature is constant, what was the ground level pressure in kPa (1 atm= 101.3 kPa)? GIVEN: V 1 = 735 ml : 0.735 L P 1 = ? V 2 = 1286 ml: 1.286 L P 2 = 0.8 atm WORK: P 1 V 1 = P 2 V 2 P 1 (0.735L)=(0.8 atm)(1.286L) P  1  = 1.028 /0.735 P   1   =1.39 atm Convert to kPa (1 atm=101.3 kPa) 1.39 x 101.3 = 140.87kPa

15. V T B. Charles’ Law- Jacques Charles

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

17. CHARLES’ LAW SAMPLE PROBLEM 1 A gas has a volume of 4.0 L at a temperature of 250 K . If its volume expands to 6.0 L , what is the final temperature of the gas, assuming the pressure remains constant? GIVEN: V 1 = 4.0 L T 1 = 250K V 2 = 6.0 L T 2 = ? WORK: T 1 ​V 2 ​=T 2 ​V 1 ​ CHARLES’ LAW SAMPLE PROBLEM 2 CHARLES’ LAW T  V  (250K)(6.0L) = (T 2 ) (4.0L) T 2 = ( 250K)(6.0L )/ 4.0 L T 2 = 375 K

18. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: T 1 ​V 2 ​=T 2 ​V 1 ​ CHARLES’ LAW SAMPLE PROBLEM 2 A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW T  V  (309K)(V 2 )= (367 K)(473 cm 3 ) V 2 = (367K X 473 cm 3 ) / (309K) V 2 = 561.78cm 3