1. The Gas Laws The Behavior of Gases

2. STPSTP b Standard Temperature and Pressure: b 273 K and 760 mm Hg b Or 0 C and 1atm

3. Temperature Conversions b [°C] = K − 273 b [°C] = 5/9(°F-32) b [K] = °C + 273 [K] = (°F + 459.67) × 5 ⁄ 9 b [°F] = K × 9 ⁄ 5 − 459.67 b [°F] = °C × 9 ⁄ 5 + 32

4. A. Boyle’s Law P V PV = k

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

6. A. Boyle’s Law

7. Boyle’s Law Practice Problem GIVEN: V 1 = 1.53 L P 1 = 5.6 x 10 3 Pa V2 = ?V2 = ? P 2 = 1.5 x 10 4 Pa WORK: P 1 V 1 = P 2 V 2 5.6 x 10 3 Pa x 1.53 L 1.5 x 10 4 Pa = 0.57L b Given 1.53 L of a sample of SO 2 at a pressure of 5.6 x 10 3 Pa. If the pressure is changed to 1.5 x 10 4 Pa at a constant temp, what will be the new volume of the gas.

8. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 = P 2 V 2 Gas Law Problems Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

9. V T B. Charles’ Law

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

11. B. Charles’ Law

12. GIVEN: V 1 = 2.58 L T 1 = 15 C V2 = ?V2 = ? T 2 = 38 C WORK: V 1 /T 1 = V 2 /T 2 2.58 L = V 2 288 K 311 K = 2.79 L b A sample of gas at 15 degrees C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38 degree C and 1 atm? (Hint** convert to K first) Charles’ Law Practice

13. P T C. Gay-Lussac’s Law

14. P T b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P 1 /T 1 = P 2 /T 2

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

16. GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 T 2 = P 2 T 1 E. Gas Law Problems b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(296K) T 2 = 217 K = -56°C

17. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 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) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

18. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: V 1 T 2 = V 2 T 1 Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

19. F. The First 4 Gas Laws b The Gas Laws Table