1. Entry Task: March 27-28 Block 2 QUESTION: P= 30 atm V= 50 L T= 293K R= 0.0821 n= X Solve for the number of moles (n)

2. Agenda Go over Combined and Ideal ws HW: Pre-Lab Proving gas law

4. P 1 V 1 = P 2 V 2 T1T1 T2T2 * Provide the equation for the combined gas law.

5. 1. If a gas occupies a volume of 100 cm 3 at a pressure of 101.3 kPa and 27  C, what volume will the gas occupy at 120 kPa and 50  C? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 100cm 3 101.3 kPa 27 + 273 = 300K 120 kPa X cm 3 50 + 273 = 323K

6. (101.3 kPa)(100cm 3 ) 300 K 323 K = (120 kPa) (X cm 3 ) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 100cm 3 101.3 kPa 27 + 273 = 300K 120 kPa X cm 3 50 + 273 = 323K

7. GET X by its self!! (101.3 kPa)(100 cm 3 )(323K) (300 K)(120 kPa) = X cm 3 (101.3 kPa)(100cm 3 ) 300 K 323 K = (120 kPa) (X cm 3 )

8. DO the MATH 3271990 cm 3 36000 = 90.9 cm 3 (101.3)(100 cm 3 )(323) (300 )(120) = X cm 3

9. 2. A toy balloon has an internal pressure of 1.05 atm and a volume of 5.0 L. If the temperature where the balloon is released is 20˚ C, what will happen to the volume when the balloon rises to an altitude where the pressure is 0.65 atm and the temperature is –15˚ C? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 5.0 L 1.05 atm 20 + 273 = 293K 0.65 atm X L -15 + 273 = 258K

10. (1.05 atm) (5.0L) 293 K 258 K = (0.65 atm) (X L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 5.0 L 1.05 atm 20 + 273 = 293K 0.65 atm X L -15 + 273 = 258K

11. GET X by its self!! (1.05 atm)(5.0L)(258K) (293 K)(0.65 atm) = X L (1.05 atm) (5.0L) 293 K 258 K = (0.65 atm) (X L)

12. DO the MATH 1355 190.5 = 7.11L (1.05)(5.0L)(258) (293)(0.65) = X L

13. 3. A closed gas system initially has volume and temperature of 2.7 L and 466 K with the pressure unknown. If the same closed system has values of 1.01 atm, 4.70 L and 605 K, what was the initial pressure in atm? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.7L X atm 466K 1.01 atm 4.70 L 605K

14. (X atm) (2.7L) 466 K 605 K = (1.01atm) (4.70L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = 2.7L X atm 466K 1.01 atm 4.70 L 605K

15. GET X by its self!! (466K)(1.01 atm)(4.70L) (2.7L)(605 K) = X atm (X atm) (2.7L) 466 K 605 K = (1.01atm) (4.70L)

16. DO the MATH 2212 1634 = 1.35 atm (466)(1.01 atm)(4.70) (2.7)(605 ) = X atm

17. 4. A closed gas system initially has pressure and temperature of 153.3 kPa and 692.0°C with the volume unknown. If the same closed system has values of 32.26 kPa, 7.37 L and -48.00°C, what was the initial volume in L? P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X L 153.3 kPa 692.0+ 273= 965K 32.26 kPa 7.37 L -48.0 + 273= 225K

18. (153.3 kPa)(XL) 965 K 225 K = (32.26 kPa) (7.37L) P 1 = V 1 = T 1 = P 2 = V 2 = T 2 = X L 153.3 kPa 692.0+ 273= 965K 32.26 kPa 7.37 L -48.0 + 273= 225K

19. GET X by its self!! (965K)(32.26kPa)(7.37L) (153.3 kPa)(225 K) = X L (153.3 kPa)(XL) 965 K 225 K = (32.26 kPa) (7.37L)

20. DO the MATH 229435 34493 = 6.65 L (965)(32.26)(7.37L) (153.3)(225) = X L

21. PV=nRT * Provide the equation for the Ideal gas law.

22. 5. At what temperature (in Kelvin) would 4.0 moles of Hydrogen gas in a 100 liter container exert a pressure of 1.00 atmospheres? P= V= T=R= n= X 100L 1.00 atm 0.0821 4.0 mol (1.00 atm)(100L) =(4.0 mol)(0.0821)(X)

23. Get X by itself! (1.00 atm)(100L) =X (4.0) (0.0821) 100 0.3284 = 305 K (1.00 atm)(100L) =(4.0 mol)(0.0821)(X)

24. 6. An 18 liter container holds 16.00 grams of O 2 at 45°C. What is the pressure (atm) of the container? P= V= T=R= n= 45+273= 318K 18 L X atm0.0821 0.5 mol (X atm)(18 L) =(0.5 mol)(0.0821)(318K)

25. Get X by itself! (0.5)(0.0821)(318) =X (18) 13.1 18 = 0.73 atm (X atm)(18 L) =(0.5 mol)(0.0821)(318K)

26. 7. How many moles of oxygen must be in a 3.00 liter container in order to exert a pressure of 2.00 atmospheres at 25 °C? P= V= T=R= n= 25+273= 298 K 3.00 L 2.00 atm 0.0821X mol (2.00 atm)(3.00L) =(X mol)(0.0821)(298K)

27. Get X by itself! (2.00)(3.00) = X (0.0821)(298) 6.00 24.47 = 0.245 mol (2.00 atm)(3.00L) =(X mol)(0.0821)(298K)

28. 8. A flashbulb of volume 0.0026 L contains O 2 gas at a pressure of 2.3 atm and a temperature of 26  C. How many moles of O 2 does the flashbulb contain? P= V= T=R= n= 26+273=299K 0.0026L 2.3 atm 0.0821X mol (2.3 atm)(0.0026L) = (X mol)(0.0821)(299)

29. Get X by itself (2.3)(0.0026) = X (0.0821)(299) 0.00598 24.55 = 0.00024 mol of O 2 OR 2.4 x 10 -4 mol (2.3 atm)(0.0026L) = (X mol)(0.0821)(299)

30. In-class Ch. 14 sec. 3 worksheet

31. Homework: Combo and ideal #2 ws