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Understanding the Ideal and Combined Gas Laws in Real-World Applications

This guide details the principles of the Ideal and Combined Gas Laws, providing step-by-step solutions to common problems encountered in gas law scenarios. It covers calculations relating to pressure, volume, temperature, and moles, demonstrating how to use these laws to find unknown variables in various contexts, including changes in altitude, temperature effects, and more. Readers will gain insights into practical applications, with sample problems illustrating the formulation and manipulation of relevant equations, such as PV=nRT and P1V1/T1 = P2V2/T2.

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Understanding the Ideal and Combined Gas Laws in Real-World Applications

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  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

  3. Provide the equation for the combined gas law. P1 V1 = P2 V2 T1 T2

  4. 1. If a gas occupies a volume of 100 cm3 at a pressure of 101.3 kPa and 27C, what volume will the gas occupy at 120 kPa and 50C? P1 = V1= T1= 100cm3 27 + 273 = 300K 101.3 kPa 50 + 273 = 323K X cm3 P2 = V2 = T2 = 120 kPa

  5. P1 = V1= T1= 100cm3 27 + 273 = 300K 101.3 kPa 50 + 273 = 323K X cm3 P2 = V2 = T2 = 120 kPa (101.3 kPa)(100cm3) = (120 kPa) (X cm3) 323 K 300 K

  6. GET X by its self!! (101.3 kPa)(100cm3) = (120 kPa) (X cm3) 323 K 300 K (101.3 kPa)(100 cm3)(323K) = X cm3 (300 K)(120 kPa)

  7. DO the MATH (101.3)(100 cm3)(323) = X cm3 (300 )(120) 3271990 cm3 = 90.9 cm3 36000

  8. 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? P1 = V1= T1= 20 + 273 = 293K 5.0 L 1.05 atm P2 = V2 = T2 = 0.65 atm -15 + 273 = 258K X L

  9. P1 = V1= T1= 20 + 273 = 293K 5.0 L 1.05 atm P2 = V2 = T2 = 0.65 atm -15 + 273 = 258K X L (1.05 atm) (5.0L) = (0.65 atm) (XL) 258 K 293 K

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

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

  12. 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? P1 = V1= T1= 466K 2.7L X atm 4.70 L P2 = V2 = T2 = 1.01 atm 605K

  13. P1 = V1= T1= 466K 2.7L X atm 4.70 L P2 = V2 = T2 = 1.01 atm 605K (X atm) (2.7L) = (1.01atm) (4.70L) 605 K 466 K

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

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

  16. 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? P1 = V1= T1= 692.0+ 273= 965K X L 153.3 kPa 7.37 L P2 = V2 = T2 = 32.26 kPa -48.0 + 273= 225K

  17. P1 = V1= T1= 692.0+ 273= 965K X L 153.3 kPa 7.37 L P2 = V2 = T2 = 32.26 kPa -48.0 + 273= 225K (153.3 kPa)(XL) = (32.26 kPa) (7.37L) 225 K 965 K

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

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

  20. Provide the equation for the Ideal gas law. PV=nRT

  21. 1.00 atm X 100L 0.0821 4.0 mol (4.0 mol)(0.0821)(X) (1.00 atm)(100L) = 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=

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

  23. 0.5 mol 18 L 45+273= 318K X atm 0.0821 (0.5 mol)(0.0821)(318K) (X atm)(18 L) = 6. An 18 liter container holds 16.00 grams of O2 at 45°C. What is the pressure (atm) of the container? P= V= T= R= n=

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

  25. 3.00 L 2.00 atm 0.0821 X mol 25+273= 298 K (X mol)(0.0821)(298K) (2.00 atm)(3.00L) = 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=

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

  27. 26+273=299K 2.3 atm 0.0026L 0.0821 X mol (X mol)(0.0821)(299) (2.3 atm)(0.0026L) = 8. A flashbulb of volume 0.0026 Lcontains O2 gas at a pressure of 2.3 atm and a temperature of 26C. How many moles of O2 does the flashbulb contain? P= V= T= R= n=

  28. Get X by itself (X mol)(0.0821)(299) (2.3 atm)(0.0026L) = (2.3)(0.0026) = X (0.0821)(299) 0.00598 24.55 = 0.00024 mol of O2 OR 2.4 x 10-4mol

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

  30. Homework: Combo and ideal #2 ws

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