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Episode I ATTACK OF THE GAS

Not so long ago, in a chemistry lab far far away…. May the FORCE/area be with you. Episode I ATTACK OF THE GAS Gas, being of upmost importance to the entire galaxy and your life, is in constant battle due to the ruthless variations of temperature, pressure, and the amount of particles.

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Episode I ATTACK OF THE GAS

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  1. Not so long ago, in a chemistry lab far far away… May the FORCE/area be with you Episode I ATTACK OF THE GAS Gas, being of upmost importance to the entire galaxy and your life, is in constant battle due to the ruthless variations of temperature, pressure, and the amount of particles. It is imperative that you understand properties of gases and how those ruthless variations affect the gases. You will need to make some minor assumptions to conquer this topic. However, there is certainty that you will prevail…

  2. Gas Laws: Finding “R”At the conclusion of our time together, you should be able to: Use the Ideal Gas Law to find the value of “R”.

  3. My Piggy Bank after my last stop at the gas station:

  4. Partial Pressure of CO2 Carbon Dioxide gas over water at 17.0 degrees C and 97.932 kPa. What is the pressure of the CO2 gas? 1.90 kPa 97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm

  5. Determine the Moles of CO2 Used: 0.575 g CO2 1 mol CO2 44.01 g CO2 0.0131 mol CO2

  6. Using Ideal Gas Law solve for “R” (0.948 atm) (0.347 L) (0.0131 mol) (R) (290 K) 0.0868 atm*L/mol*K

  7. % Error 0.08206 atm*L/mol*K (standard) 0.0868 atm*L/mol*K (experimental results) - 0.00474 atm*L/mol*K (experimental error) - 0.0048 atm*L/mol*K 0.08206 atm*L/mol*K x 100 - 5.85 % error

  8. Problems with Technology:

  9. Gas Laws: Finding “R”Let’s see if you can: Use the Ideal Gas Law to find the value of “R”.

  10. Determine “R” if: Butane gas was collected over water at 22.0 degrees C and 97.65 kPa. Determine the “R” value and the percent error if 0.33 g of butane was used and the gas collected was 155.0 mL of volume.

  11. Partial Pressure of C4H10 Butane gas over water at 22.0 degrees C and 97.65 kPa. 2.60 kPa 97.65 kPa – 2.60 kPa = 95.05 kPa = 0.938 atm

  12. Determine the Moles of C4H10Used: 0.33 g C4H10 1 mol C4H10 58.14 g C4H10 0.0057 mol C4H10

  13. Using Ideal Gas Law solve for “R” (0.938 atm) (0.1550 L) (0.0057 mol) (R) (295 K) 0.087 atm*L/mol*K

  14. % Error 0.08206 atm*L/mol*K (standard) 0.087 atm*L/mol*K (experimental results) - 0.00494 atm*L/mol*K (experimental error) - 0.005 atm*L/mol*K 0.08206 atm*L/mol*K x 100 - 6 % error

  15. Hittite Innovations:

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