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OB: review combined gas law math, continue in group work with the problem set.

OB: review combined gas law math, continue in group work with the problem set. Reference tables, calculators, and lots of paper. We’ve seen previously that pressure and volume of gases are inversely proportional.

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OB: review combined gas law math, continue in group work with the problem set.

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  1. OB: review combined gas law math, continue in group work with the problem set. Reference tables, calculators, and lots of paper.

  2. We’ve seen previously that pressure and volume of gases are inversely proportional. We’ve also determined that pressure and temperature are directly proportional. Finally we’ve seen that volume and temperature are also directly proportional. These are relationships that can be outlined with the combined gas law, found on the back of your reference tables. Look now.

  3. P1V1T1 P2V2T2 = The original conditions of pressure, volume, + temperature will equal The new conditions of pressure, volume, and temperature

  4. Problem #1 for today… Your balloon is filled on the ground. It’s 45.6 liters in size, it is filled with helium gas to a pressure of 1.20 atm, and the temperature of the gas is 20.0°C. The balloon rises into the atmosphere to 1000 yards, where the temperature dropped to just 5.00°C, and the pressure drops to 1.05 atm. What is the new volume of your balloon? Gas problems have lots of words, but they’re easy. Next slide, write the combined gas formula and fill in what we know.

  5. Your balloon is filled on the ground. It’s 45.6 liters in size, it is filled with helium gas to a pressure of 1.20 atm, and the temperature of the gas is 20.0°C. The balloon rises into the atmosphere to 1000 yards, where the temperature dropped to 5.00°C, and the pressure drops to 1.05 atm. What is the new volume of your balloon? FILL IN NOW P1V1T1 P2V2T2 = (1.20 atm)(45.6 L)293 K (1.05 atm)(V2)278 K = Solve for P2 by cross multiplying, cancel all units as you go. NOTE: temperature is always Kelvin, why????

  6. (1.20 atm)(45.6 L)293 K (1.05 atm)(V2)278 K = Becomes… (1.20 atm)(45.6 L)(278 K) = (1.05 atm)(293 K)(V2) Which changes to… (1.20 atm)(45.6 L)(278 K) (1.05 atm)(293 K) = V2 Cancel all the units you can, then do the math…

  7. (1.20 atm)(45.6 L)(278 K) (1.05 atm)(293 K) = V2 This becomes… 15212.16307.65 L = V2 49.4 liters = V2 With this combined gas law, as long as you know your starting conditions, you can change 2 conditions and calculate the third one. Remember, it’s always Kelvin, but any other units can be used for volume or pressure

  8. Problem #2… At constant temperature, a sample of (H2S) dihydrogen monosulfide (stink gas) of 50.0 cm3 and 125 kPa is put into a much larger container and it expands to 595 cm3. What is the new pressure of this gas? Before we get too far, let’s think about this. What temperature do we use? Do we use temperature? Think…

  9. Problem #2… At constant temperature, a sample of dihydrogen monosulfide gas (stink gas) of 50.0 cm3 and 125 kPa is put into a much larger container and it expands to 595 cm3. What is the new pressure of this gas? Let’s just choose a temperature to use, say standard temp, and write in the formula first: P1V1T1 P2V2T2 = (125 kPa)(50.0 cm3)273 K (P2)(595 cm3)273 K = We can automatically just cancel out the 273 K on both sides first, then do the math with less numbers and units. Or…

  10. With constant temperature, we can re-write the combined gas law without the temperature at all, and just use the rest of it, like this: P1V1 = P2V2 Either way, the math works out to the same answer. Try it yourself. So… At constant temperature, a sample of dihydrogen monosulfide gas of 50.0 cm3 and 125 kPa is put into a much larger container and it expands to 595 cm3. What is the new pressure of this gas? P1V1 = P2V2 (125 kPa)(50.0 cm3) = (P2)(595 cm3) 10.5 kPa = P2 Does this make sense? As volume goes up, pressure goes???

  11. The combined gas law is great for EVERY gas problem in our class. If one of the three (P, V, or T) is a constant, you can cancel it out of the formula and do simple math. For instance: At constant temperature, use: P1V1 = P2V2 At constant pressure, use: At constant volume, use: Or, just choose one number to fill in on both sides, it always works out! V1T1 V2T2 = P1T1 P2T2 =

  12. 12 +1 =

  13. You’re sitting in groups, let’s start the 54 gas problems set. This will take several days in class and at home. You should finish 12 today, here or at home. Paper is cheap, knowledge is valuable.

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