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

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Presentation on theme: "OB: review combined gas law math, continue in group work with the problem set. Reference tables, calculators, and lots of paper."— Presentation transcript:

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

2 Weve seen previously that pressure and volume of gases are inversely proportional. Weve also determined that pressure and temperature are directly proportional. Finally weve 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 P1V1T1P1V1T1 = P2V2T2P2V2T2 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. Its 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 theyre easy. Next slide, write the combined gas formula and fill in what we know.

5 P1V1T1P1V1T1 = P2V2T2P2V2T2 Your balloon is filled on the ground. Its 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 (1.20 atm)(45.6 L) 293 K (1.05 atm)(V 2 ) 278 K = Solve for P 2 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)(V 2 ) 278 K = Becomes… (1.20 atm)(45.6 L)(278 K) = (1.05 atm)(293 K)(V 2 ) Which changes to… (1.20 atm)(45.6 L)(278 K) (1.05 atm)(293 K) = V 2 Cancel all the units you can, then do the math…

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

8 Problem #2… At constant temperature, a sample of (H 2 S) dihydrogen monosulfide (stink gas) of 50.0 cm 3 and 125 kPa is put into a much larger container and it expands to 595 cm 3. What is the new pressure of this gas? Before we get too far, lets 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 cm 3 and 125 kPa is put into a much larger container and it expands to 595 cm 3. What is the new pressure of this gas? Lets just choose a temperature to use, say standard temp, and write in the formula first: P1V1T1P1V1T1 = P2V2T2P2V2T2 (125 kPa)(50.0 cm 3 ) 273 K (P 2 )(595 cm 3 ) 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: P 1 V 1 = P 2 V 2 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 cm 3 and 125 kPa is put into a much larger container and it expands to 595 cm 3. What is the new pressure of this gas? P 1 V 1 = P 2 V 2 (125 kPa)(50.0 cm 3 ) = (P 2 )(595 cm 3 ) 10.5 kPa = P 2 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: P 1 V 1 = P 2 V 2 At constant pressure, use: At constant volume, use: Or, just choose one number to fill in on both sides, it always works out! V1T1V1T1 = V2T2V2T2 P1T1P1T1 = P2T2P2T2

12 12 +1 =

13 Youre sitting in groups, lets 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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