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. We’ve also seen 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.

P1V1T1P1V1T1 = P2V2T2P2V2T2 The original conditions of pressure, volume, + temperature will equal The new conditions of pressure, volume, and temperature Always use Kelvin, other units do not matter as long as they will cancel each other out in the math

1. 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 293 K (20.0°C). The balloon rises into the atmosphere, and the temperature drops to 278 K (5.00°C); the pressure drops to 1.05 atm. What’s the new volume of your balloon? Gas problems have lots of words, but they’re easy.

P1V1T1P1V1T1 = P2V2T2P2V2T2 1. 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 293 K (20.0°C). The balloon rises into the atmosphere, and the temperature drops to 278 K (5.00°C); the pressure drops to 1.05 atm. What’s the new volume of your balloon? (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????

(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…

(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, it’s always Kelvin, but any other units can be used for volume or pressure

#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, let’s think about this. What temperature do we use? Do we use a temperature? Think…

Let’s 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… P 1 V 1 = P 2 V 2 Cancel the temperature before you put in all of those numbers, it will cancel out either way!

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 H 2 S 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???

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! Put these on the back of your reference table too. V1T1V1T1 = V2T2V2T2 P1T1P1T1 = P2T2P2T2

You’re sitting in groups, let’s start the Gas Problems Set. This will take several days in class and at home. You should finish at least 12 today, here or at home. Paper is cheap, knowledge is valuable.