For all editions read the sections labeled:

Gas Stoichiometry

Dalton's Laws of Partial Pressures

The Kinetic Molecular Theory of Gases
(You can skip the derivation of PV=nRT)

Real Gases

Remember because our constant (R) has units of atm·Liter/mole·K, we need the psi converted to atm and the 77°F turned into Kelvin.

Problem 2: What is the number that goes into C3?
Problem 3: What is the weight of air in the 16 liter tank (N2)?
Problem 4: What is the formula that goes into N2?
Problem 5: If you wanted to figure the total weight of the tank at 3500 psi, what cell do you change?
Problem 6: If the temperature was 92°F instead of 77°F, what would the new formula in I3 become?

The fire piston was invented by people in the southeast asia and Indonesia over a thousand years ago. The principle is that compression of gas will cause the gas to heat up. The same principle is used in diesel engines to ignite the fuel. If done quickly, the heat doesn't have time to be lost and is retained in the cylinder.

My colleagues and I debated about how this compression actually makes the air hotter. I used a spreadsheet to calculate the speed one molecule would speed up if bouncing back and forth from the top to bottom of the cylinder as a piston came down. Each time the molecule hit the moving piston it bounced back a little faster. By the time the piston moved to about 1 cm from the bottom, that molecule had bounced off the moving piston many times making the molecule move at speeds equivalent to temperatures over a 1000°F.

The pressure in the fire piston will go up at the point the tender attached to the bottom of the pistion catches fire. (I used a piece from a cotton ball in my fire piston). When the cotton burns, it will consume the oxygen but will produce carbon dioxide and water vapor and higher temperatures. So the pressure should go up due to more gases and higher temperature.
Cotton is cellulose, which has the formula of
(C₆H₁₂O₅)_n. The "n" means it is a long chain of these glucose molecules. But we can treat it like it was burning C₆H₁₂O₅. Here's the balanced equation.

2C₆H₁₂O₅+13O₂ --> 12CO₂ + 12H₂O

We can't ignore the nitrogen gas, which is 5 times the number of oxygen molecules (5 x 13=65). So we can add that to the reaction.
2C₆H₁₂O₅ +13O₂ + 65N₂--> 12CO₂ + 12H₂O + 65N₂

Looking at this we see that we start with 78 moles (12+65) of gases and end with 84 moles (12+12+65) of gases. Also, the burning will increase the temperature. The yellow flame indicates a temperature around 3,000 Kelvin.

This problem is similar to the last one but the intitial conditions are the final conditions in the above problem.
That was P₂V₂=n₂RT₂

After the flame heats up the air and creates the extra gases, the condition is different. Let's use P₃V₃=n₃RT₃ for the new final condition.

Like before we can solve for R on both and set them equal to each other. This looks like the last time we did it:
P₂V₂/n₂T₂=P₃V₃/n₃T₃

This time the moles are changing, but the volume is the same. So we need to keep the moles (n₂ and n₃) but we can drop the volumes. That simplifies it to:
P₂/n₂T₂=P₃/n₃T₃

Solving for the final pressure (P₃) by multiplying both sides by n₃ and T₃ gives us:
P₂n₃T₃/n₂T₂ = P₃

Even though we don't know the exact number of moles, we do know the ratio of moles, which works fine when you have one divided by the other. So the 78 moles for n₂ and the 85 moles for n₃ that we got from the balanced equation works fine.

We can check the units to see if they cancel and we can check the logic. In the above spreadsheet we see that we have 85 moles over 78 moles. So that's 85/78, which will make the pressure larger as expected. We see the temperature ratio of 3000 over 873 or 3000/873, which will also make the pressure larger. So these fractions are doing what we expect should happen to the pressure which is to become larger when there's more moles and higher temperatures.
Problem 11: What is the pressure now after some of the cotton burns (L2)?
Problem 12: What is the pressure of L2 in atmospheres?