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Showing posts with label Science. Show all posts
Showing posts with label Science. Show all posts

Chemistry and Science Explanation of Luthang Gapuyan or Fire Piston


Fire piston was invented in "Dipag" now Dipolog City in the modern Zamboanga del Norte, island of Mindanao, the Philippines over 1,500 years ago by a Subanen tribe teenager named “Anlangan”.

Scientific Explanation of Fire Piston


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?

Problem 7: What cells need to be updated to do that?

Problem 8: What is the final pressure in mm of mercury (mm of Hg)?

Problem 9: What is the formula that goes in L2?

Problem 10: What would be the final pressure if the end volume was 0.7mL?

Problem 11: What is the pressure now after some of the cotton burns (L2)?

Problem 12: What is the pressure of L2 in atmospheres?

The volume of the air in the fire piston cylinder starts at 10.0 mL and then is squeezed to 1.0 mL. The temperature started as 25°C (room temp) and ended up at 600.°C. The pressure before being squeezed was 740mm of mercury (Note 760mm Hg is 1 atmosphere). We want the final pressure. If temperature had remained the same, this would be easy. The volume went down to 1/10 its original size, which would make the pressure be 10 times larger (740mm x 10=7,400mm mercury). However, the temperature changed, so it's more complicated. We can start with PV=nRT; however, there are two conditions. One at the beginning and one at the end. So we need two PV=nRT formulas. Let P1V1=n1RT1 be the values before it was compressed. Then P2V2=n2RT2 would be the values after compression and becoming hot. Since R is constant, it is the same in both equations. We can exploit that fact. Let's solve both equations for R. In the first R=P1V1/n1T1, and the second is R=P2V2/n2T2. Since both are equal to R, they are equal to each other. So, P1V1/n1T1=P2V2/n2T2. Since the moles of the gas didn't change, n1=n2, we can multiply both sides by n1, which would cancel out both n1 and n2. Our equation now reads:


We know all of these values except for P2 (final pressure). So lets solve for P2 by dividing both sides by V2 and multiplying both sides by T2. We now get

P1V1T2/(T1V2)=P2, which can also be written as P1x V1x T2 /T1 /V2=P2. This looks like a good job for a spreadsheet. Since there's no R in the formula, we don't need pressure measured in atmospheres or volume in liters, but we do have to use Kelvin. So 273 gets added to the Celsius degrees.

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:


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:


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. Read more at Chemistry land