What you learn today
- Pressure, volume, temperature and moles
- rgas
- Why heating raises pressure
A full oxygen tank stands next to the furnace. An hour later the analyzer shows 20 % more pressure although nobody refilled it. No leak, no bug – just physics. If you control gases you need to know this relationship.
Four quantities
| Quantity | Unit | LogicType |
|---|---|---|
| Pressure p | kPa | Pressure |
| Volume V | litres | Volume (pipe analyzer) |
| Temperature T | kelvin | Temperature |
| Amount n | mol | TotalMoles |
A straight gas pipe holds 10 litres, a liquid pipe 20 litres. The pipe analyzer reports the volume of the whole pipe network it sits on – tanks included.
The gas law
All four are linked by one formula:
p · V = n · R · T
R is the gas constant, on the chip the constant rgas (≈ 8.314 J/(mol·K)). The units fit without conversion: kPa times litres gives joules. Solved for the amount:
mul n p v # p * V
div n n rgas # / R
div n n t # / TWith that you can work out, for example, how many moles of oxygen are in a tank when you only know pressure, temperature and volume.
Why heating raises the pressure
In a closed tank V and n stay the same. Then the pressure is proportional to the temperature: double the kelvin, double the pressure. The new pressure is
p₂ = p₁ · T₂ / T₁
div factor tNew tOld
mul pNew p factorFrom 20 °C (293.15 K) to 50 °C (323.15 K) the pressure therefore rises by a good 10 %. A tank already filled close to its limit can burst when it warms up.
The gas law always takes kelvin. With Celsius the formula is wrong, and at 0 °C you even divide by zero. A sensor delivers Temperature in kelvin anyway – convert to Celsius only for the display.
Internally the game uses R = 8.3144. The constant rgas has a few more digits; the difference is below one in a hundred thousand and does not matter for any control loop.
Exercise
Task
A pipe analyzer “Tank” sits on the pipe network of a tank. Two LED displays are named “Mol” and “Druck50”.
- “Mol” shows the amount of gas in the network: n = p · V / (rgas · T)
- “Druck50” shows the pressure the tank would have at 50 °C (323.15 K)
- If the analyzer is gone, both displays show −1
The readings are already there; the calculation and the output are missing.
Hints
Hint 1
Amount: mul n p v, then two div – by rgas and by t.
Hint 2
Pressure at 50 °C: first div pNew TARGET t, then multiply by p.
Hint 3
Without the analyzer t is NaN, so ok is 0. select n ok n -1 then sets −1.
Quiz
4 questions. You pass from 60 % – the server does the grading.
Complete the day
- Passed the simulator exercise
- Passed the quiz with at least 60 %
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