# Lifting gas pressure inside rigid airship

I guess I'm asking the wrong question. Let me rephrase with another example.

Assume I have a bottle of liquid hydrogen. It's connected to a non-porous balloon with no elasticity made of, say mylar. The maximum volume of that balloon is 1,000 cubic meters. I open the valve on my hydrogen and the balloon begins to fill. Assuming the balloon never breaks, for every liter of liquid hydrogen I let escape, what is the pressure inside that balloon? How does that pressure change based on temperature? And assuming the maximum volume of that balloon won't increase or decrease, based on altitude will the internal pressure change (I assume it won't)?

Please, I need a simplified mathematical formula I can plug into game code that gives me this pressure. This formula will represent the filling of gas bags in a rigid airship and I basically need to know when to make them rupture based on over/under inflation.

I know this has to do with the ideal gas law. But, for the past 2 days I've been researching and trying to get a working formula with no success.

Thank you.

• More info is required. How rigid are the walls of the gas cell? Does the gas cell expand as the gas pressure increases? At what difference in pressure between the gas cell and the atmosphere does the gas cell rupture? Do you already have equations for pressure vs. height and temperature vs. height for the simulated atmosphere surrounding your airship? This information is needed in order to use the ideal gas law (PV=nRT) to answer your questions. – David White Aug 5 '18 at 15:52
• For simplicity's sake, assume the gas cells do not expand or contract and have a constant max volume. The rupture point can be any arbitrary constant but will change with the size/volume of the gas bags. Atmospheric pressure and temperature based on altitude are already in the game. As I understand it, the pressure in a non-rigid airship would remain mostly constant as the gas bags expand or contract and I could calculate a rupture point based on the strechability of the material. The rigid is what's causing me headaches as the player could put it in a vacuum. – uPrompt Aug 5 '18 at 16:28
• Thinking about it, if this makes the calculations easier, assume the gas bag volume can increase or decrease up to the maximum gas bag volume. The entire airship will need to be able to handle pressures up to 5+ atmospheres and down to a vacuum. If the internal gas pressure is too much higher or too much lower than the external pressure, they need to rupture. – uPrompt Aug 5 '18 at 16:51
• How can your airship be "rigid" if the gas cells expand? Also, use the ideal gas law to calculate the pressure of the lifting gas inside the airship. – David White Aug 5 '18 at 17:18
• Again, whatever works. Assume it's a balloon inside a box if need be. I've been trying to use the ideal gas law and the results I get don't make sense. I've update the OP to show what I've been trying. – uPrompt Aug 5 '18 at 17:36

Despite the title, the balloon in question seems to be unstretchable but flexible, not rigid. Which means that, for an increasing amount of gas, the pressure $P$ inside the balloon equals the external pressure $P_\text{ext}$ up to the point where its volume $V$ reaches the full volume of the balloon, $V_\text{max}$. In particular, if placed in vacuum, any amount of gas will fully inflate the balloon.
In this situation, for $n$ moles of an ideal gas at constant temperature $T$, we have:
$$P = \begin{cases} P_\text{ext} &\text{for } V<V_\text{max}\\ nRT/V_\text{max} &\text{for } V\ge V_\text{max}, \end{cases}$$
where $R$ is the ideal gas constant. In practice, this means:
$$P = \max(P_\text{ext},\, nRT/V_\text{max}).$$
• @uPrompt Yes, you'll need the number of moles $n$, which is about half the number of grams of gas (for molecular hydrogen, H$_2$). – stafusa Aug 6 '18 at 12:52