How can the force of gases be calculated for explosive devices?

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The discussion revolves around calculating the force of gases produced by explosive devices, drawing parallels to the kinetic energy of bullets. Key considerations include the density, velocity, and mass of gases compared to bullets, with an emphasis on how these factors influence the energy delivered upon impact. The conversation highlights the challenges of applying Newton's laws to gas molecules due to their vast numbers and interactions, suggesting that while gases can be approximated as ideal, real-world conditions may cause deviations. The complexities of molecular motion, including translational, rotational, and vibrational energy, further complicate the analysis of explosive gases. Ultimately, understanding the behavior of gases in explosions requires a nuanced approach that accounts for both ideal and non-ideal conditions.
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Hunting season has arrived where I am located and I have had to choose a bullet/cartridge combination to hunt with. To do this I had to decide what range I would likely shoot, how much drop or drift I was willing to tolerate, the amount of recoil I could handle and lastly would the kinetic energy of the bullet be high enough to efficiently/ethically take down the game I was hunting. The last question deals with the amount of energy transferred to the target and essentially comes down to two things, how heavy is the bullet and at what velocity is it traveling when it hits the target. Hunters call this hitting power and the formula got me thinking, mainly about explosive's and how much of this is transferable to gasses. Now I know this is a touchy topic so right of the start I want to say I have no plans to build a bomb I am just wondering how something like this would be calculated.

With a bullet it's fairly simple. Heavier and faster equals more energy delivered on target, density and aerodynamics allow you to deliver that energy at larger range's because a dense bullet with a sleek profile bleeds off it's velocity slower. But what about a gas? A bullet's density, volume and weight are all set so all that is needed to calculate hitting power is a known initial velocity. In a explosion the density of the expanding gasses their velocity and heat start off high and dwindle. So I wondered If you don't know mass how can the energy being delivered be calculated?

I've been thinking about it for a little while and the best thing that I can come up with is molecules themselves, say an explosive detonates and produces a rapidly decompressing volume of co2, If you knew the detonation velocity could you not think of an individual CO2 molecule in the same sense as a bullet? it has a set mass does it not? how does this work? what if you have 2 types of explosives one rapidly decomposes into say helium and the other into co2, If they both expanded outward at the same velocity the CO2 explosion should impart more force then the helium explosion, right?

could some one explain this? lastly I know that explosions move outward in all directions expanding in volume spherically unlike a bullet which only travels in one direction. So I have been trying to think of it in terms of a gas expanding down a rifle barrel (or engine cylinder), where the expanding gas wave has a fixed face (say 1 square inch).

Can anybody explain this?, do Newton's laws apply to molecules? and are these two formulas in anyway related?
 
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Well you can approximate the molecules as particles however you're going to run into a problem. A gas doesn't consist of single molecules/atoms but on the order of 10^23 different particles so it going to be hard to keep track of every single one and then you have to take into account the interactions of all of the particles and you can see why this is a (kind of) bad way to approach the problem.

Approximately, you can treat both gases as ideal in which case they exert that same force given by the ideal gas law.

In reality though at high pressures (which I'm guessing you'd get in an explosion) would cause the ideal gas law to deviate, perhaps significantly.

In which case you can talk about compressibility so you are kind of right. Since CO2 can be compressed more than He.

But from a purely Newtonian approach you would come to the ideal gas law conclusion that both gases should be the same.
 
I suspect you couldn't just look at the properties of a single, idealized molecule. As Feldoh said, you would have to consider the system and how ideal it is to being with. Even if you just look at single gas particles you have to realize that it's not only the mass of the molecules that is different, but also the moment of intertia. Part of the energy is going to go into translational motion and some of it will be rotational motion. Molecules also vibrate, which uses energy as well. So there are several factors here that would affect how "fast" a molecule will be moving. For example, in your typcial beer sign you will have molecules with huge energies because the vibrational modes of the molecules are really excited, yet they won't be moving much faster than molecules at room temperature. This indicates that they are not at equilibrium. So when we want to know how fast a molecule moves relative to how it rotates and vibrates, we typically assume that the system is at equilibrium, which is achieved after a few nanoseconds when molecules collide with one another. Because of this, you can't really just consider a single molecule. At any rate, any kind of explosive device will probably not expel molecules with any kind of equilibrium distribution.

On the other hand, it's not all that difficult to consider monatomic gases since they behave like ideal particles.
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
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