Pressure caused by explosion in closed container

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Homework Help Overview

The discussion revolves around calculating the excess pressure generated by an explosion of EGDN (C2H4NO6) in a sealed container, where only 20% of the container is filled with the explosive. The problem involves applying the ideal gas law and understanding the implications of the initial conditions and properties of EGDN.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between the volume of the explosive and the resulting gas pressure, questioning how the 20% fill affects calculations. There is also discussion about the state of the density provided (liquid vs. gas) and its relevance to the problem.

Discussion Status

The discussion is ongoing, with participants raising questions about the assumptions made regarding the density of EGDN and the state of the gas produced after the explosion. Some guidance has been offered regarding the interpretation of the density, but no consensus has been reached on how to proceed with the calculations.

Contextual Notes

Participants note that the problem is part of a competition, indicating that there may be constraints or specific rules that apply, though these are not fully detailed in the discussion.

Reluctant Atom
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Homework Statement


(C2H4NO6, EGDN) is an explosive with perfect attributes. We fill 20% of a sealed container with an EGDN. Assume that EGDN explodes and the created gas bahaves according to ideal gas law. What is the excess pressure in the sealed container?

My problem: I don't know where to utilize the fact that only 20% of the container has been filled with the explosives.

Known data
ρ = 1.49 g/cm3
T = 1006 °C
maximum pressure of a container: 395 Bar
M(EGDN) = 152 g/mol
RSpecific = 8.314 J / K . mol

Homework Equations


pV = nRT
n = m/M
ρ = m/V

The Attempt at a Solution


pV = nRT / substitute n = m/M
pV = m . R/M . T / get V to the right side
p = m/V . R/M . T / substitute ρ = m/V
p = ρ . R/M . T

ρ = 1.49 g/cm3 = 1490 kg/m3
R = 8.314 J / K . mol = 8.314 kg m2 / (s2 . K . mol)
M = 152 g/mol = 0.152 kg/mol
T = 1006 °C = 1 279.15 K

p = 104 249 546.8 kg / (m . s2) = Pa
p = 1042.495468 Bar

1042.495468 Bar - 395 Bar = 647,495468 Bar

I don't know where to apply the information that only 20% of the container has been filled with EGDN.
 
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Hello, Reluctant Atom and welcome to PF!

Reluctant Atom said:

The Attempt at a Solution


pV = nRT / substitute n = m/M
pV = m . R/M . T / get V to the right side
p = m/V . R/M . T / substitute ρ = m/V

Is the red volume the same as the blue volume? (Is the given density of EGDN the density for the liquid state or the gas state?)
 
TSny said:
Hello, Reluctant Atom and welcome to PF!
Is the red volume the same as the blue volume? (Is the given density of EGDN the density for the liquid state or the gas state?)

I believe the density is of the EGDN in liquid state. It hasn't been clarified in the assignment either. Just stated that the density of EGDN is 1.49 g/cm3 and I have no further information about what gas does the EGDN produce after explosion or its density.
 
Reluctant Atom said:
I believe the density is of the EGDN in liquid state. It hasn't been clarified in the assignment either. Just stated that the density of EGDN is 1.49 g/cm3 and I have no further information about what gas does the EGDN produce after explosion or its density.
Yes, the density given is for the liquid state. This would be the mass divided by the liquid volume. The volume in the liquid state is different than the volume in the gas state.
 
This is a problem from competition that is not over yet.
 
Patrikp said:
This is a problem from competition that is not over yet.
OK, thanks for the information.
I will not provide any further help on this question.
 

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