Methane Cylinder Pressure Under Fire: 1080atm/1568psi

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SUMMARY

The discussion focuses on calculating the pressure of a methane gas cylinder under extreme temperature conditions, specifically from 200 atm at 0°C to 1080 atm at 1200°C. The ideal gas law, represented as PV=nRT, is utilized to derive the final pressure, confirming that the relationship P1/T1 = P2/T2 holds true. The calculations indicate that the pressure significantly increases due to the rise in temperature, demonstrating the behavior of methane as an ideal gas under these conditions.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV=nRT)
  • Knowledge of unit conversions between atmospheres and psi
  • Familiarity with temperature scales (Celsius to Kelvin)
  • Basic principles of thermodynamics related to gas behavior
NEXT STEPS
  • Study the implications of the Ideal Gas Law in real-world applications
  • Learn about gas behavior under varying temperature and pressure conditions
  • Explore advanced thermodynamic concepts such as non-ideal gas behavior
  • Research safety protocols for handling high-pressure gas cylinders
USEFUL FOR

Engineers, safety professionals, and anyone involved in the handling or storage of high-pressure gases, particularly those working with methane and other hydrocarbons.

ElectricMile
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Consider a high pressure gas cylinder of methane which exhibits a pressure of 200 atmosphers (absolute) at a temp. of 0*C in a small storage building. the building catches fire whcih causes the temp of the cylinder to rise to 1200*C. What would the pressure in the cylinder be then? express your answer in both atmopheres and in psi(lbf/in^2), assume methane behaves as an ideal gas.

ok. seems easy. but I am confused about this whole ideal gas thing. do i always use (0.08205 atmos. liters/ mole K) for R?

if so then would i do...

(P1)(V)=(R)(T1) = (200)(V)=(0.08205)(273)
so, V = ((0.08205)(273))/(200) = 0.1119
but this number seems unlikley. but if so
(P2)(V)=(R)(T2) = (P2)(0.1119)=(0.08205)(1473)
so. P2 = 1080atm? am i doing this somewhat right?
 
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you don't need to evaluate the volume in this problem, since you don't have the molar number.
[tex]PV=nRT[/tex] implies [tex]P/T=nR/V[/tex]
whereas the right hand side is constant... therefore, you have
[tex]P_{i}/T_{i}=P_{f}/T_{f}[/tex]
 
The result looks good now pay attention with unit conversion...

Daniel.
 

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