Is the Gas in the Tank Helium or Oxygen?

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SUMMARY

The discussion centers on determining whether the gas in a closed 2-ft³ tank is helium or oxygen based on given parameters: 0.30 lb of gas, a pressure of 12 psi, and a temperature of 80°F. The specific gas constants are Roxygen = 1554 ft-lb/slug*Rankine and Rhelium = 12419 ft-lb/slug*Rankine. The calculated gas constant R was found to be 477.391 ft²/(s²*Rankine), indicating a discrepancy with the known values for both gases. The discussion highlights the importance of unit consistency and proper application of the ideal gas law.

PREREQUISITES
  • Understanding of the ideal gas law and its application
  • Familiarity with specific gas constants for common gases
  • Knowledge of unit conversions, particularly between psi and lb/ft²
  • Basic thermodynamics concepts, including pressure, volume, and temperature relationships
NEXT STEPS
  • Review the ideal gas law and its application in real-world scenarios
  • Study unit conversion techniques between different pressure and volume units
  • Learn about specific gas constants for various gases and their implications in calculations
  • Explore thermodynamic properties of gases, focusing on density and specific weight calculations
USEFUL FOR

Students in engineering or physics, particularly those studying thermodynamics, gas laws, and fluid mechanics, will benefit from this discussion.

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Homework Statement


A 2-ft^{3} closed tank is filled with 0.30 lb (Weight) of gas, which is thought to be either oxygen or helium. A pressure gage attached to the tank reads 12 psi at a temperature of T=80F. Is it He or O_{2}? Why? Jeopardizing safety?


Homework Equations



\gamma=weight/volume

\gamma=\rho*g

where \rho is the specific density.

\rho=p/RT

where p is pressure, R is engineering gas constant, and T is temperature in Rankine.

The Attempt at a Solution



In this case, I'm looking for R. I know R_{oxygen}=1554 ft-lb/slug*Rankine and R_{helium}=12419 ft-lb/slug*Rankine.

So I know \gamma, which is 0.30-lb/2-ft^{3}=0.15-lb/ft^{3}

so, since \gamma=\rho*g=Pg/RT

and T(R) = 539.6 Rankine

R=Pg/\gammaT=((32.2 ft/s^{2})*(12psi*144 lbs/ft^{2}/psi))/((0.15-lb/ft^{3})*539.6 Rankine)

R=477.391 ft^{2}/(s^{2}*Rankine)

_____________________________________
I'm not sure where I'm going wrong with this, but it should either be 1554 or 12419... Oh, and my units are off too.
 
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Yes, your units are way off. However, the answer lies inside the units. One unit for the gas constant is;

(psia*ft^3)/(lbm*R)

Compare these units to the ones stated in your problem and try for the solution again using the ideal gas equation.

Also, you don't need to convert 0.30 lbs to slugs.

Thanks
Matt
 
Last edited:

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