What is the Mass of Oxygen in a Leaking Tank?

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SUMMARY

The discussion focuses on calculating the mass of oxygen in a leaking tank with a volume of 8.00 x 10^-2 m^3, a gauge pressure of 2.80 x 10^5 Pa, and a temperature of 39.0°C. The correct approach involves using the ideal gas law, pV = nRT, to find the number of moles (n) and subsequently the mass (m) of oxygen. The initial calculation yielded an incorrect mass due to a misunderstanding of gauge pressure and an error in the tank volume. The final mass of oxygen calculated is 276 grams, or 0.276 kg.

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  • Familiarity with gauge pressure vs. absolute pressure
  • Basic unit conversions (e.g., cubic meters to liters)
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Homework Statement



A welder using a tank of volume 8.00X10^-3 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol) at a gauge pressure of 2.80×10^5 Pa and temperature of 39.0 Celsius. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.3 Celsius, the gauge pressure of the oxygen in the tank is 1.90×10^5 Pa.

1) Inital Mass of the Oxygen.
2) Find the mass of the oxygen that has leaked out.

Homework Equations



pV = nRT
n = pv/RT
m = Mn; M = Molar Mass

The Attempt at a Solution



1)
n = pv/RT
n = (2.80*10^5)(.08)/(8.314)(312) = 8.635 mol

m = Mn
m = (32 g/mol)(8.635 mol) = 276g = .276 kg

I got this answer and it was incorrect, I don't know where I went wrong...

2) I need the answer to 1) before I can attempt this.
 
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The method you applied is correct, but the gauge pressure is the pressure relative to the local atmospheric or ambient pressure, and you need the absolute pressure of the gas. The other mistake is that 8.00X10^-3 m^3 is not 0.08 m^3.

ehild
 
Ah yes, I'm sorry I didn't update the question, I figured it out already. And also, it was 8.0 x 10^-2 not -3, that was a mistake. Thank you for your time, I do appreciate it.
 

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