What is the Mass of Oxygen in a Leaking Tank?

  • Thread starter Thread starter CJIME017
  • Start date Start date
  • Tags Tags
    Mass Oxygen Tank
AI Thread Summary
The discussion centers on calculating the mass of oxygen in a leaking tank, initially filled at a gauge pressure of 2.80×10^5 Pa and a temperature of 39.0 Celsius. The correct volume of the tank is 8.0 x 10^-2 m^3, not 8.00 x 10^-3 m^3, which affects the calculations. The method for determining the number of moles using the ideal gas law is correct, but the absolute pressure must be used instead of gauge pressure. After correcting these errors, the initial mass of oxygen can be accurately calculated. The conversation highlights the importance of using absolute pressure and the correct volume in gas calculations.
CJIME017
Messages
2
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top