Clarification on PV=nRT question.

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The discussion revolves around calculating the initial mass of oxygen in a tank using the ideal gas law (PV=nRT). The welder's tank has a gauge pressure of 3.00×10^5 Pa, but to find the correct mass, the absolute pressure must be used, which involves adding atmospheric pressure (approximately 100,000 Pa) to the gauge pressure. After correcting the pressure to 4.0×10^5 Pa and using the given volume and temperature, the calculation yields an initial mass of approximately 0.374 kg of oxygen. Participants clarify the importance of using absolute pressure for accurate calculations and confirm that adding atmospheric pressure is necessary for gauge readings. The final consensus is that the correct initial mass of oxygen is indeed 0.374 kg.
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A welder using a tank of volume 7.50×10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.00×10^5 Pa and temperature of 36.5 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C, the gauge pressure of the oxygen in the tank is 1.85×10^5 Pa . Find the initial mass of oxygen. correct answer : .374 kg
I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT
P = 3.0e^5 - in pascals
V = 7.5e^-2
R = 8.3144621
T = (36.5 C + 273.15)= 309.65 K
(3.0e^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles
mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.
So yeah I am puzzled.
 
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Rizke said:
A welder using a tank of volume 7.50×10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.00×10Pa and temperature of 36.5 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C, the gauge pressure of the oxygen in the tank is 1.85×105 Pa . Find the initial mass of oxygen. correct answer : .374 kg
I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT
P = 3.0e^5 - in pascals
V = 7.5e^-2
R = 8.3144621
T = (36.5 C + 273.15)= 309.65 K
(3.0^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles
mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.
So yeah I am puzzled.
Do you know the definition of gauge pressure?
 
oh that's right a gauge will read zero, but there is actually the ATM pressure that is on the gauge. which is like 15 right? so i have to subtract that pressure after i convert it to pascals. or is this not right? Thanks for the quick reply.
 
Rizke said:
oh that's right a gauge will read zero, but there is actually the ATM pressure that is on the gauge. which is like 15 right? so i have to subtract that pressure after i convert it to pascals. or is this not right? Thanks for the quick reply.
Add 100000 Pa to the gauge pressure to get the absolute pressure.
 
P = (3.0e^5)+100000 = 4.0e^5
(4.0e^5)(7.5e^-2) / (8.3144621)(309.65) = 11.6524 moles => 11.65*32 = 372.89 grams => .373 kg
Awesome Thank you!
p.s can i always add 100000 Pa for absolute pressure? that is if i already convert ATM to PA.
 
Rizke said:
P = (3.0e^5)+100000 = 4.0e^5
(4.0e^5)(7.5e^-2) / (8.3144621)(309.65) = 11.6524 moles => 11.65*32 = 372.89 grams => .373 kg
Awesome Thank you!
p.s can i always add 100000 Pa for absolute pressure? that is if i already convert ATM to PA.
If the actual atmospheric pressure is known (rather than 100000 Pa), you add that to get the absolute pressure. Otherwise, as an estimate, you use 100000 Pa.
 
Okay i think i finally understand why we add to the pressure. Assuming the welder in this problem is at a location where there is one ATM pressure, the gauge does not account for this. So we need to add that 1 ATM, or 101325 PA to the final pressure reading. i get .374 kg
 
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