# Clarification on PV=nRT question.

1. Apr 22, 2017

### Rizke

• Does not use the template because it was originally posted in a non-homework forum
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 cant 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 im puzzled.

Last edited: Apr 22, 2017
2. Apr 22, 2017

### Staff: Mentor

Do you know the definition of gauge pressure?

3. Apr 22, 2017

### Rizke

oh thats 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.

4. Apr 22, 2017

### Staff: Mentor

Add 100000 Pa to the gauge pressure to get the absolute pressure.

5. Apr 22, 2017

### Rizke

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.

6. Apr 22, 2017

### Staff: Mentor

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.

7. Apr 22, 2017

### Rizke

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