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Homework Help: Clarification on PV=nRT question.

  1. Apr 22, 2017 #1
    • 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. jcsd
  3. Apr 22, 2017 #2
    Do you know the definition of gauge pressure?
  4. Apr 22, 2017 #3
    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.
  5. Apr 22, 2017 #4
    Add 100000 Pa to the gauge pressure to get the absolute pressure.
  6. Apr 22, 2017 #5
    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.
  7. Apr 22, 2017 #6
    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.
  8. Apr 22, 2017 #7
    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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