Clarification on PV=nRT question.

  • Thread starter Rizke
  • Start date
  • #1
5
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:

Answers and Replies

  • #2
21,489
4,867
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 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
(3.0^5)(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 im puzzled.
Do you know the definition of gauge pressure?
 
  • #3
5
1
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
21,489
4,867
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.
Add 100000 Pa to the gauge pressure to get the absolute pressure.
 
  • #5
5
1
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
21,489
4,867
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.
 
  • #7
5
1
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
 
  • Like
Likes Chestermiller

Related Threads on Clarification on PV=nRT question.

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
5K
  • Last Post
Replies
2
Views
23K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
Top