How Does Adding Nitrogen Affect Pressure in a Sealed Tank?

  • Thread starter Thread starter Myr73
  • Start date Start date
  • Tags Tags
    Storage Tank
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 14K views
Myr73
Messages
120
Reaction score
0

Homework Statement

[/b][/B]


a Storage tank at STP contains 18,5kg of nitrogen (N2). a) What is the volume of the tank? b) What is the pressure if an additional 15.kg of nitrogen is added without changing the temperature.


Homework Equations


n= m/molar mass
P= nRT/V

The Attempt at a Solution



For a) I got n= m/molar mass --> 18.5 X 1000 / 28 = 660.71 mol
1 mole occupies 22.4 L
--> V= 660.71 X 22.4 = 14 799.99 L 14.8 m^3

For b) I have T= 273.15 n= (18.5 +15) x 1000 /28= 1196 mol
i think for some reason V is assumed to be constant and doesn't change so ,
--> P= nRT/V, which gives me 183.44 N/m^2

However in b the answerr is suppose to be 1.83 X10^5 Pa, or 183 X 10^ 3 Pa, ans I thought n/m^2 is the same units as Pa. I would like clarification, thanks
 
Physics news on Phys.org
Myr73 said:
For b) I have T= 273.15 n= (18.5 +15) x 1000 /28= 1196 mol
i think for some reason V is assumed to be constant and doesn't change so ,
--> P= nRT/V, which gives me 183.44 N/m^2
"for some reason"??
Don't you know?!
i.e. does the tank change size?

However in b the answerr is suppose to be 1.83 X10^5 Pa, or 183 X 10^ 3 Pa, ans I thought n/m^2 is the same units as Pa. I would like clarification, thanks
So you are out by a factor of 1000?

Take a closer look at your numbers. Do a dimensional analysis to check the units.
i.e. show me the exact calculation - you missed a term or divided by 1000 when you shouldn;t have or forgot an exponent.

Note: the calculation is easier is you compare the state equations before and after the extra mass is added.
 
Oh I see, I should have used V in m^3 not in L. Thank you