How Does Adding Nitrogen Affect Pressure in a Sealed Tank?

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SUMMARY

The discussion focuses on calculating the volume and pressure of nitrogen in a sealed tank using the ideal gas law. Initially, 18.5 kg of nitrogen results in a calculated volume of approximately 14.8 m³. When an additional 15 kg of nitrogen is added at constant temperature, the pressure is computed to be 183.44 N/m², which is incorrect as the expected answer is 1.83 x 10^5 Pa. The discrepancy arises from a misunderstanding of unit conversions and the application of the ideal gas law.

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  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of molar mass calculations (e.g., nitrogen's molar mass is 28 g/mol)
  • Familiarity with unit conversions (e.g., converting liters to cubic meters)
  • Basic principles of dimensional analysis
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  • Practice unit conversion techniques, especially between different volume measurements
  • Study dimensional analysis to ensure accuracy in calculations
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Students in chemistry or physics, engineers working with gas systems, and anyone interested in thermodynamics and pressure calculations in sealed environments.

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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
 
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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
 
Regarding Part b: From the ideal gas law, if the temperature and volume are held constant, the pressure is proportional to the number of moles.

Chet
 
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