How Does Adding Nitrogen Affect Pressure in a Sealed Tank?

[Myr73]

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

 

[Simon Bridge]

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.

 

[Myr73]

Oh I see, I should have used V in m^3 not in L. Thank you

 

[Chestermiller]

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

 

[HalfLostAndSearching]

.

I would like to clarify that the units for pressure are indeed Pascal (Pa), which is equivalent to N/m^2. In your calculation, you have correctly used the units of N/m^2, but the answer given in the problem is likely in scientific notation, where 1.83 X 10^5 Pa would be equivalent to 183,000 Pa or 183 kPa. It is important to pay attention to the units and the correct use of scientific notation in order to accurately interpret and communicate scientific data.