Solving a Nitrogen Can Homework Statement

  • Thread starter Thread starter songoku
  • Start date Start date
  • Tags Tags
    Homework Nitrogen
AI Thread Summary
The discussion revolves around solving a homework problem involving a nitrogen can with specific pressure, volume, and temperature conditions. Key points include the need to convert volume to cubic meters for accurate calculations, with the corrected volume at STP being approximately 2.73 x 10^-3 m^3. The mass of nitrogen was calculated correctly, but units were not specified, and the change in pressure over temperature requires re-evaluation based on the corrected volume. The energy per molecule of nitrogen was clarified to be calculated using the formula E = (3/2)kT, emphasizing the importance of using Avogadro's constant for precise results. Overall, the thread highlights the importance of unit consistency and accurate formula application in thermodynamic calculations.
songoku
Messages
2,467
Reaction score
382

Homework Statement


A can filled with nitrogen has pressure 10 Pa, volume 10 cm3, and temperature 300 K.
(i) Find the volume at STP
(ii) Find the mass of the nitrogen
(iii) Find the change of pressure over temperature at 300 K
(iv) Find the energy of each molecule of nitrogen

Homework Equations


PV = nRT

The Attempt at a Solution


(i)
\frac{P_1 V_1}{P_2 V_2}=\frac{n_1 R T_1}{n_2 R T_2}

Assuming n1=n2 : ------> Is this right ?

\frac{10*10}{10^5*V_2}=\frac{300}{273}

V_2=2.73 x 10^{-3}\; m^3(ii)
P1V 1= nRT1
10*10-5=n *8.31* 300
n = 4.011 x 10-8 mole

m = n*Mr = 4.011 x 10-8*28 = 1.12308 x 10-6(iii) Is the question asking the change at 300 K compared to STP ?

\frac{\Delta P}{\Delta T}=\frac{P_2-P_1}{V_2-V_1}=\frac{10-10^5}{10^{-5}-2.73 x 10^{-3}}\approx 3.68 x 10^7 \frac{Pa}{m^3} ??(iv)
E=\frac{3}{2}nRT

Not sure about using T = 300 K or T = 273 K (STP)

Thanks
 
Physics news on Phys.org
(i) Incorrect input value for volume - should be in m3. Also, the value "2.73" is suspicious. The ratio 300/273 is close to one and the other side is all powers of 10.

(ii) Mass calculation looks OK, but no units are given.

(iii) Method OK but calculation needs to be redone because it depends on answer in (i).

(iv) 300 K is not much different from 273 K. Expression is incorrect. You are asked to find the energy per molecule, not the total energy.
 
Hi kuruman

I've revised my answer for the (i), (ii), and (iii) according to your correction :)

For the last one :

\text{Energy per molecule}=\frac{3}{2}\frac{nRT}{N_A} ??

where NA = Avogadro's constant

Thanks
 
For the total energy of an ideal gas, I prefer the form

E = \frac{3}{2}N k T

where N is the number of molecules and k the Boltzmann constant. Then the energy per molecule is simply
\epsilon = \frac{3}{2}k T
 
Hi kuruman

Ok I get it now.

Thanks a lot for your help :smile:
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top