Spherical gas distributed symmetricaly over space

AI Thread Summary
The discussion centers on the application of thermodynamic principles to a spherical gas distribution. It begins with the energy equation for an infinitesimal spherical layer and the ideal gas law, leading to a derived equation involving pressure and volume changes. The main issue raised is the invalidity of integrating the derived equation over the entire gas volume due to changes in the scenario, particularly concerning pressure gradients and thermodynamic equilibrium. The participants emphasize that the number of moles, when altered, affects the validity of the equations used. Ultimately, the integration approach must consider the uniformity of pressure and temperature to maintain accuracy in the calculations.
Moara
Messages
43
Reaction score
5
Homework Statement
Since the problem is long, and my doubt is very specific, I'll show it in a picture, it is the problem 6. But summing up, the question is in the item ii) where it is asked to calculate the total thermal energy of the gas.
Relevant Equations
U=nCvT ; PV=nRT;
Screenshot_2019-11-09-13-17-08-541_com.google.android.apps.docs.jpg
.

We have that energy in a infinitesimal Spherical layer with number of mols dn is:
dU=Cv.T.dn (1)
But by the ideal gas law:
PV=nRT (2)
Differentiation gives:
PdV+VdP=RTdn (3)
(3) in (1) and using CV=3R/2 (monoatomic)
gives:
dU=3/2.(PdV+VdP) (4)
Integration of (4) over the whole gas will not give the result expected. Where is my mistake?
 
Physics news on Phys.org
For (2) ##PV=nRT ##, the ## n ## in that formula is the total moles in a volume ## V ## at pressure ## P ## and temperature ## T ##. Changing the number of moles by ## dn ## at temperature ## T ##, (##dn ## uniformly distributed throughout the volume ##V ##), will result in the formula you obtained when you differentiated both sides. The integration of both sides of the equation, where you are completely changing the scenario to which the equation applies, is simply invalid.
 
Last edited:
You say that equation 3 is valid, but it is not valid to integrate it over the whole volume ? This change of the scenario would be caused by what? Is it because of the gradient pressure, like if you were assuming all gas to be at the same thermodynamic equilibrium?
 
In the manner that you are using ## dn ##, you have a system at constant uniform pressure (and constant temperature), and ## n ## is proportional to ## V ##, so that ## dn=\frac{P \, dV}{RT} ##. Integrating both sides gives ##n=\frac{PV}{RT} ##.
 
  • Like
Likes Moara
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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