Is the Joule-Kelvin Expansion Reversible?

AI Thread Summary
The Joule-Kelvin expansion, also known as the Joule-Thompson effect, involves the free expansion of gas through a throttle valve. This process is quasistatic but is not reversible due to the necessity of a finite pressure difference across the throttle valve. The expansion relies on the presence of attractive forces between gas molecules, which is essential for the cooling effect to occur. Since throttling inherently involves irreversibility, the Joule-Kelvin expansion cannot be considered a reversible process. Thus, the conclusion is that the Joule-Kelvin expansion is not reversible.
bon
Messages
547
Reaction score
0

Homework Statement



Is the Joule Kelvin expansion reversible or not?



Homework Equations





The Attempt at a Solution



It\'s definitely quasistatic but I\'m not sure if it is reversible or not? I have a feeling it\'s not but can;t explain why...any ideas?
 
Physics news on Phys.org
bon said:

Homework Statement



Is the Joule Kelvin expansion reversible or not?

...

It\'s definitely quasistatic but I\'m not sure if it is reversible or not? I have a feeling it\'s not but can;t explain why...any ideas?
The Joule Kelvin or Joule-Thompson cooling effect requires letting the gas expand freely using a throttle valve. (It only works with gases having attractive forces between molecules.). Throttling is not a reversible process since it requires more than an infinitessimal pressure difference between the two sides of the throttle valve.

AM
 
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