What Is the Equilibrium Temperature in an Adiabatically Isolated System?

AI Thread Summary
An adiabatically isolated system with 1 mol of diatomic gas at 100K and 2 mol of solid at 200K reaches an equilibrium temperature of 171K after the removal of a rigid adiabatic wall. The calculations involve using the Dulong-Petit law to relate the energy of the solid to its temperature change. The gas's heat capacity is expressed as Qgas = 5/2 nRT, considering its degrees of freedom. Some participants noted a lack of clarity in the calculations and the application of the Dulong-Petit law. Understanding the relationship between the solid's energy and temperature change is crucial for solving similar problems.
sodaboy7
Messages
81
Reaction score
0

Homework Statement


An adiabatically isolated system consists of 1 mol of a diatomic gas 100K and 2 mol of a solid at 200K separated by a rigid adiabatic wall. Find the equilibrium temperature of system after adiabatic wall is removed, assuming that the solid obeys the Dulong-Petit law.


Homework Equations


mgascgas(T-100) = msolidcsolid(200-T)

mgascgas = Qgas = 5/2 nRT ( 5 is degree of freedom, I guess so)

The Attempt at a Solution



5/2 nRT = Q = msolidcsolid(200-T)

4. Answer

171K
 
Physics news on Phys.org
sodaboy7 said:

Homework Statement


An adiabatically isolated system consists of 1 mol of a diatomic gas 100K and 2 mol of a solid at 200K separated by a rigid adiabatic wall. Find the equilibrium temperature of system after adiabatic wall is removed, assuming that the solid obeys the Dulong-Petit law.


Homework Equations


mgascgas(T-100) = msolidcsolid(200-T)

mgascgas = Qgas = 5/2 nRT ( 5 is degree of freedom, I guess so)

The Attempt at a Solution



5/2 nRT = Q = msolidcsolid(200-T)

4. Answer

171K
I agree with that answer, though you didn't really show how you got it. You haven't shown how you used the Dulong-Petit law, and msolid shouldn't be part of the calculation.
 
The answer was given on back pages of the book. I have not got any answer yet. I don't know how they got that answer.
 
What does your book or class notes say about the Dulong-Petit law? You should look that up, it relates the energy of the solid and its temperature change.
 
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