Entropy Calculation Homework: Isothermal Free Expansion

AI Thread Summary
The discussion centers on a thermodynamics homework problem involving isothermal free expansion in a cylinder with a frictionless piston. Participants clarify that the final pressure after the piston is released is 1.5 atm, with the temperature remaining constant at 300 K. There is confusion regarding the calculation of entropy change, with one participant noting a discrepancy in their entropy calculation compared to the expected result of 0.0566 J/K. It is established that the process is not a free expansion since the piston remains in the cylinder, leading to the conclusion that it is a spontaneous expansion. Participants explore whether the gases can reach different temperatures and discuss the implications of temperature changes on the system's entropy.
cowcow8866
Messages
10
Reaction score
0

Homework Statement


A thermally insulated cylinder, closed at both ends, is fitted with a frictionless heat-conducting piston that divides the cylinder into two parts. Initially, the piston is clamped in the center with 1 liter of air at 300 K and 2 atm pressure on one side and 1 liter of air at 300K at 1 atm pressure on the other side. The piston is released and reaches equilibrium in pressure and temperature at a new position. Compute the final pressure and temperature and increase of entropy if air is assumed to be the ideal gas. What irreversible process has taken place?


Homework Equations


\DeltaS = \DeltaQ/T

The Attempt at a Solution


This is an isothermal free expansion. So temperature remains constant. I know the final pressure will be 1.5 atm (intuitively, (1+2)/2), but how can I compute this out by ideal gas law?

By PV=nRT,
(101325)(1000/1000000)=n (8.31)(300)
n1(the compartment with 1atm)=0.040643802
n2(the compartment with 2atm) = 0.081828
TOTAL number of moles = 0.12193

When I search through the Internet, I find this equation
\DeltaS=nRln(Vf/Vi) = 0.12193(8.31)ln2 = 0.702323...

but the answer is 0.0566, what's wrong with my calculation?
 
Physics news on Phys.org
Why did you use V_f/V_i=2? That ratio is supposed to represent the final and initial volumes.
 
Actually, from the phase'1 liter of air', i am not quite sure about it is about the volume or number of moles. Moreover, when I think this question again, it is not free expansion as the sentence "The piston is released and reaches equilibrium in pressure and temperature at a new position." provides the fact that the piston is still in the cylinder. Then another question comes, if it is not free expansion, then, what irreversible process has taken place.

Together with this question, i have one more question to ask...
Answer: 1.5 atm. 300K, 0.0566 J/K
For the first answer, the piston stops moving when the pressure of the two compartments are equal, that is 1.5 atm.
From the second answer, it seems that the temperature is assumed to be unchanged. However, we know that PV=nRT, it may be the case that the temperature increase.
Anybody can help me to clarify? It may be crucial in solving the increase in entropy.
Thanks You.
 
cowcow8866 said:
Actually, from the phase'1 liter of air', i am not quite sure about it is about the volume or number of moles.

The liter is a measure of volume. By applying an assumption (ideality), and knowing the pressure and temperature, one can calculate the amount of gas in moles.

cowcow8866 said:
Moreover, when I think this question again, it is not free expansion as the sentence "The piston is released and reaches equilibrium in pressure and temperature at a new position." provides the fact that the piston is still in the cylinder. Then another question comes, if it is not free expansion, then, what irreversible process has taken place.

It is not a free expansion (because the gas does not expand into a vacuum), but it is a spontaneous expansion.

cowcow8866 said:
From the second answer, it seems that the temperature is assumed to be unchanged. However, we know that PV=nRT, it may be the case that the temperature increase.
Anybody can help me to clarify? It may be crucial in solving the increase in entropy.
Thanks You.

Can the gases end up at different temperatures, considering the boundary between them?
Can the gases end up at an average temperature different from 300K, considering the boundary between the gases and the outside environment?
 
I can figure out the calculation of the increase in entropy now. However, I still cannot understand why the temperatures remain unchanged. Thank You.
 
I find it most useful to take a reductio ad absurdum approach here: assume the temperature does change in one or both of the chambers. What are the implications? Do these implications make sense, considering what we know about the barrier properties, the system boundary conditions, and the energy of an ideal gas as a function of temperature?
 
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