Entropy Calculation Homework: Isothermal Free Expansion

In summary, when the piston is released, it expands to equilibrium at 1.5 atm. This process increases the entropy of the system.
  • #1
cowcow8866
11
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


[tex]\Delta[/tex]S = [tex]\Delta[/tex]Q/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
[tex]\Delta[/tex]S=nRln(Vf/Vi) = 0.12193(8.31)ln2 = 0.702323...

but the answer is 0.0566, what's wrong with my calculation?
 
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  • #2
Why did you use [itex]V_f/V_i=2[/itex]? That ratio is supposed to represent the final and initial volumes.
 
  • #3
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.
 
  • #4
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?
 
  • #5
I can figure out the calculation of the increase in entropy now. However, I still cannot understand why the temperatures remain unchanged. Thank You.
 
  • #6
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?
 

What is entropy and why is it important in thermodynamics?

Entropy is a measure of the disorder or randomness in a system. In thermodynamics, it is an important concept that helps us understand the direction and efficiency of energy transfer and transformations. It also plays a crucial role in the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time.

What is isothermal free expansion and how is it related to entropy?

Isothermal free expansion is a process in which a gas expands into a vacuum without any change in temperature. This type of expansion is considered reversible and there is no change in internal energy. However, the entropy of the system does increase, as the gas molecules are now spread out and in a more disordered state.

How do you calculate the change in entropy during an isothermal free expansion?

The change in entropy during an isothermal free expansion can be calculated using the equation ΔS = nRln(V2/V1), where n is the number of moles of gas, R is the gas constant, and V2/V1 is the ratio of the final volume to the initial volume.

What factors can affect the change in entropy during an isothermal free expansion?

The change in entropy during an isothermal free expansion can be affected by the initial and final volumes of the gas, as well as the temperature. A larger volume expansion will result in a larger change in entropy, while a lower temperature will result in a smaller change in entropy.

Can the change in entropy during an isothermal free expansion ever be negative?

No, the change in entropy during an isothermal free expansion can never be negative. This is because the expansion always results in an increase in the disorder of the system, leading to a positive change in entropy. However, the magnitude of the change in entropy can be very small if the expansion is not significant.

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