Thermodynamic entropy

In summary, the change in entropy of a system when a partition is given depends on the distinguishability of the gases in each compartment. If the gas molecules are indistinguishable, there will be no change in entropy when the partition is removed. If the gases are different, the change in entropy can be calculated by finding a reversible path from the initial to final states and solving for ##\int dQ/T## along that path for each gas on both sides of the partition."
  • #1

Homework Statement


Box divided by a partition into two equal compartments containing ideal gas.
Each compartment is having volume V ,temp T and pressure P
1.entropy of the system when the partition is given?
2.entropy of the system when the partition is removed.
[/B]

Homework Equations

The Attempt at a Solution


For the first part I have tried from 1st law Tds=Cvdt+Pdv
And solving S becomes CvlnT +RlnV
Is it right?
For the second part I need some hints to actually proceed
Thanks[/B]
 
Physics news on Phys.org
  • #2
The correct answers depend on whether the gases in each compartment are distinguishable. If the gas molecules are indistinguishable there is no change in entropy when the partition is removed. If there is, the gas in each side effectively undergoes adiabatic free expansion, each doubling its volume and halving its pressure (but doing no external work, so T remains constant).

Assuming they are different gases on each side of the partition, to calculate the change in entropy one must find a reversible path from the initial to final states. The change in entropy is defined as ##\int dQ/T## along a reversible path between these states. An example of a reversible path would be a reversible isothermal expansion to twice the initial volume (ie. very slow expansion in which work is done very slowly on the surroundings and there is gradual heat flow into the gas to keep the gas at constant T). Then it is just a matter of working out ##\int dQ/T## along such a path for the gas on each side.

AM
 
Last edited:
  • #4
Andrew Mason said:
The change in entropy is defined as ##\int TdS## along a reversible path
Typographical error for the integral expression?
 
  • #5
rude man said:
I don't think there's a change in entropy. Chet?
The problem statement is not very clear. If the same gas is present on both sides of the partition but the temperatures and/or the pressures are initially different on the two sides of the partition, then there will be a change in entropy. If there are different gases on both sides of the partition, but the temperatures and pressures are initially the same on the two sides of the partition, then there will also be a change in entropy. So, what the OP needs to do is provide an exact statement of the problem, and not just his interpretation of the problem statement.
 
  • #6
TSny said:
Typographical error for the integral expression?
Yes. Sorry about that. I should have said ##\Delta S = \int dS = \int dQ/T## over a reversible path.

AM
 
  • #7
Chestermiller said:
The problem statement is not very clear. If the same gas is present on both sides of the partition but the temperatures and/or the pressures are initially different on the two sides of the partition, then there will be a change in entropy. If there are different gases on both sides of the partition, but the temperatures and pressures are initially the same on the two sides of the partition, then there will also be a change in entropy. So, what the OP needs to do is provide an exact statement of the problem, and not just his interpretation of the problem statement.
same gas is present with initially same temp T and same pressure P
 
  • #8
Andrew Mason said:
The correct answers depend on whether the gases in each compartment are distinguishable. If the gas molecules are indistinguishable there is no change in entropy when the partition is removed. If there is, the gas in each side effectively undergoes adiabatic free expansion, each doubling its volume and halving its pressure (but doing no external work, so T remains constant).

Assuming they are different gases on each side of the partition, to calculate the change in entropy one must find a reversible path from the initial to final states. The change in entropy is defined as ##\int dQ/T## along a reversible path between these states. An example of a reversible path would be a reversible isothermal expansion to twice the initial volume (ie. very slow expansion in which work is done very slowly on the surroundings and there is gradual heat flow into the gas to keep the gas at constant T). Then it is just a matter of working out ##\int dQ/T## along such a path for the gas on each side.

AM
the exact statement of the problem is given below
15275694239441270586883.jpg
 

Attachments

  • 15275694239441270586883.jpg
    15275694239441270586883.jpg
    52.3 KB · Views: 696
  • #9
Chestermiller said:
The problem statement is not very clear. If the same gas is present on both sides of the partition but the temperatures and/or the pressures are initially different on the two sides of the partition, then there will be a change in entropy. If there are different gases on both sides of the partition, but the temperatures and pressures are initially the same on the two sides of the partition, then there will also be a change in entropy. So, what the OP needs to do is provide an exact statement of the problem, and not just his interpretation of the problem statement.
the exact statement of the problem is given below
 

Attachments

  • P_20180529_101959.jpg
    P_20180529_101959.jpg
    57 KB · Views: 426
  • #10
Apashanka das said:
the exact statement of the problem is given below
If that is the problem statement, then there is no change in entropy between the initial and final states of the system.
 
  • #11
Apashanka das said:
the exact statement of the problem is given belowView attachment 226320
It can be confusing unless you keep in mind that entropy is a state function. Change in entropy requires a difference between two states of thermodynamic equilibrium. If there is no difference between the "before" and "after" thermodynamic equiiibrium states, there is no change in entropy.

A thermodynamic equilibrium state is a statistical concept. Although the molecules are always flying around, when a gas is in thermodynamic equilibrium the fact that individual molecules constantly changing positions and energies within a fixed volume (at a constant temperature and pressure) does not affect the state of the gas. There is no statistically material change due to "mixing" of the molecules of the gas amongst themselves (volume, temperature and pressure not changing). It remains in the same thermodynamic state.

So there is no change in the thermodynamic state of the gas when the partition is removed. The mixing of the two halves of the container does not change the thermodynamic state of the gas as a whole. One can return to the initial state by effecting no thermodynamic change to the state of the gas - just insert the partition again!

That is not the case if the gas molecules on each side of the partition are different. When the partition is removed, the two gases suddenly fill the whole volume and mix together. Reinserting the partition does not restore the initial states of the two gases. They stay mixed! The volume of each gas has doubled!

AM
 
Last edited:
  • Like
Likes Chestermiller

What is thermodynamic entropy?

Thermodynamic entropy is a measure of the disorder or randomness in a system. It is a fundamental concept in thermodynamics, which is the study of energy and its transformations.

How is thermodynamic entropy different from information entropy?

Thermodynamic entropy and information entropy are two different concepts, although they are related. Thermodynamic entropy refers to the disorder in a physical system, while information entropy refers to the uncertainty or randomness in a set of data or information.

What causes an increase in thermodynamic entropy?

An increase in thermodynamic entropy can be caused by processes such as heat transfer, chemical reactions, or mixing of substances. These processes lead to an increase in the number of possible microstates of a system, resulting in a higher level of disorder.

Can thermodynamic entropy be reversed?

No, thermodynamic entropy cannot be reversed. According to the second law of thermodynamics, the total entropy of an isolated system always increases over time. While local decreases in entropy may occur, the overall trend is towards an increase in entropy.

How is thermodynamic entropy related to the concept of energy?

Thermodynamic entropy is related to the concept of energy in that it represents the amount of energy in a system that is unavailable to do work. As entropy increases, the energy in a system becomes more dispersed and less able to do useful work.

Suggested for: Thermodynamic entropy

Back
Top