Thermodynamic Entropy Change Upon Partition Removal

[Apashanka das]

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=C_vdt+Pdv
And solving S becomes C_vlnT +RlnV
Is it right?
For the second part I need some hints to actually proceed
Thanks[/B]

 

[Andrew Mason]

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

 

[rude man]

I don't think there's a change in entropy. Chet?

 

[TSny]

The change in entropy is defined as $\int TdS$ along a reversible path

Typographical error for the integral expression?

 

[Chestermiller]

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.

 

[Andrew Mason]

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

 

[Apashanka das]

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

 

[Apashanka das]

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

 

[Apashanka das]

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

 

[Chestermiller]

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.

 

[Andrew Mason]

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