I This gas work violates 2nd law of thermodynamics?

[Chestermiller]

Fine, the states $A$ and $B$ are not the same. But do they have the same entropy, namely, is $S(B)=S(A)$? You seem to be saying that one entropy can be larger, e.g. $S(B)>S(A)$. But then in the path $B\to A$ we have $dS<0$, which contradicts the 2nd law. So how can such a path exist?

What process are you considering here?

 

[Demystifier]

What process are you considering here?

Any process with an alleged reversible cycle you referred to.

 

[Chestermiller]

Let’s focus on a reversible isothermal expansion or compression of an ideal gas, ok @Demystifier?

 

[Demystifier]

Let’s focus on a reversible isothermal expansion or compression of an ideal gas, ok @Demystifier?

OK. So can you answer my question in #50 now?

 

[Chestermiller]

OK. So can you answer my question in #50 now?

$$S_B-S_A=R\ln(V_B/V_A)$$$$S_A-S_B=R\ln(V_A/V_B)$$$$\Delta S_{cycle}=0$$So what's the problem?

 

[Demystifier]

$$S_B-S_A=R\ln(V_B/V_A)$$$$S_A-S_B=R\ln(V_A/V_B)$$$$\Delta S_{cycle}=0$$So what's the problem?

During the contraction (i.e. when the volume decreases), the entropy decreases. This contradicts the 2nd law during the contraction. Of course, it doesn't mean that the gas cannot be contracted. But it means that there is some additional change of entropy (in the environment) not captured by the equations above, so that $\Delta S_{\rm total}> 0$ even during the contraction.

You are really saying that entropy of a subsystem (the gas) can decrease. I agree, but I point out that the entropy of the full system (gas + environment) cannot decrease.

What really is important is the following. Suppose that the full system consists of several subsystems, each with its own temperature $T_i$. Then for each subsystem we have $dQ_i=T_idS_i$. The second law is
$$\sum_i dS_i\geq 0$$
or in terms of heats
$$\sum_i \frac{dQ_i}{T_i} \geq 0.$$
In particular, if all subsystems have the same temperature $T_1=T_2=\cdots=T$, then the last inequality simplifies to
$$\sum_i dQ_i\geq 0.$$
In this case we also have
$$\sum_i dQ_i=T\sum_i dS_i$$
or shortly
$$dQ=TdS$$
where $dQ=\sum_i dQ_i$ and $dS=\sum_i dS_i$.

 

[Chestermiller]

During the contraction (i.e. when the volume decreases), the entropy decreases. This contradicts the 2nd law during the contraction.

This is not a contradiction of the 2nd law. It satisfies the Clausius inequality for a reversible process applied to the system.

Of course, it doesn't mean that the gas cannot be contracted. But it means that there is some additional change of entropy (in the environment) not captured by the equations above, so that $\Delta S_{\rm total}> 0$ even during the contraction.

The > sign applies to an irreversible process path. For a reversible process path, the entropy change of system plus surroundings is zero.

You are really saying that entropy of a subsystem (the gas) can decrease. I agree, but I point out that the entropy of the full system (gas + environment) cannot decrease.

Correct. So what.

What really is important is the following. Suppose that the full system consists of several subsystems, each with its own temperature $T_i$. Then for each subsystem we have $dQ_i=T_idS_i$. The second law is
$$\sum_i dS_i\geq 0$$
or in terms of heats
$$\sum_i \frac{dQ_i}{T_i} \geq 0.$$
In particular, if all subsystems have the same temperature $T_1=T_2=\cdots=T$, then the last inequality simplifies to
$$\sum_i dQ_i\geq 0.$$

The = sign applies only to a reversible change in all the subsystems. If the path is irreversible, then the temperature in one or more of the subsystems must be non-uniform and the T in dQ/T for that subsystem must be taken as the temperature at the interface between the subsystem and its neighbors through which the heat dQ flows (see Fermi, Thermodynamics and Moran et al, Fundamentals of Engineering Thermodynamics).

 

[Demystifier]

@Chestermiller I still have a feeling that I'm missing something. Let me try this way. You said that the formula $dQ=TdS$ is valid only for reversible processes. If so, is there a generalization of this formula valid for all processes?

 

[Chestermiller]

@Chestermiller I still have a feeling that I'm missing something. Let me try this way. You said that the formula $dQ=TdS$ is valid only for reversible processes. If so, is there a generalization of this formula valid for all processes?

For a closed system, the entropy change of the system for an irreversible or a reversible process path is $$\Delta S=\int{\frac{dQ}{T_B}}+\sigma$$where $T_B$ is the temperature at the interface between the system and surroundings and $\sigma$ is the entropy generated within the system during the process due to irreversibility; for an irreversible path $\sigma>0$ and, for a reversible path, $\sigma=0$. Also, for a reversible path, the temperature within the system is uniform at all points along the path and equal to the interface temperature at all points along the reversible path. Furthermore, for a reversible path, the heat variations dQ along the path do not have to match those along irreversible path, nor do the boundary temperature variations along the path.

Here is how to find the entropy change for an irreversible process path: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

 

[Philip Koeck]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

View attachment 328132

I've tried to simplify the problem, hopefully retaining its essence, and do an analysis, but I ran into some difficulties. Maybe someone can tell me how to proceed.

If I have a closed cylinder as in the sketch but only 1 piston between gas A and gas B, so no vacuum, then I don't need to think about collisions between pistons.
Initially I have the following situation:
Gas A occupies 1/3 of the cylinder on the left, gas B occupies 2/3 on the right and they are separated by a piston. The number of moles (n) is the same for both and remains constant.

For the starting values of volume, pressure and temperature we have:
V_B = 2 V_A
P_A = 2 P_B
T_A = T_B = T

There is no heat transfer between a gas and anything else (piston, cylinder or the other gas).
The piston is extremely heavy so that expansion and compression will be quasi-static.

If I release the piston gas A will expand very slowly and gas B will be compressed. The sum of volumes is constant.

This process should stop when the pressures of gas A and B are the same.
Gas A will have cooled off and gas B will have heated up.

Here's my problem:
If I treat the expansion of A and the compression of B as reversible adiabatic, then the whole process will be reversible since nothing happens in the surroundings.
(I checked that analytically, just to be sure, and if I use expressions for reversible adiabatic processes ( P V^γ = constant, for example), I get ΔS = 0 for both gases (with some rounding errors).

However, by just looking at it, I would swear the process must be irreversible.

How would you calculate the final temperatures and volumes of the gases and the entropy-changes?

 

[Chestermiller]

I've tried to simplify the problem, hopefully retaining its essence, and do an analysis, but I ran into some difficulties. Maybe someone can tell me how to proceed.

If I have a closed cylinder as in the sketch but only 1 piston between gas A and gas B, so no vacuum, then I don't need to think about collisions between pistons.
Initially I have the following situation:
Gas A occupies 1/3 of the cylinder on the left, gas B occupies 2/3 on the right and they are separated by a piston. The number of moles (n) is the same for both and remains constant.

For the starting values of volume, pressure and temperature we have:
V_B = 2 V_A
P_A = 2 P_B
T_A = T_B = T

There is no heat transfer between a gas and anything else (piston, cylinder or the other gas).
The piston is extremely heavy so that expansion and compression will be quasi-static.

If I release the piston gas A will expand very slowly and gas B will be compressed. The sum of volumes is constant.

This process should stop when the pressures of gas A and B are the same.
Gas A will have cooled off and gas B will have heated up.

Here's my problem:
If I treat the expansion of A and the compression of B as reversible adiabatic, then the whole process will be reversible since nothing happens in the surroundings.
(I checked that analytically, just to be sure, and if I use expressions for reversible adiabatic processes ( P V^γ = constant, for example), I get ΔS = 0 for both gases (with some rounding errors).

However, by just looking at it, I would swear the process must be irreversible.

How would you calculate the final temperatures and volumes of the gases and the entropy-changes?

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

 

[aliinuur]

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

There are some irreversible path problems (most) that can't be solved without solving the PDE's for continuity, Navier Stokes, and differential thermal energy balance internal to the system for the spatial variations of temperature, pressure, and velocity. This is one such problem.

I think gas A and gas B volumes oscillate between 1/3V and 2/3V, so yeah, this is reversible process.

 

[Chestermiller]

This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.

 

[Herman Trivilino]

TL;DR Summary: Gas by doing negative work(losing heat), say accelerating a piston violates 2nd law of thermo, because kinetic energy of piston can be used to make other gas hotter, thus in summary one gas gets colder, other one gets hotter and this by simple piston!!!!

There is no transfer of heat in this scenario. There is work being done, and temperatures are changing, but neither of those things need be associated with a transfer of heat.

Heat is a transfer of energy due to a temperature difference. Just because there's a temperature change you can't assume there's heat energy being transferred. There could instead be (and indeed there is) work done. That is the essence of the 1st Law of Thermo.

 

[Philip Koeck]

This is incorrect. Gas viscosity damps the motion so that, in the final state, the pistons are no longer moving.

I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?

 

[Chestermiller]

I assume even an ideal gas would have a certain viscosity due to random collisions between molecules.
Do you agree?

Definitely. See chapter 1, Transport Phenomena by Bird el al.