Can adiabatic process be isothermal?

[passingthru]

The OP does not state what the contents of the system are or whether or not those contents do any work.

I do not understand free expansion. For example, question 47 in the GRE Physics bulletin describes the volume being doubled by removing a divider where one side initially contains an ideal gas and the other is evacuated. The answer is given that the change in entropy is nR ln2. If this is true, then the change in Q is nRT ln2, and not zero. I can arrive at that answer, but not with the change in Q and W being zero. All the texts state that the change in Q and the change in W are zero. When the divider is removed, the molecules continue to travel with the same kinetic energy, so it would seem that the temperature remains constant. The molecules still apply the same amount of force, but to a larger inner surface area, so wouldn't the pressure go down? And, the gas does no work on the container, but an outside mechanical force has changed the volume constraint. Hasn't it done work on the system by increasing the volume . If this is true, then energy has been added to the system, therefore it isn't really adiabatic. Help.

 

[Andrew Mason]

The OP does not state what the contents of the system are or whether or not those contents do any work.

I do not understand free expansion. For example, question 47 in the GRE Physics bulletin describes the volume being doubled by removing a divider where one side initially contains an ideal gas and the other is evacuated. The answer is given that the change in entropy is nR ln2. If this is true, then the change in Q is nRT ln2, and not zero. I can arrive at that answer, but not with the change in Q and W being zero. All the texts state that the change in Q and the change in W are zero. When the divider is removed, the molecules continue to travel with the same kinetic energy, so it would seem that the temperature remains constant. The molecules still apply the same amount of force, but to a larger inner surface area, so wouldn't the pressure go down? And, the gas does no work on the container, but an outside mechanical force has changed the volume constraint. Hasn't it done work on the system by increasing the volume . If this is true, then energy has been added to the system, therefore it isn't really adiabatic. Help.

The gas does no work on the surroundings. Since the kinetic energy of the expanding gas does not follow a Maxwell-Boltzmann distribution, its temperature is undefined while it is expanding. You could say that it does work on itself (causing the gas molecules to gain non-thermal kinetic energy), but ultimately the distribution of kinetic energy of the gas molecules approaches a Maxwell-Boltzmann distribution as things settle down.

So, the pressure goes down, volume expands and temperature remains the same. No work is done. There is no heat flow from or to the surroundings so it is adiabatic. dQ = dU = W = 0.

AM

 

[SpectraCat]

The OP does not state what the contents of the system are or whether or not those contents do any work.

I do not understand free expansion. For example, question 47 in the GRE Physics bulletin describes the volume being doubled by removing a divider where one side initially contains an ideal gas and the other is evacuated. The answer is given that the change in entropy is nR ln2. If this is true, then the change in Q is nRT ln2, and not zero.

the relation dQ=TdS only holds for a reversible process .. in an irreversible process (such as a free expansion) you can increase entropy without producing any heat.

 

[passingthru]

I appreciate your help, both AM and SC. I'm rereading my thermo text. It says basicly the same thing, that dQ = TdS only when in equilibrium. I'm still wondering how to arrive at ETS's answer with dQ= dW = 0. I know that dS is an exact differential, and so is dV. If we wait until the system reaches equilibrium after the irreversible process, then does it matter how we get the change in entropy? I guess, if both dQ and dW equal zero, they are still equal to each other, which is why it's okay to say dS = PdV/T, and since the T cancels out of the problem, it doesn't matter that it is not defined during the relaxation period. When it's through, the volume is twice as much, no matter what the temperature does, and since we have exact differentials, the path doesn't matter either.

Although, in reading the posts, another question comes to mind. Pardon my ignorance, I just want to learn. Is it irreversible because work would be required to put it back into the smaller volume, and heat removed to counter the added energy of the work done on the system, which means the reverse would not be adiabatic? I know this sounds confusing. It's because I don't know what I'm talking about.

Thank you.

 

[peeyush_ali]

yessss...why not?? pv^y =const for an adiabatic process...if u can make y=1 then pv=const which means isothermal...

 

[Andrew Mason]

I
Although, in reading the posts, another question comes to mind. Pardon my ignorance, I just want to learn. Is it irreversible because work would be required to put it back into the smaller volume, and heat removed to counter the added energy of the work done on the system, which means the reverse would not be adiabatic?

That is one way of looking at it. A reversible adiabatic path is adiabatic in either direction. But it may be clearer if you stick to definition of reversible.

The path is not reversible because the direction cannot be reversed by an infinitessimal change in conditions. In other words, the system and the surroundings with which it is in thermal contact are not in (arbitrarily close to) thermodynamic equilibrium during the process.

With a free expansion of gas, you cannot change the direction (from rapid escape of gas to a compression of the gas) by an infinitessimal change in pressure of the gas or surroundings.

AM

 

[Count Iblis]

I think it is nonsensical to say that no heat can be created inside the system in an adiabatic process. Doing so would force one to change the notion of adiabaticity to isentropicity, defeating the purpose of a notion of adiabaticity. Let me explain why.

Suppose you have some isolated system. It can perform work but it is perfectly insulated, so no heat can flow into or out of it. This is how we have defined what it means for the system to be adiabatic. The whole point if the notion of adiabaticity is that only the heat flow across the system boundary counts and you do not want consider the details of what happens inside the system.

Suppose the system undergoes some non-isentropic adiabatic process. This obviously implies that the system evolves through states that are not in thermal equilibrium. But this does necessarily mean that a thermodynamic description is not at all possible. If the system is evolving slowly enough, you can in practice describe the system thermodynamically by introducing more parameters. In practice this means chopping the system up in small parts and assuming that each subsystem is close to thermal equilibrium.

Whithin such a finer description the irreversible non-isentropic process can be described thermodynamically. The entropy increase can then be atributed to heat flows within the system.

So, it should be clear that one cannot object to a process being adiabatic just because you can give a finer description in which you can see a heat flow within the system like in the example I gave below. One can always play this game unless the system is so violent that no local thermodynamic treatment can even in principle be given. When I following an astrophysics course about the Sun, the Prof. told that local thermal equilibrium is still good enough for many of the extremely violent process in the Sun.


What this whole diascussion points to i.m.o., is that thermodynamics is not taught correctly outside of theoretical physics. Many engineers learn only thermodynmaics in a phenomenological way and then they think that concepts like heat have an absolute meaning leading to this confusion about adiabaticity.

The theoretical physics point of view takes into account, right from the start, that in general you cannot describe a system with 10^23 degrees of freedom in terms of only a few degrees of freedom. Thermodynamics has to be understood as resulting from integrating out almost all of the degrees of freedom of the system leaving only few degrees of freedom. Here one makes assumptions about the statistical behavior of all the degrees of freedom.

The arbitrary nature of the thermodynamical description is thus due to the choice on has for the thermodynamic variables one chooses to describe the system with. This is completely arbitrary. The more variable on choses, the larger the thermodynamic state space will be.