Can adiabatic process be isothermal?

[Andrew Mason]

Consider a gas in a volume in which there is a heat source. The heat cannot leak out, so whatever happens to the system, it will always be an adiabatic process. The produced heat is controlled by a thermostat which keeps the temperature the same. If the initial pressure of the system is larger than the outside pressure, the system will expand adiabatically and isothermally until the inside pressure is the same as the outside pressure.

This is not an adiabatic process. Heat is still flowing into the gas. Adiabatic means that there is no heat flow into or out of the system.

AM

 

[Count Iblis]

This is not an adiabatic process. Heat is still flowing into the gas. Adiabatic means that there is no heat flow into or out of the system.

AM

This depends on where you put the system boundary. I can take my system to be the gas plus the heat source. Then, even though the heat source produces heat, this does not count as heat in the thermodynamic sense.

This is analogous to free expansion process, throttling process etc. There you do not count the heat due to internal friction, because it stays in the system and does not flow across the system bondary.

 

[Andrew Mason]

This depends on where you put the system boundary. I can take my system to be the gas plus the heat source. Then, even though the heat source produces heat, this does not count as heat in the thermodynamic sense.

This is analogous to free expansion process, throttling process etc. There you do not count the heat due to internal friction, because it stays in the system and does not flow across the system bondary.

There is no such thing as internal friction in an ideal gas. All collisions at the molecular level are elastic.

If you take the system to be the gas and the heat source, there can be no creation of heat from other energy forms within the heat source (eg. chemical energy). If heat flows into the system (by converting chemical energy to heat) it is not adiabatic.

But you do point out an important issue: the conversion of heat (thermal kinetic energy) into mechanical energy (non-random kinetic energy) - in effect the gas doing work on itself. The gas that exits from a throttle in a free expansion has kinetic energy that does not follow a Maxwell-Boltzmann distribution. This results in the temperature of the gas being undefined while it is escaping. It eventually reaches a new equilbrium, however, in the expanded volume, in effect converting that non-thermal kinetic energy back into thermal energy.

AM

AM

 

[Andy Resnick]

See? I told you! I'm not the one to give you those kinds of answers... lol

I think one thing that stands out to me that seems even more interesting is that the work the gas does is not going anywhere, or more to the point, it's going back into the gas as near as I can tell.

For example; consider a well insulated cylinder fitted with a piston that's placed inside a pipe at a location where the pressure is 100 psig. Assume air is in the cylinder and also in the pipe. The cylinder will be allowed to travel along the length of the pipe to where the pipe discharges to atmosphere at 0 psig. The work done by the air in the cylinder must go into the flow stream. If the cylinder is well insulated (adiabatic) and the expansion is isentropic as we would expect it to be, the air in the cylinder cools significantly. In comparison, air in the pipe won't cool significantly at all. It's pretty close to being an ideal gas, and the change in internal energy for the air traveling down the pipe is essentially zero. There is no change unless there's heat/work added to or removed from the flowstream.

So in the first case, we have a traveling cylinder that does work PdV, and in the second case we have air that does no work on the surroundings but undergoes an identical change in pressure with no change in its internal energy.

Does the air flowing through the pipe do work on itself? If so, it does work exactly equal to the work needed to keep the internal energy change zero. At least, that's what I've always assumed is going on. It just seems strange that the work the air does exactly equals that needed to keep dU=0. The molecules always seem to 'know' how to space themselves out so that internal energy (and enthalpy) doesn't change. I guess I shouldn't be so amused by that - it's a simple conservation of energy.

This is an excellent puzzle...

 

[Count Iblis]

If you take the system to be the gas and the heat source, there can be no creation of heat from other energy forms within the heat source (eg. chemical energy). If heat flows into the system (by converting chemical energy to heat) it is not adiabatic.

The mu dN terms count as heat?

 

[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 basically 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.