1. Nov 25, 2004

kitsos

Hi there, after years of blissful unawareness, I suddenly managed to find a simple yet, annoyingly perplexing question regarding our old friend, the adiabatic process. Let's look at the facts.

Any textbook will tell us that in an adiabatic process no heat is exchanged between the system and its surroundings (1).

Any textbook will also tell us that the equation that governs such a process is PV^gamma = constant (2).

Finally, that textbook will also tell us that the internal energy of a gas is dependent on its temperature ONLY (3).

We have all the actors on stage, now let's hear their story.

2 tells us that, as volume expands, pressure will drop. A quick calculation involving PV=NRT will show that temperature follows suit. This is the most classic textbook case of basic thermodynamics, an expanding gas cools down.

Now, 3 will tell us that, since temperature drops, surely, the internal energy drops as well. So, since energy cannot be destroyed, it must have gone somewhere right?

But 1, says that in the adiabatic process, no heat is exchanged between the system and its surroundings. Now, the first law leaves a loophole for energy to leave in the form of work. But an expanding gas does not always do work. Imagine the case of a volume of gas freely expanding in vacuum for example, no work done yet temperature drops. Why?

As far as I can see, it seems that PV^gamma=constant seems to be in direct contradiction with the first law since it implies a drop in internal energy of an expanding gas with no visible route for this energy to leave the gas. This is silly but try as I might, I cannot see where the flaw in my logic is. If any of you guys can point out the obvious to me I 'll be very grateful.

2. Nov 25, 2004

Bystander

Well, let's see --- you've mixed irreversible (free expansion) and reversible (PVgamma = const) processes, ideal and real gases --- what else? Thermo is about attention to detail. If it involves reading the book a second, third, ..., hundredth time, do so.

3. Nov 26, 2004

kitsos

I see your point about free expansion. However, consider this scenario. A baloon of gas is burst inside a huge vacuum dome. This is free expansion all right? There is no heat exchange by conduction until the gas touches the walls and radiation can be disregarded as insignificant for the brief period it takes for the gas to fill the whole volume. From the definition of an adiabatic process, I really don't see why this isn't adiabatic, as wel as free expansion but anyway, let's look at this transient period of expansion into the whole volume.

What happens to the temperature? If we can't use PV^gamma = constant, then temperature should be assumed to remain constant. Is it just me that cringes at the thought of a gas expanding its volume a million times and keeping its temperature constant? If that is true, we should be able to fill a whole vacuum filled stadium with gas at a cosy room temperature (but very low pressure) by bursting a pellet of gas at the required temperature. I can't see where the mistake in the logic is but I wouldn't bet any money on it.

As for confusing real and ideal gases, I don't see where I 've done that and how it would affect the basic argument anyway. Real gases behave pretty close to ideal and here we are talking about very gross differences. These should be noticeable from ideal gases to semi-vaporised petrol fumes.

4. Nov 26, 2004

Bystander

Real gases may cool, remain at a constant temperature, or warm during free expansion. This can be contrasted with the definition of an ideal gas which includes (explicitly or implicitly depending upon your reference) the property of constant temperature during free expansion.

Burst a balloon? Sure. Adiabatic? Sure. Reversible adiabatic? NO. PVg applies only to reversible adiabatic processes. Got it?

5. Nov 26, 2004

krab

Yes.

You are confusing temperature and energy density. Burst a balloon in outer space and the (ideal) gas molecules will continue on to infinity with the same speeds they had while they lived happily in the balloon. Temperature relates to average energy per particle, not average energy per unit volume.

The difference with a reversible process is that there, the particles are bouncing off the moving wall that's increasing the volume. This slows them down.

6. Nov 27, 2004

Clausius2

1) I'm not able to seeing the contradiction between PV^gamma and the first principle, because it is derived from the first principle:

$$dU=\delta W=-PdV=nc_vdT$$

Therefore:

$$-PdV=nc_v \frac{d(PV)}{nR}=\frac{c_v}{R}(PdV+VdP)$$
$$-VdP=(\frac{R}{c_v}+1)PdV=-VdP=\gamma PdV$$
and integrating:
$$PV^{\gamma}=cte$$

How is it possible that something derived from another thing gets into contradiction with this last thing?

2) Imagine an adiabatic cylinder filled with ideal gas. The surroundings are the vacuum, and the cylinder has a mobile wall (a piston). Once you release the piston, the gas is expanded. So that:
$$\delta Q=0$$
$$\delta W=0$$ and therefore
$$dU=0\rightarrow dT=0$$

On the other hand, you can check that entropy increases, because it is an irreversible process.

7. Nov 29, 2004

kitsos

Thanks for all the feedback guys. I understand how energy density is not equivalent to temperature. I also see the point about reversible and non reversible adiabatic processes.

The reason for the query was a simple discrepancy in modeling an adiabatic contraction in an internal combustion engine. The adiabatic compression segment can be modelled either via the first law with dQ=0 or straightforward with PV^gamma = constant.

I have chosen the first law, since it is a more general equation and will hold for the whole compression/expansion, not just the adiabatic compression. However, after checking the curves, it seems that the pressure and temperature given by the ODE solver solving for P,T,V derivatives based upon the first law and the equation of state result in considerably higher values than the curves given by PV^gamma = constant. At the moment I am unsure why that might be and (unfortunately) it seems that my working out of the derivatives is correct.

Anyway, thanks again for all the help guys. I 'll check my algerba for the millionth time and if it still looks OK I 'm in trouble.