# Combined gas law vs. adiabatic relations

1. Nov 12, 2009

### gothamxi

Why is there a difference when calculating pressure temperature and volume using the combined gas law, or when using adiabatic relations? As an idiot I am very confused. Why must I use a very similar equation to calculate the final temperature of an ideal gas but resulting in very different answers. Is the combined gas law only useful in non-adiabatic situations?

I will have a follow up question to this most likely, as I am trying to determine the work done on a gas being compressed in this situation: Air being pumped (amazingly adiabatically) under water into a diving bell. I need to find the work done to accomplish this.

As far as I can tell (keeping in mind that I am an idiot),
Compression Work=(5/2)nR(T2-T1)

I'm probably going about this all wrong, but I assumed the bell to be at a depth of about 30.5 meters, so roughly 4 atm. absolute pressure. Then I used Charles law and Boyles law (probably my problems started there?), to calculate the final temperature using that final pressure and an initial pressure of 1 atm, and I got crazy wrong answers. But using the adiabatic formula t2=t1(p2/p1)^(.283) gave more reasonable answers.

Which is all good and fine, except I dont understand why, and so I dont know if I'm still wrong, or if I was ever even close. From what I can gather, it has something to do with specific heats and degrees of freedom, but I thought ideal gas relations could be calculated very proportionately without even dealing with those. This is just confusing me. I've tried researching this online, but I get lost in the abstract thermodynamics.
Anybody help? Am I on the right track, and why?

Last edited: Nov 12, 2009
2. Nov 13, 2009

### Q_Goest

Hi gothamxi, welcome to the board. Adiabatic is a term that only means that some process has no heat transfer. The first law is generally written: dU = Q + W, so the Q term drops out for an adiabatic process and you're left with dU = W. But work can be zero also. All adiabatic means is that there's no heat transfer. So flow through a valve for instance, which is an isenthalpic process and the work is essentially zero, is a different process than an isentropic one in which work is done. But both processes are adiabatic. All those equations you learn in college are simply different ways of describing the first law of thermodynamics. Some include work and some don't.

3. Nov 13, 2009

### gothamxi

right, there is no heat transfer, so we can use adiabatic equations to calculate compression variables for maximum efficiency, correct? but i still dont understand, why the adiabatic calculation for final temperature of a compressed ideal gas is different than when you do the same calculation, changing the volume in a PV/T=PV/T(final) combined gas law equation. am i just making myself really confused, and using this gas law equation improperly?

4. Nov 13, 2009

### Q_Goest

Not sure what you mean by that. An adiabatic process can have less than a 100% isentropic efficiency, such as a process wherein part of the process is isenthalpic. I guess the point to remember is only that adiabatic processes don't have heat transferred into or out of the fluid, but the process is not defined by saying it is adiabatic. Saying a process is adiabatic is generally not enough to define what process the fluid undergoes. You generally need some more information that just "adiabatic". For example, you might say an adiabatic expansion of gas through a valve, in which case you can infer that no work is done so the process is isenthalpic. But you need to know if work is done or not and how efficiently the work is being done in order to determine what the process is and what the final state is.
The ideal gas equation is simply an equation of state, meaning it tells you how P, V, n, and T are related. As you say, PV/T initial is equal to PV/T final, because PV/T = nR and nR is a constant. So the initial state and final state are related by PV/T initial = PV/T final. But that equation has 3 variables on both sides, so even if you know the initial state, you have too many unknowns to determine the final state unless someone hands you 2 of the 3 variables. The final state could be arrived at through an isenthalpic process or an isentropic process for example, but the final state would differ. Therefore, you need an equation that determines the final state, and that equation will differ depending on what process the gas undergoes. Hope that helps.

5. Nov 15, 2009

### gothamxi

That actually helps a lot. I can clearly see why what I was doing before was wrong. And now I can realize just how deep of water I've stepped into. I'd better learn to swim a little.

"For example, you might say an adiabatic expansion of gas through a valve, in which case you can infer that no work is done so the process is isenthalpic. But you need to know if work is done or not and how efficiently the work is being done in order to determine what the process is and what the final state is."

So, if I'm understanding you right, adiabatic just defines any process that does not transfer heat in or out of the system (one adjective of a few that could describe a given process.

And what I should have said or been looking at, for the work of compressing of a gas in my ideal situation , was an 100% efficient isentropic process? Which by definition, would be also be adiabatic.

Correct?

Now these terms, isentropic and isenthalpic are also not exclusive?
I've read articles that lead me to believe that isentropic expansion is isenthalpic.

So, next question...because I'm getting confused again. The work done is equal to the change in internal energy. And the change in internal energy is dependent on the process that brings about this change...like isentropic, isenthalpic.
So the question: entropy is basically measuring efficiency of work is it not, actually measuring energy that doesn't go toward work? And enthalpy is the amount of work done by the system (i.e. molecular kinetic energy converted into macroscopic kinetic energy during expansion)?

Then isentropic means no change in unused energy (no energy loss), and isenthalpic means no change in ...work done by the gas to expand? There I get lost. Isnt the expansion of a gas itself work?

6. Nov 15, 2009

### Q_Goest

Yes, exactly.

Sounds like you're confusing isentropic and isenthalpic. An isentropic process is not an isenthalpic one. Maybe this will help:

Isenthalpic refers to a process in which enthalpy does not change. The initial and final states of the process have the same enthalpy. This is the case for an adiabatic process in which there is no work done on or by the fluid. (dH = 0 or U2-U1 = -(p2V2-p1V1) )

Isentropic refers to a process in which entropy does not change, so the initial and final states have the same entropy. This is the case for an adiabatic process in which only work is done by or on the fluid. (dS = 0 and dU = W)

For a better overview, I'd suggest the MIT web site that discusses this:
http://web.mit.edu/16.unified/www/FALL/thermodynamics/chapter_4.htm

I'd like to point out that the isentropic efficiency of a compressor or expander (ie: a process which is characterized by dU = W plus some losses) can be calculated as follows:

For a compressor:
isentropic efficiency = isentropic work / actual work = dhisentropic / dhactual

For an expander:
isentropic efficiency = actual work / isentropic work = dhactual / dhisentropic

In either case, the process the fluid undergoes may be adiabatic and largely isentropic, but it could also partly undergo an isenthalpic process, in which case the isentropic efficiency is less than 100%. The process might also have heat added or removed, in which case, the process isn't adiabatic any more. But the point I want to make is that an adiabatic process may be a process with less than 100% isentropic efficiency.

Also, it might be worth pointing out that the first law can have potential and kinetic energy added to it. We sometimes neglect these two forms of energy because they're often negligible when compared to the internal energy, but for a converging/diverging nozzle for example, kinetic energy is often significant. See for example, this web site: