Calculate Work for Reversible and Irreversible Expansion/Compression
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- เผยแพร่เมื่อ 29 มี.ค. 2016
- Organized by textbook: learncheme.com/ Shows graphically the areas on a pressure-volume diagram that are proportional to work for reversible and irreversible expansions and compressions of a gas in a piston/cylinder. Made by faculty at the University of Colorado Boulder, Department of Chemical & Biological Engineering.
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Thank you so much for this video. I had a very specific conceptual issue that has been dogging me for the last few days and this video cleared it up for me. Thank you.
Thanks a lot!! My teacher didn't explain this so nicely..
Constantly thankful for CU Boulder. Great videos that are straight to the point. Great simulations for my Fluids class. So many helpful resources!! Thank you!
thanks for this video it really made me understand the reversible and irreversible process completely
Immaculate explanation of the problem at my hand. I have a patent for decompression of exhaust gases in additional cylinder(s).
Could u show what if the weight w is reduced in steps (w/2, w/4,..) in expansion work.
Sum of the irreversible work areas may be shown to converge to the final rectangular irreversible work area , that may make the understanding intuitive.
Pressure cannot drop instantly from p1 to external pressure.
It should take some time.
The irreversible work done is less than reversible & is understandable.
The reduced Inside pressure will do the expansion work. That is a different story if the inside pressure is equal to the atmospheric work.
By the way are u adding heat during the expansion to keep the temperature constant? Or the expansion is exclusively due to the weight taken off & release of potential energy?
Thanks
Helped allot, topic not covered well by the rest of the internet
Thanks. It helped Me a lot.
Thank you so much for this video
Thank you so much!
Thanks for the video. External vs internal should be defined here. For example, if the piston is frictionless and no external force on it, the work done by the gas is still not zero and equal the pressure at the internal face of the piston times the volume change. I think the word average is a better term. The average pressure of the gas is higher than the pressure at the face of the piston.
If the piston is frictionless and has no external force is on it (and the piston has no weight), then the work done by the gas is indeed zero. The work is not equal to the pressure at the internal face of the piston times the volume change.
I don't follow that. Why are we assuming that the piston has no weight. What if it did? Work is force times distance isn't it and force is pressure times area.
It’s a thought exercise. Free expansion of a gas into a vacuum does no work, therefore if the piston has no mass and is frictionless, no work is done by the gas. If the piston had mass, then work would be done.
Very helpful
So does that mean reversible work always equal to actual work by the gas?
W(usable) = W(actual) - W(surrounding)
Is it W(reversible) = W(actual) ?
Thanks a lot!!
Then by this uncompensated heat produced in compression is negative ...which defies clausius inequality .how come
Thanks a lot ❣️❣️
In the irreversible compression case, the surroundings will have to exert pressure P to do work, so it looks like that the surroundings have done work amounting to -∫PdV. Then, this quantity and the energy gained by the system through compression (-∫PexdV) do not look the same in magnitude. How can this discrepancy be accounted for?
@r did you get the answer, I have same doubt? In irreversible compression(or expansion),the work done by surrounding on the system should be equal in magnitude to work done by system on surrounding but opposite in sign. Am I wrong? Or why don't we do this: find the work done by system on surrounding and vice versa and then subtract their magnitude to find net work done??
What can we say about Final temperature of reversible and irreversible ?
yes you are right why only pressure no volume or temperature
it's isothermal
thank u
Great job thanks sir
Hi, several questions :
What is the kinetic energy definition for the compression of air? so that the kinetic energy variation + potential compression energy is conserved ?
And if it is 1/2 * mass moved * speed, what does the speed represent? First speed, average speed?
Final question, how long will it take for the compression to be completed?
I think it would make more sense if you refer to a quasi-static (left) and non quasi-static (right) process, and finally you end up concluding that only the quasi-static process could be reversed (both system and sorroundings), therefore that's by definition reversible.
i understand the concept. but.. in real life how do we know which one is reversible or irreversible process? the pump, the turbine and all heat involve
Reversible processes are hypothetical, hence all the process are irreversible in nature
What does negative indicates in pdv ?
Refer lecture TD010C on my channel
Love u sir
why irreversible work done by the sorroundings on the system is lower than the irreversible work done by the system on the sorroundings? , my logic tells me they should be equal. Can you please explain this?.
David Glass because it's an irreversible process, Wa->b =/= Wb->a
If irreversible work is done by the system on surroundings, and expansion occurs, you lose internal energy from heat transfer with friction, sound etc.. a non infinitessimal change. Mathematically, that is why you need to use Pext in the integral, because Psystem equilibrates to equal Pext "instantly".
For compression, work is done by surroundings onto system, P1 goes to P2. For irreversible, this means plugging P2 into Pexternal for the integral, as it doesn't infinitessimally change. P2 > P1
In reversible, P1 will slowly and more efficiently be approaching approaching P2 (and vise versa) and that's why it is the area under the curve from the integral, since you integrate Pdv= RT/V dv
David Glass basically remember that Q and W depend on path, and irreversible makes a different path, while reversible makes the same path, just reversed
nicely explained sir.
DavidG. He has taken external pressure as final pressure at 1.43 min.
Work done is volume gradient ie integral of volume . Great job. Thanks UC Boulder.
Absolutely spot on. Loved it!
Thanks a lot
Wow. It need not be an adiabetic/isothermal or somewhat in-between. At 2.42. wow. That answers my problem of my patent.
whT is ideal work
But *why* is the work less for irreversible process??
Irreversible process is less beacuse some amount of work is lost due to the irreversability of the porcess
If we expand the beaker slowly then the beaker has time to absorb the heat from surrounding which keeps the pressure high and so the work done( note that it must absorb heat from outside to keep the internal and external pressure same that is to keep the reaction reversible). While if beaker expands irreversibly( sudden expansion) then the pressure will be dropped drastically in the starting of the expansion( because of no heat entering into beaker and no pressure rise) so the work done by piston will be less.
Why is it smaller when expanding and larger when compressing?
Do you mean the amount of work?
LearnChemE Yes! However I have just completed my chemistry exam and think some extra study has helped me understand this better
helpfull !!