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work done by a gas/on a gas

Discussion in 'MCAT Study Question Q&A' started by doxycycline, Jan 4, 2009.

  1. swamprat

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    So work done by a gas is positive and work done on the gas is negative correct? So I was looking at a question and it showed a chart and asked what happens to the net work when you go from A to B. B has a greater volume than A and they are both at the same pressure so you can either calculate the area under the line or use W=PV. Since the volume is decreasing doesnt that mean that the gas is having work done to it, so the energy would be negative?

    To add this is coming from EK 1001 chemistry and the question #336 and it uses A and F, but same deal. The explanation is :

    "Volume is decreased on the gas, so work is done on the gas."
     
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  3. LaCasta

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    This is my understanding of it

    If the gas is expanding, and you imagine a piston on top of the container, then the gas would move the piston up. The gas would be expending energy to move the piston up, so the work is done by the gas and is negative since the gas is losing energy. An expanding gas also cools, so you can think of it realted to temperature, since T is proportionate to internal energy.

    So I guess to answer your question, if the volume is decreasing, then the piston is doing work on the gas, and the work would be positive.
     
  4. swamprat

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    That does make sense but in one of the kaplan books it states:

    "Work done by the system is positive, but work done on the system is negative."

    I even did another problem in the book with a simple triangle and as you decrease volume from 1 point to another while keeping pressure constant the energy was negative.


    Also just wondering, if a gas is expanding why does that decrease temperature. When I think of it, you decrease temp to go from gas->liquid->solid so you want to contract the volume.
     
  5. rocuronium

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    The signs that you are referring to are just a convention, and you might see them written opposite to what you are expecting. What's important is to be consistent with how you approach these problems. Learn (and understand) what is happening and why and then the signs should not be an issue.

    As for your question regarding temperature here is my understanding. If a gas is expanding adiabatically it is still doing work on the surroundings. This means that it is losing internal energy which is manifested as a decrease in temperature.

    I think what may be confusing is looking at this concept as an application of the ideal gas law (if volume goes up, temperature should go up as well). In my understanding at least the ideal gas law refers to a system that is at constant energy. This system is not at a constant energy as work is being done on/by the gas.

    Does that help? Let me know if you find any flaws in my logic.
     
  6. swamprat

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    In regard to the first part:

    So you have U=Q-W

    Q = heat so if you add heat to the system its (+) and if you lose heat from the system its (-).

    W = work so if the system does work its (+) and if it has work done to it(as in volume contraction) its (-)

    Thats what i have always went by and it has always worked for me until i opened up the EK book and did that problem. I don't really understand what you are saying with sign convention. I mean I understand what that means(such as picking the right side of the x-axis to be positive and staying consistant w everything else throughout the problem), but I am being consistant. It really just doesn't add up.

    The second part of what you said well I'll get back to I really just would like to figure this part out lol
     
  7. LaCasta

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    I think the confusion stems from the fact that the equation is sometimes written

    U = Q + W

    Which would them mean that when the system does work, like moving a piston, that the work is negative, because U has to decrease.

    So its a matter of convention.

    If you are using:

    U = Q - W

    then you would designate the work being done by the gas moving the piston as positive, since you want energy to decrease. So its a matter of convention.
     
  8. swamprat

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    Yeah that makes sense but still doesn't explain the diagram discrepancy. Example:


    (picture this a chart)


    A __________________ B

    Now the x-axis is volume and y-axis is pressure. Lets say the different between the 2 points is 5 and the pressure is 2.

    What is the net work done when going from B to A?

    So W=PV=5*2=10. But since the gas is losing volume, work is being done on it so its -10 .

    But if you use U=Q+W then ur saying by convention its +10 ? Thats a problem then right? Which one is it really lol
     
  9. Ibn Rushd

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    ^ Exactly (referring to LaCasta). I've studied thermodynamics using EK, TPR, and Kaplan materials. The concepts are the same, but each company teaches a certain convention. Actually, I think EK and TPR taught the q+w convention, and Kaplan taught it as q-w. It's been a while since I read about thermo, so pelase correct me if I'm wrong.
     
  10. LaCasta

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    Its still a matter of convention. You have found the magnitude of the work done by finding the area under the line. So if you are using the U = Q + W equation, you would make the work negative, so that U would decrease.

    U = Q + (-W)
     
  11. swamprat

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    Yeah but it was a multiple choice question w both magnitudes listed just 1 was + and one was -. I chose the negative one which gave me the wrong answer(in the EK book) but a diff question same concept i chose the - one(in kaplan book) and i got it right.

    So if something like that shows up in the mcat as a strictly what is the work done and they dont just ask for magnitude what do you do?
     
  12. Ibn Rushd

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    I would recommend choosing a convention and just sticking with it. Choose one, review it, review it some more, and do your best to disregard the other convention so as to avoid confusion. I think TPR and EK explain things clearly, they use the q+w equation. It really comes down to what you designate positive and/or negative. I learned the following using TPR and EK:
    Heat transfer INTO a system is positive (+q)
    Heat transfer OUT OF a system is negative (-q)

    Work done BY the system is negative (-w)
    Work done ON the system is positive (+w)
    I just looked at my Kaplan book and they use the q-w equation instead and they designate work done BY the system as being positive ... the other companies say it's negative. You're working questions that were written by two different companies that attempt to teach two different conventions, thus you noticed a discrepancy. Again, I would recommend choosing one and sticking to it. If this topic comes up on the real MCAT, then I think we'd be told the convention in the passage. Otherwise, any questions asked would be beyond the sign one uses for work, rather, the question would ask about internal energy. Regardless of the convention one uses, they ought to arrive at the same answer when solving for internal energy.
     
    #11 Ibn Rushd, Jan 4, 2009
    Last edited: Jan 4, 2009
  13. rocuronium

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    You would never have to choose between + or - on the MCAT unless they gave you some additional information. They're not trying to trick you. It's the concepts that are important and it seems that you understand them. If you stick with one convention as Ibn suggested, you should be fine!

    Good luck with the studying!
     

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