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Homework Help: Calculate (a) the total work done by the gas in the process

  1. May 23, 2003 #1
    Hello.

    I am totally lost on this one. There seems to be something missing from the problem. Here goes.

    A perfect gas is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L. Heat is added to the gas holding the volume constant, and the pressure and temperature are allowed to rise until the temperature reaches its original value. Calculate (a) the total work done by the gas in the process, and (b) the total heat flow into the gas.

    Thanks
    MSDJ:wink:
     
    Last edited by a moderator: Feb 5, 2013
  2. jcsd
  3. May 23, 2003 #2
    The rules of the forum state that you have to show us where you got stuck before we can help :smile:.

    Why do you think there is something missing? What formulae do you know for Isobaric and Isochoric processes?
     
  4. May 23, 2003 #3
    Well, shouldn't there be some kind of stated temperature? From all my notes that seems to be the missing piece. Not sure what are Isobaric and Isochoric processes. This was given as a research project.
     
  5. May 23, 2003 #4
    OK - this is interesting 'cause I'm studying this topic for my upcoming final.

    "Slowly compressed" tells you that heat is being removed while the volume decreases -- otherwise 2 Atm. pressure would not be enough to compress the gas to 1/5 of its volume.

    Then, holding the volume constant, enough heat is added back to raise the temp. to its original value (while the pressure also increases).

    Think about the implications of that. Does that help?
     
  6. May 23, 2003 #5
    Ok, Isobaric - constant pressure
    Isochoric - constant volume - both which you've got here. Looking up these two terms would definitely be a good place to start. P-V diagrams could also be useful.

    You should be able to find formulae/derivations for the work done in each of these processes.

    You'll also need to know the first law of thermodynamics to do any of it.

    Oh and is 10.0 L 10 litres? So you have the pressure and volume and from the ideal gas law you can work out some temperatures if you need to. Again, definitely look this up if you haven't done anything on it.
     
  7. May 23, 2003 #6
    I thought I knew how to do this, but I realize now that I can't unless I assume some quantity, such as 1 mole (or any other specific quantity), of gas.

    Can it be done if the quantity and starting temperature are both unknown?
     
  8. May 23, 2003 #7
    Gnome,

    I am looking up this thing but don't see how it can be done without starting temperature, or as you said, to assume a the quantity. Unless I am missing a formula somewhere along the line that helps us to arrive at the temperature.
     
  9. May 23, 2003 #8
    It's easy to determine the work; actually it's negative work. For that I get -1.62 x 103 J. That's just a function of the pressure and change in volume -- independent of the type and amount of gas. But then, some unknown amount of heat energy is removed while the gas is compressed. Since the pressure is constant, the temp at the end of the first phase must be 1/5 of the initial temperature (in Kelvins). But that doesn't seem to be enough information.

    Did you post the question exactly as it was printed in your book?

    (Do you know what the answer is supposed to be?)


    edit: heat is DEFINITELY being removed during the first phase. Some to lower the temperature, and more to reduce the volume. But I don't see how we can determine how much is removed if we don't know any more details about the gas; and if we don't know what was removed, I don't see how we can determine how much had to be added back.
     
    Last edited: May 23, 2003
  10. May 23, 2003 #9
    I posted it exactly as it was given to us. From his post, Mulder seems to think that you can work out the temperatures. Not sure how he arrived at that.
     
  11. May 23, 2003 #10
    Ok, I can work out the new pressure, from which you should be able to work out the heat flow. I'm not going to post the working (quite a few lines) till I get up tomorrow though since it's 3.45am here, when I've had a check through it, incase I've made a huge mistake :wink:.

    Will post again in a few hours.
     
  12. May 23, 2003 #11
    gnome,

    I still don't see how you arrived at the work. Sorry, but I'm a dud at Physics. This is my last class before graduation and I am so lost. What formula did you use to get the work. Nothing I use
     
  13. May 23, 2003 #12
    Pressure is constant so work is given by
    w = p * (vf - vi)
    w = 2 atm. * 1.013*105 Pa./atm. * (-8)*10-3 m3
    = -1.62 * 103 J

    if I haven't blown the arithmetic somewhere...:)
     
  14. May 24, 2003 #13
    hmm sorry msdjohnson without knowing, say the number of moles of the gas (unless theres some trick I'm missing). I thought with it returning to it's original temperature, some unknowns might cancel and disappear, but this didn't happen. Unless the answer for the heat transfer can actually only be worked out in terms of the number of moles or something, you might need some input from one of the many people cleverer than me on here, though I'll have another look at it over the weekend :smile:
     
  15. May 24, 2003 #14
    Thanks Mulder. Thank you too gnome. Your help is much appreciated. I will contact the professor to see if there is something missing.
     
  16. May 24, 2003 #15
    Thanks for working on this, Mulder. I'm still playing around with this but not getting very far. Something like this will probably be on my final exam; I hope my prof. isn't this nasty.

    If we assume there is 1 mole of gas, then we can say
    T1 = P1v1/nR
    T1 = 2.026*105Pa. * 10-2m3/8.314
    T1 = 244 K

    T2 = P1v2/nR
    T1 = 2.026*105Pa. * 2*10-3m3/8.314
    T2 = 48.7 K

    P2 = nRT1/v2
    P2 = 8.314*244/(2*10-3m3)
    P2 = 1.014*106 Pa. = 10 Atm.

    But I still can't see any way to determine how much heat is added in the last step without knowing more about the gas -- monatomic, diatomic, etc. or unless I am told how much heat was removed during the compression phase.
     
  17. May 24, 2003 #16
    um, lol I think the answers been staring me straight in my face (as usual :wink:).

    Looking at the constant pressure step;

    [del]Q = nCp(T2 - T1)

    But T1 = PV1/nR and T2 = PV2/nR. Try that, could be this simple. No?
     
  18. May 24, 2003 #17
    Yes, but we don't know what Cp is.
    monatomic/diatomic/something else? An ideal gas can be any of these.

    I've been trying to combine this with the CV phase using Cp = Cv+R, trying to get the Cv to drop out, but all I get is hopelessly messy expressions & no apparent progress. :frown:
     
  19. May 24, 2003 #18
    Well I don't think that's that important. msdjohnson I'm sure can check it with his teacher/prof., I was just assuming that by 'perfect' it meant ideal monatomic (Cp = 5R/2).
    I was thinking that if he wasn't familiar with terms like Isobaric then, although it's a research project (and near graduation = high school?), it won't be extremely complicated. Just my thoughts :smile:

    Incidentally I also have a thermal and kinetic exam next week which covers a lot of this.
     
    Last edited: May 24, 2003
  20. May 24, 2003 #19
    Thanks for all your help guys. We all agree that there is something missing from the project. I will wait to hear from my professor.
     
  21. May 24, 2003 #20
    You're very welcome. Thanks for posting an interesting question. It helped me review for my final.

    Mulder, good luck on yours.

    I'll post if I get stuck on anything; you do the same.
     
  22. May 25, 2003 #21
    Hello guys,

    Heard from my professor and he said the question is good as is. So somewhere in there we must find T for ourselves, or figure out the nature of the gas. Now I am even more at a loss.
     
  23. May 25, 2003 #22
    Ok, well check out my post at the top of this page (page 2) and see if you can see that you don't even need to know the temperature or number of moles of gas. I would assume that it's an ideal monatomic gas too. Try a text book and see if you can come up with anything at all to start you off - formulae etc.
     
  24. May 25, 2003 #23
    Well, I've given this a lot of thought. If anyone can improve on this I'd really like to see it. Really. But
    Mulder had the right idea, and I think this takes it as far as it can go.

    As I said yesterday, in the constant-pressure phase the work done depends ONLY on the change in volume so
    W = p * (vf - vi)
    W = 2 atm. * 1.013*105 Pa./atm. * (-8)*10-3 m3
    W = -1.62 * 103 J

    Just hold that aside for the moment.

    Now, we also know this about the heat that had to be removed during that phase:
    Qcp = nCp(T2-T1)

    And, during the constant-volume phase, the heat that had to be added was:
    Q = nCv(T1-T2) (this is what we are ultimately looking for)

    So applying the 1st Law of Thermodynamics to the 2 phases:
    ΔU1 = nCp(T2-T1) - W
    ΔU2 = nCv(T1-T2) => -ΔU2 = nCv(T2-T1) (work = 0 in this phase)

    and we also know that the final internal energy equals the initial internal energy so
    ΔU1 = -ΔU2
    nCv(T2-T1) = nCp(T2-T1) - W

    now substituting Cp = Cv + R
    nCv(T2-T1) = nCv(T2-T1) + nR(T2-T1) - W
    nR(T2-T1) = W
    n(T2-T1) = W/R
    now multiply both sides by -Cv:
    nCv(T1-T2) = -(W/R)Cv
    Q = -(W/R)Cv

    That's it. W and R are constants, Cv depends on the particular gas. It might be (3/2)R, it might be (5/2)R, etc. I am very confident that there is not enough information given to determine the actual value. W tells us nothing about the gas: it depends only on the pressure and the change in volume. All we know is that some amount of some kind of gas was cooled and compressed at constant pressure, and then reheated at constant volume back to its original temperature. That can be done to any amount of any gas, so there is no information there that lets us determine how much or what kind.

    So, using W = -1.62 * 103 J and R = 8.314 J/K*mol we have
    Q = 195*Cv Joules
     
  25. Jun 2, 2003 #24
    Hello Guys,

    Here is my professor's response on this matter.

    (a) Work is pressure times change of volume.
    (b) Because the process is isothermal (no change in
    temperature). Change of internal energy is zero and
    work will be equal to heat.

    gnome, I think you had the first part right. Thanks again for your help. Blessings.
     
  26. Jun 3, 2003 #25
    (Groan).

    Thanks for your professor's completely trivial answer. You'll find that answer in my equations here:

    Mulder and I were trying to answer your question
    In other words, we were trying to state the amount of heat flowing IN (during the heating phase) separately from the amount flowing out during the cooling phase. Your professor answered the question, "What is the net difference between the amount of heat flowing into the gas and the amount of heat flowing out of the gas?" Or, to put it in terms of that equation, we were looking for
    nCv(T2-T1)
    he is just giving you
    nCp(T2-T1) - nCv(T2-T1) = W

    The problem is English, not Physics.
     
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