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A PV thermodynamics question

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data

    See attachment
    I am having trouble getting my brain to understand what is happening here. See attachment.

    Part a is easy since pv = nrt, Ta = pv/nr = (1E5)(.002)/(.05)(8.31) = 481K



    It is not asked but likewise Tb = 2406 K, and Tc = 1203 K



    Part b is the problem…. I need to fill in the blanks



    I know,

    for Pab there is no work since V is constant so heat added, Q, adds to U.

    for Pbc, isothermal, there is no change in U, so Q must be equal to W

    for Pca, W = p(Vc-Va) = (1E5)(.005 - .002) = 300 , problem states 150J of energy removed so Q = -150



    I am confused by the signs of the variables in the equation deltaU = Q – W.



    I assert the following.

    Heat added results in a +Q while energy removed is a –Q. So Pab should have a positive Q and a +U.



    Work done by the gas in say moving a piston should be +W. Since Pbc is isothermal, the Q must be equal to W resulting in no change in U.



    Can someone straighten me out.


    2. Relevant equations


    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 17, 2014 #2
    I see an obvious problem. The process is not isothermal since the temperature at C is lower than at B. Not sure how this enters into the solution.
     
  4. Nov 17, 2014 #3

    rude man

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    Right, but I'm not checking your math
    B-C is isotermal so Tb = Tc!
    Unfortunately, this problem looks like it was posted by a chemist or whoever, since positive work is here defined as work done ON ther gas. This is unconventional in physics and engineering. WE say dU = δQ - δW = δQ - pdV.
    Anyway, that makes your last statement wrong.
    See my statement above. You have been victimized by chemists!
    (Sorry, @chet, if you'r re reading. Just my perverse sense of humor ... :D ).
    Again, see my observation. I'm sorry you got off on the wrong foot by the problem taking W as negative if p*deltaV is positive.
    EDIT:
    To sum up: do the table over with the first law reading dU = δQ + δW.
    Better yet, stick to dU = δQ - δW and then change all your W's to -W.
     
    Last edited: Nov 17, 2014
  5. Nov 17, 2014 #4
    I am still confused but do you agree that Pbc is NOT isothermal? pv/t =pv/t should still apply correct? To be isothermal, additional heat would have to be added during Pbc, yes?
     
  6. Nov 17, 2014 #5

    rude man

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    Pbc IS isothermal. The problem says that explicitly.
    Yes, heat is added during b-c. Q is not zero for process b-c.
     
  7. Nov 17, 2014 #6
    Then are you saying that pv/t at point B is not equal to the pv/t at point c? I calculate the temperature at C to be less than a b, 1203K at c vs 2406K at b.
     
  8. Nov 17, 2014 #7

    rude man

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    pV/T is the same at all three points. That's just the ideal gas law with unchanging number of moles n: pV = nRT so pV/T = constant = nR. BTW I recommend you use T for temperature, p for pressure, and V for volume. By convetion, v is specific volume. t is almost always time and p should be lower case since it's an intrinsic variable.
    To repeat, since b-c is isothermal, Tb = Tc. That's by definition of "isothermal". So show us how you calculated Tb and Tc.
     
  9. Nov 17, 2014 #8
    I started at point A and used pV = nrT. p = 1E5 Pa, V= 0.002, n = 0.05, r = 8.31. This gives the temperature at A = 481K.
    Then using pV/T for point A = pV/T for point B I get the temperature at B to be 2406K. Then using pV/T at B = pV/T at c I get the temperature at point C to be 1203K. This is not isothermal so I am thinking that the graph showing Pbc to be isothermal is incorrect.
     
  10. Nov 17, 2014 #9

    rude man

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    OMG I just realized that the given "isotherm" is not an isotherm! No wonder you 're still confused!
    You can change the .005 reading on the V axis to .01 for example, or change the p axis such that pV = constant at b and c. I'd go with the .01.

    EDIT: see how much trouble chemists can get you into? :rolleyes:
    Again, my apologies @chet ...
     
    Last edited: Nov 17, 2014
  11. Nov 17, 2014 #10
    Agree, if I change 0.005 to 0.01 then it IS isotherm. Assuming we do this, then we can assume that since the process is isothermal, there is no change in U and the added heat during Pbc = work, yes?
     
  12. Nov 17, 2014 #11

    rude man

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    Yes! Work done BY the gas, not ON it though.
    And I'm sorry I kept ignoring you when you claimed the curve was not an isotherm. One should never trust the problem setters, I should know that by now. Nice work.
     
  13. Nov 17, 2014 #12
    OK, now, based on the fact tht the 0.005 is changed to 0.010 then..

    during Pcb dU = 0, yes?
    during Pab w = 0, yes?

    if for the cycle the work is -700 (per problem) and150J of energy is removed during Pca, is there enough information to fill in the chart?

    Specifically, Does this mean Pca,Q = -150J? and Total,W = -700J ?
    Also Pbc,U = 0 and Pab,W = 0 yes?

    So how do I figure Pab,U and Pab,Q ?

    As you can see, I am still confused,
     
  14. Nov 17, 2014 #13

    rude man

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    Yes.
    Next noticed problem-setter screw-up: delta U for c-a should be negative, not positive. This problem must have been construed after at least 4 drinks.

    Tell you what, Barry, with the change we already had to make (.005 to .01 m3, and with the above screwup, I have to make up a new table for you. Up to you if you want to pursue the problem that way:

    process
    a-b: W = 0 0
    b-c: delta U = 0
    c-a: delta U = -150, Q = -950, W = -800.
    TOTAL W = 700

    Wca has to be correct per our redefined graph with V going from 0.002 to 0.01 cu.m.
    Reason: area under the ca line is -800 = -(.01 - .002)1e5 = Wca.
    Then Qca has to be -950 if delta Uca = -150.
    This could be rearranged to make delta U more negative and Q more positive, as long as Q - delta U = -800.
    Then the other entries can be deduced.

    For starters: total delta U = 0 (why?)
    Also, total W = Wab + Wbc + Wca
    total Q = Qab + Qbc + Qca
    So 0 = Q - W
    Try this if you want. I'd read this guy the riot act for making up so sloppy a problem for you.
     
  15. Nov 17, 2014 #14
    I definitely want to continue this until I thoroughly understand. I will use the above info and attempt to understand the chart and will resubmit it tomorrow. I hope you will be around.

    I am a high school math and science tutor, from pre-alg through calculus, chemistry and physics. Thermo is my weakest subject. This problem came from one of the physics teachers here.I am trying to improve my thermo skills while using this problem as an example. Trying to learn using a bad example is tough. However, when I understand, I doubt I will forget. I have often found bad problems from the teachers in other subjects and when I call them on it, they really get defensive and rarely answer my emails. Thank goodness for physics forum!
     
  16. Nov 17, 2014 #15
    A quicky, the problem says that during Pca 150J is removed so why wouldn't Q = -150?
     
  17. Nov 17, 2014 #16

    rude man

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    I should add that we have one or two VERY high-powered thermodynamicists on this forum, well ahead of me, if you need advanced info later. Or if I slip up they will probaly waste little time in pointing that out to us!
     
  18. Nov 18, 2014 #17

    rude man

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    Because as I said, W = -800 is a certainty if we go with the revised p-V graph. And we have
    delta U = Q - W = Q + 800. But we know that delta U has to be negative since the temperature dropped in going from c to a. So Q < -800 is required, absolutely.

    OK Barry, I have constructed the real table for this process. (I can do this because I can look up specific heats; see comment below). The basis is our new graph with Vc = 0.01, n = 0.05 and gas is Helium. So I will now give you actual entries in lieu of your original table:

    process a-b: W = 0
    process b-c: delta U = 0
    process c-a: delta U = -1202J, Q = -2002J, W = -800J
    Total W = 809J
    I'm still not sure that the table can be filled without using a specific heat but I'll look into that tomorrow (Tues.). But I now have all the cells computed.

    Have you covered heat capacities Cp and Cv? This problem tried to get around invoking these constants by partially filling in the table, but you should still understand that Q = CvΔT at constant volume, and Q = CpΔT at constant pressure. (Ideal gas assumed). In specific heat terms, Cv = n*cv and Cp = n*cp where cv and cp are the molar specific heats of the gas. You can look up cv and cp in handbooks etc. There are also approximate formulae for these constants depending on the molecular nature of the gas.

    I will be happy to help you thru the rest of it. You just need one or two well-directed hints I think, for each of the three processes.
     
    Last edited: Nov 18, 2014
  19. Nov 18, 2014 #18
    Good morning Rude Man. I am back on the problem. I have learned a bit about Cv.

    To summarize, Recall the problem is a pV process Pabca. To make the problem isothermal, we changed the volume from 0.005 to 0.010 and left everything else the same. Using pV=nRT , Ta = 481K, Tb = 2406K and Tc = 2406K. We have a 0.05 mole of Helium.

    Pab is constant volume so no work is done. However since the temperature is increased I calculated'
    Q = nCv(Tb-Ta) = (0.05)(12.5)(2408-481) = 1204J. So, Q added is 1204 J and dU is +1204J. (Cv for Helium is 12.5)

    Pbc is isothermic so dU = 0. The work done, per the Internet, can be caldulated as W = nRTln(Vc/Vb) = 1608J. Is this correct???
    Since dU = 0, we must add an additional Q of 1608J to account for the work done if the temperature is constant.

    Pca is a bit confusing to me. Work ON the gas should be p(Vb-Va) = (1E5)(0.010-0.002) = 800J. Since we already have an internal energy of 1204J and we do work of 800J we must remove heat Q = (1204+800) = 2004 to make the internal energy return to 0 at point A.

    I noticed that Cp = Cv + R = 12.5 + 8.31 = 20.81. Finding Q = nCp(Tc-Ta) = 2004, same as above.

    So, what are we left with. During the complete cycle, we added (1204 + 1608) = 2812 Joules of heat, our net work is (1608 - 800) = 808J so we must remove the excess heat o (2812-808) = 2004 J.At this point, the internal energy is back where we started.

    Comments????
     
    Last edited: Nov 18, 2014
  20. Nov 18, 2014 #19
    See my previous post.
    FYI, I have attached a copy of the original problem I was given. Looking at your previous post I think our numbers are converging.
     

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  21. Nov 18, 2014 #20

    rude man

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    Totally correct including all numbers.
    All OK except we do not talk of "having an internal energy". We talk of changes in U, not U.
    Actually, there are charts giving U (and other parameters such as entropy S and enthalpy H) as functions of the thermodynamic parameters of pressure, specific volume and temperature, and at some refence point of those coordinates (probably standard temperature and pressure and unit specific volume), U, S and H are arbitrarily called "zero". But I'm not sure what the reference point is so don't take that as gospel. The important point here is that U, S and H are state variables which means they are unique functions of the state (p,V and T). This is NOT true of W and Q.

    Not sure why you are still confused. Everything you say is correct except for the punctilio mentioned above.
    What can I say? I am giving you a straight A! You confirmed all the numbers I came up with late last night.

    Saxophone player, eh? Don't know if you're into classical music but maybe you know that the saxophone features in important classical music too. Prokofiev, Villa-Lobos and others.
     
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