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First law of thermodynamics & state functions

  1. Mar 23, 2015 #1
    1. The problem statement, all variables and given/known data
    1 kg air at the pressure ##10^6##Pa and the temperature ##125^\circ C = 398K## expand until the volume is 5 times larger. The expansion is done with change in heat at every moment being ##1/4## of the work done by the gas. Calculate the end pressure.


    2. Relevant equations
    ##dU = dQ+dW##
    ##dU = nC_vdT##
    ##W = -\int pdV##

    3. The attempt at a solution
    We have that ##dQ + dW = \frac{5}{4}dW## and the two first equations can then be combined into
    ## \frac{5}{4}dW = nC_vdT##
    And if ##dW = -pdV## from the third equation we get
    ## nC_vdT = -\frac{5}{4}pdV##
    Assuming an ideal gas and for air ##C_v = \frac{5}{2}R##
    ## nC_vdT = -\frac{5}{4}nRT\frac{dV}{V} \Longleftrightarrow 2\int \frac{dT}{T} = -\int \frac{dV}{V}
    \Longleftrightarrow ln(\frac{T_2}{T_1}) = -\frac{1}{2} ln(\frac{V_2}{V_1})##
    Using that ##\frac{V_2}{V_1} = 5## we get
    ##T_2 = T_1\frac{1}{\sqrt{5}} \approx 178K##

    Using the ideal gas law again to calculate the starting volume with ##n = 1000/29 \approx 35.5 mol##
    ##V_1 = \frac{nRT_1}{p_1} \approx 0.114 m^3##
    ##V_2 = 5V_1 \approx 0.57 m^3##
    And invoking the ideal gas law a third time
    ##P_2 = \frac{nRT_2}{V_2} \approx 92 kPa##

    The correct answer according to the book should be ##123kPa##. Any ideas where i go wrong?
     
  2. jcsd
  3. Mar 23, 2015 #2
    Maybe the heat is going in the opposite direction so that it is 3/4 rather than 5/4.

    Chet
     
  4. Mar 24, 2015 #3
    Thats right! If i use 3/4 i get exactly the right answer. But why should the heat be negative? The book im using is using the sign convension
    ##dU = dQ + dW##. Our book doesn't seem to bother explaining the sign convention here but according to the wikipedia article
    any energy transfer to the system should be positive. And the heat is transfered to the system.
     
  5. Mar 24, 2015 #4

    ehild

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    The gas loses energy when doing work. In your first relevant equation, dW means the external work done on the gas. If the process is quasi-stationary, dW = - PdV where PdV is the elementary work done by the gas. So the correct form of the first law is dU = dQ-PdV, and dQ=1/4 PdV
     
    Last edited: Mar 24, 2015
  6. Mar 24, 2015 #5
    I guess you were supposed to read their minds. If the gas was expanding, then, in your sign convention, W was negative. So I guess they thought that, to assist with this, some heat should be added. In any event, the problem statement was poorly worded.

    Chet
     
  7. Mar 24, 2015 #6

    ehild

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    The wording of the problem was clear. The heat added was 1/4 of the work done by the gas, positive in case of expansion.

    In the equation dU = dQ + dW , dQ means the heat added to the gas and dW is the external work done on the gas.The work done on the surroundings by the gas is opposite to that. dWgas=-dWexternal. So dU=dQ-dWgas The gas is expanding, its work, PdV is positive, so the heat added, which is 1/4 of that work is also positive dQ=1/4 dWgas

    dU = 1/4 dWgas - dWgas= - 3/4 PdV
     
  8. Mar 24, 2015 #7
    With all due respect, ehild, I must admit that I was confused by the problem statement too. So the OP was not alone in this. The part that confused me were the words change in heat. I have no idea what that's supposed to mean. Heat is not a state function, so use of the word "change" with regard to heat is confusing (at best). To be more precise, I would have used the words "The expansion is done with the (cumulative) addition of heat at every moment..." You automatically made this change above when you said "The heat added was..." I think that whether the problem statement was clear or not is a judgement call.

    Chet
     
  9. Mar 24, 2015 #8

    ehild

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    Well, the expression "change of heat" was sloppy, as heat was not a state function, as you pointed out. As the First Law is usually written as dU = δQ + δW, both heat and work positive when added to the system, I translated the expression "change in heat at every moment" as δQ. .I am used to decipher texts translated from other languages - as English is not my native language, I am accustomed to poor English.
     
  10. Mar 24, 2015 #9
    Thanks for replying both of you! The question originally wasn't written in english so it may be that I mistranslated it myself as well adding further confusion! I think "change" may have been a bad choice of translation by me, "added" may be another more accurate translation.

    Anyway thanks for helping me out both of you!
     
  11. Mar 25, 2015 #10

    ehild

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    I am curious, on what language it was written?
     
  12. Mar 25, 2015 #11
    Swedish originally, actually I think the confusion probably was from me misinterpreting/mistranslating the question.
     
  13. Mar 25, 2015 #12

    ehild

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    That was a good translation, only the "change of heat " was not accurate. I wonder how it was written in Swedish.
    Your problem in the solution was that you mixed the work done by the gas with the work done by the environment on the gas. The work done by the gas is PdV, and you took it -PdV.
     
  14. Mar 25, 2015 #13
    A correct translation would be added but as you said my problem was confusing the sign of the work.
    I add the original swedish question in case you feel you get something out of it (not that many people would know swedish I would assume).

    1,0 kg luft med trycket ##10^6## Pa och temperaturen ##125 ^\circ## C får expandera tills volymen blivit fem gånger större. Expansionen sker så att i varje ögonblick den tillförda värmemängden är en fjärdedel av det arbete gasen uträttar. Beräkna sluttrycket. 1 kmol luft har massan 29 kg, ##C_V = 5/2R##.
     
  15. Mar 26, 2015 #14

    ehild

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    Thank you for the Swedish text! It looks a bit similar to English and German! I understood some words :) Google Translate gave the English meaning as


    1.0 kg air pressure is 106 Pa and the temperature 125∘ C must expand until the volume has been five times greater. Expansion is such that at each moment the amount of heat supplied is a quarter of the work the gas does. Calculate the final pressure.

    The first sentence is not quite English, yours is much better. But the second sentence is better than yours. It looks that Google Translate is real help for you!

    "tillförda" is feed ~ supplied, so it was not change of heat, but supplied heat.

    https://translate.google.hu/

    @Chestermiller : Chet, can you make it really English, please?
     
    Last edited: Mar 26, 2015
  16. Mar 26, 2015 #15
    1 kg. of air at a pressure of 106 Pa and 125° C expands (presumably reversibly) to 5 times its original volume. At each infinitesimal step along the expansion path, the infinitesimal amount of heat added to the gas is equal to 1/4 the infinitesimal amount of work done by the gas on its surroundings. Calculate the final pressure.

    I've done my best to make it as precise and unambiguous as possible. Hope that this helps.

    Chet
     
  17. Mar 26, 2015 #16
    I may definitively have use of google translate to check my translations. Mostly in this case I should've reread the question an extra time, sometimes you read something but lose information when rephrasing it in your mind (even in the same language). I have a lot of experience reading english but almost noone writing or speaking it myself making it hard sometimes :)

    Cheers guys!
     
  18. Mar 26, 2015 #17

    ehild

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    Thanks :oldsmile:. That infinitesimal amount makes it really precise! Also, stating that the process is reversible.
     
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