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Help solving this thermo problem.

  1. Sep 12, 2009 #1
    this problem is quite long, all im asking is some for some help by clarifying and hints so i can try to solve this question. Thanks in advance!
    Consider the combustion of liquid ethanol, C2H5OH(l), to form
    CO2(g) and H2O(g). Suppose we have 1 mole of ethanol in the presence of 4 moles of O2
    gas in balloon at a temperature of T = 298K and a pressure of 1 atm (i.e., pi = pext = 1atm).
    The ethanol combusts to completion, and assume we can neglect any changes in volume due to
    the liquid ethanol. All of the gases can be treated as ideal gases.

    a) If the combustion takes place such that the internal temperature of the gas remains constant, and the external pressuredoesn’t change, calculate ΔH, ΔE, q, w (hint: even though ΔT = 0, ΔE and ΔH are nonzero since the chemical energy and number of moles of gas have changed).

    b) Suppose instead that the balloon’s volume was fixed, and that that the balloon could
    not exchange heat with the surroundings (i.e., q = 0). Instead, the heat from combustion was
    used to only increase the temperature of the gases in the balloon. Question: what is ΔE, ΔH,
    w, and ΔT. What is the final temperature? What’s the final pressure? Hint: remember that
    ΔH = Ef + pfVf- ΔEi -ΔpiVi.

    c) Now suppose instead that the balloon was thermally isolated from the environment
    (i:e:; q = 0) and the balloon was allowed to expand the volume of the balloon against a constant
    pressure of 1 atm. What is the final temperature and volume of the balloon? What is ΔE, ΔH, w,
    and ΔT?
     
  2. jcsd
  3. Sep 12, 2009 #2
    sorry for double post:

    can you see if this is correct:

    a)
    using enthalpy of formation (info found from a table of values) of the reaction, i can say, enthalpy = delta(U) + delta (pv)
    and that delta (pv) = RT* delta n

    with that, i solve for delta U

    work is p delta v, (vfinal found from using info given using gas ideal gas law)

    therefore delta q = delta u - work.


    b was tricky for me:

    if q = 0, then delta u = delta work

    using heat of formation from part A:
    delta H = delta u + delta (pv)

    now im stuck. im left with too many unknowns, a final temperature, and a final volume.

    what im truly confused about is while Work = 0 under constant volume, i dont think it applies to molecules that have combusted. ugh
     
  4. Sep 13, 2009 #3

    Andrew Mason

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    How many moles of gas are produced? How many did we start with? What volume do/did they occupy? What work is done (on the external atmosphere) in the reaction? How is the work done related to the heat flow? to the change in internal energy?

    AM
     
  5. Sep 13, 2009 #4
    how much moles of gas was produced:

    EtOH + 4 O2 -> 2CO2 + 3 H2O + O2 (left over from the reaction)
    therefore N initial = 4 moles of gas, N final = 6 moles of gas

    using pv = nrt with initial conditions would lead to roughly 97 L

    messing around with the ideal gas law equation it leads to:
    Vfinal = (#mol final/#mol initial) V initial

    .... i explained how to get the info in my second post, ok more specifcally
    Work done on external = pext delta V = 1 atm ( (6mol/4 mol)(97 L) - (97L))
    work done on external = 48.5 atm L = 4914.26 joules

    change in enthalpy = delta (u) - delta (pv)
    and that delta pv = (delta n) RT = (6-4)RT

    enthalpy of formation (in this case combustion) = -1243.75 Kj
    therefore

    delta (u) = -1243.75 Kj - 2R(298K) = -1248.70 Kj

    delta (u) = delta q + delta work
    delta q = -1248.70 Kj - 4.91426 Kj = -1253.62

    something feels wrong about it tho :-/
     
    Last edited: Sep 13, 2009
  6. Sep 13, 2009 #5

    Andrew Mason

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    Why are you using the enthalpy of combustion? It seems to me that the heat from combustion is irrelevant, since we know the initial and final state of the gases. The change in internal energy minus the work done gives [itex]\Delta Q[/itex], which is the net heat flow. The net heat flow is all that you are concerned with in part a.

    AM
     
  7. Sep 13, 2009 #6
    i thought i could use that to work around the Delta(H) = Delta(U) + Delta(pV) so i could solve for Delta(U), and that knowing work, would just give me q.

    i guess i am wrong. how am i supposed to approach it?
     
  8. Sep 14, 2009 #7

    Andrew Mason

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    You start with 4 moles of O2 and you end up with 2 moles of CO2, 3 moles of H2O and one mole of O2 at the same temperature and pressure. What is the difference in internal energy? What is the work done? How is that related to the net heat flow according to the first law?

    AM
     
  9. Sep 14, 2009 #8
    yeah man, im at a complete halt. please help
     
  10. Sep 14, 2009 #9

    Andrew Mason

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    You have essentially converted 3 moles of O2 into 3 moles of H2O + 2 moles of CO2, all at the same T and P. What is the internal energy change? [itex]\Delta U[/itex] would be the internal energy of 3 moles of H2O + 2 moles of CO2 minus the internal energy of 3 moles of O2. Can you work that out? What is Cv for each gas? How is Cv, n, and T related to U?

    There is a slightly easier approach using Cp, since this is all at constant pressure. What is the relationship between Q (or H) and Cp, n, and T?

    AM
     
    Last edited: Sep 14, 2009
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