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Calculate the work that must be done

  1. Apr 3, 2005 #1
    calculate the work that must be done at 298.15 K and 1.00 bar against the atmosphere for the production of CO2(g) and H20(g) in the combustion of 8.50 mol C6H6(l). Calculate the change in internal energy, delta U, of the system.

    What equation do I use to solve for the work needed, and how should I go about finding the change in internal energy.
     
  2. jcsd
  3. Apr 3, 2005 #2
    Is this a question on the Enthaply Change of Combustion???

    The Bob (2004 ©)
     
  4. Apr 3, 2005 #3
    No, Enthalpy of reaction formation, rather.
     
  5. Apr 3, 2005 #4

    GCT

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    I'm not in physical chemistry yet, however you'll probably need to use an equation somewhat related to the following

    [tex] \Delta E_{formation} = q + w [/tex]

    q relates to the change in internal energy, since it is a state function of temperature while the work relates to the expansion against the atmosphere, thus indicating that enthalpy is an energy concept relating to open systems.

    I'm neither familiar with any formulas nor ingenious enough to derive anything off the top.
     
  6. Apr 3, 2005 #5

    GCT

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    actually

    [tex] \Delta E_{formation} = U + PV [/tex], every term pertains to a particular mole value I'm assuming

    I'm sure you can do the rest
     
  7. Apr 3, 2005 #6
    Is the combustion is taking place in a closed container, which is isothermal to the atmosphere ?
     
  8. Apr 3, 2005 #7

    Bystander

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    Initial state: 8.5 mol benzene + oxygen necessary for complete combustion at a specified T and P.
    Final state: combustion products at same T and P.

    You are interested in calculating the work --- constant pressure has been specified --- what other piece of information implicit in the given data do you need to calculate the work?

    You do have to make ONE assumption regarding a phase for water, or you can calculate the work for the possibilities at the T, P given.
     
  9. Apr 3, 2005 #8
    So the process is isothermal expansion done at constant pressure ? I think the questions wording is a bit vague.
     
  10. Apr 3, 2005 #9

    Bystander

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    Oops! Make that "zero" assumptions (ideal gas equation of state is a stipulation at this level of problem).

    "Vague?" The question specifies initial and final states --- short of including the answer, it can't get too much more specific.
     
  11. Apr 4, 2005 #10

    GCT

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    This is a question pertaining to enthalpy, thus it is an open system, not a closed container. For the work aspect of the question, it is isobaric (constant temperature), analogous to a container sealed with a frictionless piston.
     
  12. Apr 4, 2005 #11
    Work cannot be done on a system with an open container.
     
  13. Apr 4, 2005 #12
    Perhaps not vague, but it could have been worded better Assuming what I think its saying, I would have said "8.5 mol of benzene is combusted in a closed container at a constant pressure of 1.00 bar and constant temperature of 298.15 K." Calculate the work done and the change in internal energy of the system."
     
  14. Apr 4, 2005 #13

    GCT

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    We're talking about atmospheric pressure here, work is done by expanding against the atmosphere. The whole concept of enthalpy takes this into account thus

    [tex] \Delta E_{formation} = U + PV [/tex]

    otherwise [itex] energy=q,~related~to~U [/itex] as it is in a closed container.
     
  15. Apr 4, 2005 #14

    Gokul43201

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    This is really pretty straightforward.

    a) From PV = nRT, and from the definition of w = P(V2 - V1), you can find the work done once you write and balance the equation for the combustion of benzene.

    b) From the work done and the enthalpy change (look up std. tables, or Google it), the internal energy E (or U) follows directly from the Second Law (at constant pressure).
     
  16. Apr 4, 2005 #15
    I think we might have our semantics confused. A "closed" container is one i which there is some physical barrier between the system and surroundings. An isolated container is one in which there is physical barrier, and in addition there can be no energy flow between the system and the surroundings. For the purposes of thermodynamics, this would be a sealed, adiabatic container. An "open" container would be one in which there is no physical barrier between the system and the surroundings. It is difficult to use the laws of thermodynamics in the third case, since you cannot separate the interanal energy of the system from its surroundings.

    Expansion work cannot be done on a system with an "open" container, since there is no bulk mass to do work against(i.e., no piston). Gases would diffuse freely, and the work done would be zero.
     
    Last edited: Apr 4, 2005
  17. Apr 4, 2005 #16

    GCT

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    You need to read up on enthalpy and find out exactly what it means...the equation is self explanatory and you should know that enthalpy pertains to an open system. Once again

    [tex] \Delta E_{formation} = U + PV [/tex]

    there should be no confusion about this whatsoever, it simply states that chemical energy of a reaction (frequently pertaining to breakage and bonding of chemical bonds) is converted to the internal energy of the system U and the work done in expanding against a constant pressure atmosphere.
     
  18. Apr 4, 2005 #17

    Please define what V is for a system with an open container.
     
  19. Apr 5, 2005 #18

    GCT

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    the concept used to explain PV work is an abstraction...a ideal frictionless piston is interposed between the chemical system and the atmosphere, the amount to which the frictionless piston changes in volume corresponds to the V.

    you're taking "closed" and "open" too literally. With an open container, the system still needs to expand against the atmosphere, note existence of an atmospheric pressure.

    this site should help

    -enthalpy

    http://dbhs.wvusd.k12.ca.us/webdocs/Thermochem/Enthalpy.html

    -PV work

    http://dbhs.wvusd.k12.ca.us/webdocs/Thermochem/PV-Work.html
     
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