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A rather simple prb.

  1. Apr 25, 2006 #1
    hi!!

    Just gave a chem obj. type exam recently. I am confused about one question on thermodynamics.was hopin some1 could throw light on this problem.

    the question goes hence-
    An exothermic reaction takes place in an adiabatic process.what will happen to the temp of the system?
    A)the temp will increase
    B)the temp will decrease
    C)it will remain constant.
    D)any of the above.

    so now i wrote d answer as D.but the correct answer it seems was given as "A".

    my thought process behind it was that- the exothermic reaction "injects" a FIXED amount of energy into d system.now since an Adiabatic process is taking place, we start with internal energy = "injected" energy plus d amount of internal energy that d system already possessd bfore d commencement of d reaction.

    now since an adiabatic process is taking place, work may be done on d expense of internal energy.if the work done(BY THE SYSTEM) is equal to the energy given by the reaction, the internal energy(and hence the temperature) will remain as it was bfore.now on the other hand, had the SYSTEM done work greater than that given by the reaction, some amount of the INITIAL internal energy would have gotten used up.and hence, the temperature of the system would have decreased.lastly, had no work been done, or had d SURROUNDINGS done work on d system, d temp would have increased.

    can pls any1 comment on whether my logic is correct? or is A the correct answer?

    any help would be sincerely appreciated

    Thanking u
    Nikhil
     
  2. jcsd
  3. Apr 25, 2006 #2

    vanesch

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    You are entirely right. In fact, the question is totally silly, for several reasons. The first reason is as you mention: an adiabatic process is at constant entropy, so there can be an arbitrary amount of exchange of mechanical energy. The only thing we cannot have is an exchange of heat energy.
    But second, it is ridiculous to talk about an *exothermal reaction* during an *adiabatic process*, an exothermal reaction being irreversible!
     
  4. Apr 25, 2006 #3

    siddharth

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    Hmm.. Vanesch, what I'm learning now is that, only a reversible adiabatic process is at constant entropy.

    Besides, why is it not right to talk about an exothermal reaction during an adiabatic process? In fact, it's possible to have adiabatic reactors.

    I think the original question is poorly worded in any case. Is it an open system or closed system? As vanesch pointed out, is there any work interaction?
     
  5. Apr 25, 2006 #4
    Well in response to the siddharth's question, I asked the Ppl who set the paper, and the response that i got was that, there is "no mention" of any expansion that may occur in the system and hence they say that answer "A" is correct.

    Now my arguement is that, since they have mentioned a "adiabatic process" we can arbitarily change any of the state variables, and hence there is no need for there to be an explicit mention of "volume expansion".

    Also, since they have said so, I would also like to ask if it is possible for the SYSTEM to do work despite the temperature dropping(due to change in internal energy, which will be converted into work by the SYSTEM) WITHOUT change in VOLUME?? is this POSSIBLE??

    thanks
    Nikhil
     
  6. Apr 25, 2006 #5

    Gokul43201

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    The statement about the adiabatic process, I believe, is merely to tell you that you have an isolated system (ie: the contents of an infinitely insulating box). Now, within some part of this system, you have an exothermic reaction (which is another way of saying that PE is being converted to KE). This increase in total KE is nothing but an increase in the temperature of the system.

    To take an extreme (for the sake of illustration) case, assume the box had two H-atoms flying about at some (sufficiently low) speed v. Initially, they are far enough from each other that the PE between them can be neglected. At some point of time, they happen to approach close enough that they feel the mutual attractive force of the other atom. This causes an acceleration towards each other (in the process decreasing the PE and increasing the KE). Soon, they are close enough that a bond forms between them (they attain the minimum possible PE). The energy released during this bond formation goes into their KE - the resulting molecule has a total KE greater than the sum of KEs of the individual atoms.

    If we have a large number of such atoms, we would measure the increased KE as an increase in temperature, so long as the molecules do not lose energy through inelastic collisions with the box wall (fortunately, this is disallowed by calling the system isolated).
     
  7. Apr 25, 2006 #6
    as per what u have said(gokul), u have converted the system into just an isolated system.I completely understood your explaination but the main prob here is that u have modified the actual statement of the question, in a sense, havent u? i mean when i say that there is an "adiabatic process" taking place, i have quite a bit of freedom in changing the parameters of the systems, keeping it isolated.while ur case considers only the fact that it is isolated.and not that the parameters can be changed.

    what do u say?
     
  8. Apr 25, 2006 #7

    Gokul43201

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    I'm not sure I understand what you're saying here. How do you define an adiabatic process ?
     
  9. Apr 25, 2006 #8

    Borek

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    Adiabatic process is the one where heat is not exchanged with surroundings, but AFAIK nothing prevents it from loose energy doing some work. Work cycle in piston engine is almost adiabatic (fast enough to make heat exchange neglibile), yet it does some work loosing energy.

    IMHO question is poorly worded - the idea was that there is no heat exchange so temperature must rise, but if system is allowed to expand (without heat exchange) its temperature may be constant or even decrease.
     
  10. Apr 25, 2006 #9

    Gokul43201

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    I agree with that. I assumed the system was isochoric (else molecular speeds will not be conserved), which is unjustified.
     
  11. Apr 27, 2006 #10

    Borek

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    Heh, I have forwarded the question to some other place inhabited mostly by teachers (HS, college and Uni levels). Somebody proposed to add ice to the system to use latent heat to not allow temperature change :smile:

    And D is THE correct answer.
     
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