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Work in Enthalpy

  1. Mar 1, 2015 #1
    Hello,

    I have been trying to understand why the work in the change of enthalpy equation is cancelled out at constant pressure. How can work done by a system equal the work done on the system by atmospheric pressure. For example, when a reaction, taking place in piston-cylinder- like apparatus at standard conditions, releases gas, the piston moves up which means work done by the system.So how can the work done on the system by the atmospheric equals the pressure done by the gas from the reaction if we assume that the piston is mass-less.
     
  2. jcsd
  3. Mar 2, 2015 #2

    DrClaude

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    Staff: Mentor

    Newton's third law, maybe?

    What other value of work would you expect?
     
  4. Mar 2, 2015 #3
    The gas in the cylinder pushes the surrounding air outside the cylinder back by exerting a force on it through a distance.

    Chet
     
  5. Mar 2, 2015 #4
    If work done by atmospheric pressure equals the work done by the gas expansion then why the piston moves up ? In other words , if the pressure is applying a downward force on the piston and the expansion is applying an upward force and the piston moves upward(which I assume only expansion has done work), then why the work done by the gas in enthalpy is zero or why PV and W cancel out in the change of enthalpy equation?
     
  6. Mar 2, 2015 #5
    The gas does work on the piston in pushing it up against the atmosphere. So the work done by the gas on the piston is PΔV. If it's a closed system, then ΔU=Q-PΔV, where PΔV is the work. So, ΔU+PΔV=Q. But, expressed in terms of the enthalpy, ΔH=Q. So, if it's expressed in terms of the internal energy, the work is in there, but, if it's expressed in terms of the enthalpy, the work is not in there (or at least hidden). Both ways are equivalent.

    Chet
     
  7. Mar 4, 2015 #6
    So the system does work, but we ignore it by just calculating the heat.
     
  8. Mar 4, 2015 #7
    No. The work is included explicitly when we calculate ΔU. And, ΔH is merely defined as ΔU+Δ(PV), because it is sometimes handy to work with. So the work is always included.

    Chet
     
  9. Mar 4, 2015 #8
    Does this mean there will always be work done at constant pressure? So if I run a reaction in an open beaker there will always be work done.
     
  10. Mar 5, 2015 #9
    I'm not quite sure what you are getting at here. An open beaker is an open system, and, up to now we have been talking about closed systems (I think). Have all your questions been about applying the first law to open systems? Is that what you are learning about now? Also, running a reaction in an open beaker implies that you are dealing with a liquid. Is that the case? Is a gas given off by your reaction?

    Chet
     
  11. Mar 8, 2015 #10
    The thing that is confusing me is why exactly enthalpy=q at only constant pressure? What if the pressure was not constant,what would happen?
     
  12. Mar 8, 2015 #11
    This only happens in a situation where the gas expands by adding heat to the gas, while maintaining the external pressure of the surroundings (against which the gas does work) constant. Under these circumstances, the work done by the gas on the surroundings is w = PΔV, so that ΔU=q-PΔV. So, ΔH=q. Of course, this would also apply to removing heat from the gas, and letting it contract.

    If the pressure is not constant, then the work is ##\int{PdV}##,

    ##ΔU=q-\int{PdV}##,

    and ##ΔH=q+\int{VdP}##.

    Also, I might add that all the equations I have written here apply exclusively to reversible processes. For irreversible processes, the situation is a little more complicated.

    Chet
     
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