1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Thermodynamics: First law and work definition

  1. Jun 17, 2013 #1
    I've been having some doubts regarding the definition of work in thermodynamics and how it is related to the change of energy in the system. I have tried asking this in physics stack exchange but I was unable to obtain a satisfactory answer; now that I am trying again in this forum I will aim to do my best in being clear about my question and I will also try to motivate it with a "problem" from the begining. I apologize if this is not the right subforum for this, wasn't sure where else to post it.

    When teaching thermodynamics, the work exchanged by a system is usually given as a definition ([itex]W_{PV}=\int -p_{ext}\cdot dV[/itex]) in the context of the first principle ([itex]\Delta U=\Delta Q + \Delta W [/itex]), which is then said to be an expression of the conservation of energy. However, classical dynamics already introduce concepts of work and conservation of energy and, at least to me, the correlation between these and the thermodynamical ones is not that apparent. For example, considering an adiabatic process, it is not immediate that the change of internal energy of a system will be equal to the work of the external forces (why the external? why not the internal? why not the resultant?).

    Looking through the internet, I found this paper, which seems promising but it gets a little messy near the end and so I wasn't able to determine if it has good science; if someone is willing to check it out, it would be great.

    1. The problem statement, all variables and given/known data
    Consider a very big cilinder with a piston in the middle, dividing it into two sections (A and B), both containing a certain ammount of gas. All walls, including the piston, are adiabatic. The piston is locked and system A has a bigger preassure that system B. If we were to release the piston but lock it again before any significant change in preassure of any of the sections: calculate the work and ΔU for each section and for the whole system.

    2. Relevant equations
    [itex]dW_{PV}=p_{ext}\cdot dV[/itex] and [itex]dU=dQ+dU[/itex] (I am using the egocentric reference frame).

    3. The attempt at a solution
    Since each system is adiabatic, the change of internal energy must be equal to the work for that system, which is equal to [itex]p_{ext}\cdot\Delta V_{int}[/itex]. So then:

    [itex]\Delta U_{A}=W_{A}=-p_{B}\cdot\Delta V_{A}[/itex]

    [itex]\Delta U_{B}=W_{B}=-p_{A}\cdot\Delta V_{B}=p_{A}\cdot\Delta V_{A}[/itex]

    [itex]\Delta U_{T}=\Delta U_{A} + \Delta U_{A} =(p_{A}-p_{B})\cdot\Delta V_{A}>0[/itex]

    But, on the other hand, the total system is adiabatic and its walls don't move, so its change in internal energy must be equal to 0. What I suspect is happening here is that I should also consider the loss of potential energy of the piston, but I have never seen such a consideration in other problems of thermodynamics. Moreover, I have many times heard the reasoning that "the work done to system B must be the same as the work done by system A" without any backing argument, which in this case seems clearly not to be true.

    The reason I believe this problem is related to my conceptual doubt, is that it all has to do with not knowing clearly why is it that we relate the work of the external forces with the changes in the internal energy, what are the approximations and/or considerations behind that reasoning; and hence not knowing when other considerations need to apply.
  2. jcsd
  3. Jun 17, 2013 #2
    It is an adiabatic process and You are takink pressure constant. The total change in internall energy of the system is zero. The formula for work done is dW = pdv. When pressure is constant it becomes p*delta(V).

    There are three equations which i use in the thetmodynamics. I am able to solve most of the problems usnig these three only.

    Q = U + W
    dW = pdV
    U = fRT/2

    you are having problem with the second one i guess.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted