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Ideal Gas Work

  1. Feb 20, 2010 #1
    The expression for the work done by an ideal gas is:

    [tex]\int_{V_i}^{V_f} P dV [/tex]

    But what if [tex]V_f \prec V_i [/tex]? If [tex]W \prec 0[/tex] does that mean

    1) work is done on the gas or

    2)on this case, [tex]W= - \int_{V_i}^{V_f} P dV [/tex] so that [tex]W \succ 0[/tex] and W still means the work done by the gas?

    Note: The question may seem obvious but the derivation of the above expression only makes sense for work done by the gas
     
  2. jcsd
  3. Feb 20, 2010 #2
    If the volume decreases, then the gas does negative work on the environment, which also means that the environment does positive work on the gas.

    [tex]
    \int_{V_i}^{V_f} P dV
    [/tex]
    is always the work done by the gas (but it can be a positive or negative value). Also:
    [tex]
    W_{on gas} = -W_{by gas}
    [/tex]
    which is a direct result of Newton's third law.
     
  4. Feb 20, 2010 #3
    JaWiB,

    I think I got it. According to what you said, when we say, in thermodynamics, that work is done by the system only when

    [tex] W \succ 0[/tex]

    this is not true. Positive work done by the system is when [tex] W \succ 0[/tex] and negative work done by the system is when [tex]W \prec 0[/tex].

    Somehow, this a bit more confusing than the previous formulation: [tex]W \succ 0[/tex] - work by the system; [tex]W \prec 0[/tex] work on the system.

    Any thoughts on how to make it clearer?
     
  5. Feb 20, 2010 #4
    Yes. Some external agent must do work in order to decrease the volume. The agent does work on the gas during this process. By the way, the work done by the agent is ultimately manifest as an increase in gas temperature.
     
  6. Feb 20, 2010 #5
    The first law of thermodynamics: [tex]{\Delta}U = Q - W[/tex]
    when W is the work done by the system. If W is positive, and no heat is transfered to the system, then it loses energy (and this is often manifest as a decrease in temperature).

    The way you're thinking about it might be problematic because it is tempting to think of work as a one-way process: either the work done by the system is positive and the environment gains energy or the work done on the system is positive and the system gains energy. In reality, if the system gains energy the environment loses energy, and if the environment gains energy then the system loses energy.
     
  7. Feb 20, 2010 #6
    Not necessarily. The gas can also lose energy on the form of heat such that Q=W.


    JaWiB,

    Good explanation. Though at first it seemed confusing, I think I got it now.
     
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