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Work done by ideal gas

  1. Jan 29, 2014 #1
    Hi all,

    I was wondering if I am having a definition problem on mechanical work.
    Since dW = -PdV (as I was told in class), is it correct to say the pressure is fixed with W = -PV, since dW = d(PV) = -VdP - PdV = -PdV suggests dP = 0?

    Thanks

    S.
     
  2. jcsd
  3. Jan 29, 2014 #2

    Philip Wood

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    Gold Member

    No, it is not correct. dW is misleading notation. It suggests that dW is an infinitesimal change in some quantity, W, which the system possesses (a so-called 'function of state'). This is not the case. [For the same reason, dQ is also misleading notation, suggesting, wrongly, that some quantity, Q, is possessed by the system.] Work and heat are both energy IN TRANSIT, rather than residing in the system.

    PV is a function of state (because P and V are both functions of state), but it is a mistake to regard dW as a differential of pV (or -PV).
     
  4. Jan 29, 2014 #3
    Hi Philip,

    Thank you for the reply. However, I still think P is fixed in general, leading W = PV. Let's look at an integral form of ideal gas law. In essence, we have dT = (PdV+VdP)/R; integrating this would give the ideal gas law. But the actual integrated solution as PV = nRT implies either dV or dP to be zero.
    Another example would be the heat function in a system of fixed volume, where Q = W = E +PV (Landau textbook chapter 2, the W is a little confusing). He took the derivative of W into dP and dV separately.
    Any thoughts? Thanks again.

    S.
     
  5. Jan 29, 2014 #4
    The differential work done by the gas on the surroundings is dW=PdV, not dW=-PdV. Even if the pressure isn't constant, the differential work is still PdV. This comes from dW=Fdl, where F is the force and dl is the differential displacement. Since F = PA, dW = PAdl. But Adl = dV. So dW = PdV. See my Blog on my PF personal page.

    Chet
     
  6. Jan 30, 2014 #5

    Andrew Mason

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    The integral of dT is not PV unless you start from absolute 0.

    [tex]nRΔT = nR\int_A^B dT = \int_A^B (Pdv + VdP) = \int_A^B d(PV) = P(B)V(B)-P(A)V(A) = ΔPV[/tex]

    P could be constant in which case nRΔT = PΔV (i.e. ∫VdP = 0). Or V could be constant, in which case nRΔT = VΔP. Or P and V could both change. One cannot determine [itex]\int Pdv[/itex] or [itex]\int VdP[/itex] separately for a given process without knowing how P or V varies during the process. But from the ideal gas law, PV = nRT, we know that d(PV) = nRdT

    AM
     
  7. Jan 30, 2014 #6

    Philip Wood

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    Two of the best known ideal cases (but good approximations to real changes) are isothermal and adiabatic expansions. P varies in both of these.

    When P is constant, then W = P ΔV, not PV.
     
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