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Ideal gas equation

  1. Nov 26, 2009 #1
    PV = nRT

    why when T is constant,

    Vdp + pdV = 0?
  2. jcsd
  3. Nov 26, 2009 #2


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    If T = const., then PV = const.

    If we take the derivative of both sides with respect to "x" (which is any relevant variable), we obtain:

    [tex] \frac{d}{dx}(PV) = 0 [/tex]

    where the right hand side is zero because the derivative of a constant is zero. Now, the product rule from differential calculus says that the left-hand side becomes:

    [tex]P \frac {dV}{dx} + V\frac{dP}{dx} = 0 [/tex]
  4. Nov 27, 2009 #3
    oh. does it have any meaning?

    because somehow, it is manipulated to become the compressibility of the gas. k = -1/v (dv/dp) = 1/p.

    it reminds me of the momentum = mv which became mdv + vdm.

    so does it mean anything the equation pdv + vdp ?
  5. Nov 27, 2009 #4


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    But you don't need to use that relation to get the compressibility. You only need to find the derivative [itex](\partial V / \partial P)_T[/itex] using the ideal gas law.
  6. Nov 27, 2009 #5
    aren't they the same?
  7. Nov 27, 2009 #6
    pdV is pressure times an infinitesmal change in volume while Vdp is volume times an infinitesmal change in pressure. There's no obvious reason why they should be equal aside from the derivation.
    Since work is given by [tex]\int pdV[/tex] in some situations it may be more convenient to calculate [tex]\int Vdp[/tex]
  8. Nov 27, 2009 #7


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    Here's my interpretation of the meaning: if the temperature of the gas remains constant, then so does its total internal energy. If so, then this must mean that:

    energy lost = energy gained.

    -pdV = VdP

    We can interpret PdV as the infinitesimal work done by the gas on its surroundings (a source of loss of internal energy if the work is positive). Similarly, we can interpret the term VdP as the internal energy gained through heat flow (heating a gas at constant volume will increase the pressure, which is proportional to the energy density). So, you could interpret this equation as saying that if the temperature (and therefore internal energy) of an ideal gas is to remain constant, then any energy loss by work done on the surroundings must be exactly balanced by energy gained through heat flow.
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