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Heat specific

  1. Sep 1, 2013 #1
  2. jcsd
  3. Sep 1, 2013 #2

    Andrew Mason

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    Cv is a property of the gas.

    From Newton's second law, m = F/a. Does the mass go to zero if there is no force acting on it?

    AM
     
  4. Sep 1, 2013 #3
    I think you're tricking yourself with some algebraic manipulations instead of thinking about the actual physics.

    Here's what I think you're saying:
    "We have Cv= dU/dT = (dQ-pdV)/dT|V = dQ/dT. Now if dQ=0 for an adiabatic process, then dQ/dT=0."

    I think in this pseudo-algebra, the problem is that you can't possibly compute dQ/dT because dT is also zero if you insist there is no heat AND no work. If we have dU=dQ-pdV and both dQ and dV are zero, then dU is zero. Your equation basically changes into to 0=0 and it's not surprising you run into trouble when trying to take derivatives of this.

    Let's look at it from another angle. The equation of state for the ideal gas is PV=NkT. Now in your example we have a closed system so N is fixed (dN = 0), and if we are using the specific heat at constant volume, then we are assuming dV=0. So we have V dP=Nk dT. Without doing any work (which would require nonzero dV), the only possible way to change the temperature is by increasing the system's pressure. The only way we could increase the system's pressure is by adding heat, but for an adiabatic process this is zero. so we have Nk dT = V dP = 0. So when you write dQ/dT = 0/dT = 0/0. So the pseudo-algebra has a divide by zero error.

    Edit: Andrew Mason gave a nice analogy that is probably a better explanation than mine. But his rhetorical question "Does the mass go to zero if there is no force acting on it?" also can be interpreted in terms of the m=F/a=0/0 fallacy.
     
    Last edited: Sep 1, 2013
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