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Question about reversible adiabatic expansion

  1. Oct 12, 2015 #1
    Hello guys, I have some troubles to understand a reversible adiabatic expansion.
    we have dq=0 and dw=-pdV and for a perfect gas dU= Cv*dT. I don't really understand why dU= Cv*dT since we have a variation of volume and it is established that dU= Cv*dT when we haven't any modification of the volume. Considering a small variation dV is like constant volume ?
     
  2. jcsd
  3. Oct 12, 2015 #2
    There is a difference between how Cv is measured experimentally, and how it is used to solve problems. In a constant volume experiment, the amount of work done is zero, so ΔU=Q. So, to get the change in internal energy as a result of changing only the temperature, we measure the heat added in a constant volume experiment. The heat capacity Cv is defined to be equal to the (∂U/∂T) at constant V. This is how we measure Cv experimentally. However, for an ideal gas, we know that the internal energy U is a function only of T, but not of specific volume V. So we can then use the measured value of Cv from the constant volume experiment to calculate the change in U for any other types of process paths for an ideal gas. For materials other than ideal gases, it isn't valid to assume that the internal energy is independent of specific volume.

    Chet
     
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