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Work done by gas in adiabatic process

  1. Aug 19, 2013 #1
    In an adiabatic process PVγ=constant

    Now I thought the work done by an ideal gas in an adiabatic process was given by the equation here: http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiab.html

    But while doing a GRE question the answer was given as PfVf - PiVi / 1 - γ

    Is this correct (it must be!)? And if so why haven't I seen this equation anywhere before?
  2. jcsd
  3. Aug 19, 2013 #2


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    I think the two are equivalent. Reexpress the formula in the link using, for an adiabatic process, $$P_1 V_1^{\gamma} = P_2 V_2^{\gamma}$$
  4. Aug 22, 2013 #3

    Andrew Mason

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    Did you mean that the GRE answer was W = (PfVf - PiVi) / (1 - γ) ?

    Since it is a reversible adiabatic process (Q=0), the work done BY the gas has to be equal and opposite to its change in internal energy: W = - ΔU (first law).

    For an ideal gas, ΔU = nCvΔT = nCv(γ-1)/(γ-1)ΔT = nRΔT/(γ-1) = Δ(PV)/(γ-1) = (PfVf - PiVi)/(γ-1).

    Last edited: Aug 22, 2013
  5. Aug 25, 2013 #4


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    Alternatively, $$P_i V_i^{\gamma} = P_f V_f^{\gamma} = K \Rightarrow KV_i^{1-\gamma} = P_iV_i\,\,\text{and}\,\,KV_f^{1-\gamma} = P_fV_f$$
    Then sub in.
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