1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    CAF123

    User Avatar
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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).

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

    CAF123

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work done by gas in adiabatic process
  1. Work done by ideal gas (Replies: 5)

Loading...