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Change in internal energy due to adiabatic expansion of ideal gas

  1. Apr 14, 2013 #1
    what is the change In internal energy of an ideal gas when it is expanded Adiabatically from (p °,v°) to ( p,v )?
    The relevant equations :
    dQ= dU+ pdv ; pV^ r = K (constant). r = Cp / Cv.
    Attempted solution :
    During adiabatic process dQ=0 ; p= k v ^ -r
    Th f r : du= -k v^ -rdv
    Integrating the above relation : Uf-U i = -k [ ( v ^ -r + 1 ) / -r + 1 ] v to v° .
    = -k [ {(v ^ 1-r ) / (1-r)} -{(v° ^ 1-r ) /( 1-r) } ]
    = 1 / (1-r)[ (k v ^ -r) v-( k v° ^-r ) v° ] = 1/ (1-r) [ pv-p° v° ] .
    IS there any mistake ? Am i absolutely right ?? .
     
  2. jcsd
  3. Apr 15, 2013 #2

    TSny

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    I believe you dropped the negative sign in front of the k when writing the last line. Otherwise, it looks good to me.

    Using r = CP/Cv and CP-Cv = R , you can rewrite 1/(1-r) in terms of R and Cv if you wish.
     
  4. Apr 15, 2013 #3
    My bad .what I intended to write was 1/(r-1)( pv-p° v° )
     
  5. Apr 15, 2013 #4

    TSny

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    Good.

    There's another approach to this problem. Internal energy U is a state variable. So, the change in U between the state (Po,Vo) and (P,V) doesn't depend on whether or not the process was adiabatic. Any process connecting those two states gives the same ΔU.

    For an ideal gas, U = nCvT. Using the ideal gas law, PV = nRT, this can be written as U = (CV/R)PV. So, ΔU = (CV/R)(PV - PoVo) and you can show that CV/R = 1/(r-1).
     
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