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Obtaining gamma (γ) in an adiabatic, reversible expansion

  1. May 22, 2012 #1
    PROBLEM:

    One mole of an ideal monatomic gas initially at 300 K (T1) and a pressure of 15.0 atm (P1) expands to a final pressure of 1.00 atm (P2). The expansion occurs via an adiabatic and reversible path. Calculate q, w, ΔU, and ΔH.


    SOLUTION:

    q = 0 (adiabatic; no heat exchange occurs)
    Thus, ΔU = w = CvΔT

    ΔH = CpΔT

    ΔT = T2 - T1

    Need T2 to calculate values (not an isothermal expansion)...

    Known equation for reversible adiabatic process: P1V1γ = P2V2γ (γ = gamma)

    Using PV = nRT, substitute in and simplify to obtain expression for T2...

    T2 = T1(P1/P2)((1-γ)/γ)

    *where problems arise...

    γ = ?

    The solutions manual says γ = 5/3, but I don't know how this value is obtained.

    I do know that, as an ideal monatomic gas, the internal energy (U) of the gas (per mole?) is: 3/2RT (we assume only translations occur). But how is this related to Cv, Cp, and more importantly, γ?
     
    Last edited: May 22, 2012
  2. jcsd
  3. May 23, 2012 #2
    γ can be obtained if you know the degrees of freedom for an ideal mono atomic gas.
     
  4. May 23, 2012 #3

    [itex]\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}[/itex]

    Where f is the number of degrees of freedom of the gas.

    Taking the monoatomic case, the number of degrees of freedom are 3. So, [itex]\frac{2+3}{3} = \frac{5}{3}[/itex]

    Note that the internal energy U of one mole monoatomic ideal gas is

    [itex]\frac{3}{2}RT[/itex]

    But in general, for any ideal gas, internal energy is

    [itex]\frac{1}{2}fnRT[/itex]
     
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