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!

Homework Help: Obtaining gamma (γ) in an adiabatic, reversible expansion

  1. May 22, 2012 #1

    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.


    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


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

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook