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: Thermodynamic Application Isothermal Work

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data

    One mole of an ideal gas does 3000 J work on its surroundings as it expands isothermally to a final pressure of 1 atm and volume 25 L. Determine a) initial volume and b) Temp of gas

    2. Relevant equations 3. The attempt at a solution
    Well, its isothermal, so Temp is constant, so
    W = nRTln(vi/vf)
    It says it expands, so the initial volume is <25 L...
    that's kinda all I got right now; I'm not sure how to get the next step, since both T and vi are in the equation. It must have something to do with pressure.
     
  2. jcsd
  3. Oct 23, 2008 #2

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    How do you find the temperature of the gas? Find that and then determine what ln(vi/vf) has to be if W=3000J.

    AM
     
  4. Oct 23, 2008 #3
    PV = nRT
    (1.013 x 10^5)(10 m^3) = (1)(8.314)T
    T = Some really big, way illogical number? Unless the moles thing is wrong...

    ARGH. This one is frustrating me. I'm wondering if I don't have to find volume first??
     
    Last edited: Oct 23, 2008
  5. Oct 23, 2008 #4
    Oooh...

    3000 J = nR(PV/nR)(ln(vi/vf))
    3000 J = (10.13 x 10^5)(25 L) ln(vi/25)
    1.184 x 10^-3 = vi/25
    vi = 0.029 L??

    Possible??

    EDIT: No, cause my final temp is still 304606.7 K !!
     
    Last edited: Oct 23, 2008
  6. Oct 24, 2008 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Start with PV=nRT

    [tex]P_iV_i = nRT = P_fV_f[/tex]

    You know [itex]P_f, V_f[/itex] and n so work out T. Be careful to use the correct units for volume. Units are litres or 10^-3 m^3.


    Then, from W = nRTln(vi/vf), work out what vi is.

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