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Finding pressure increase required to impart 1J of work

  1. Oct 2, 2015 #1
    1. The problem statement, all variables and given/known data
    a) Estimate the pressure increase required to impart 1 J of mechanical work in reversibly compressing 1 mol of silver at room temperature.
    b) What pressure rise would be required to impart 1 J of work to 1 mol of alumina at room temperature? For AL2O3 take the molar volume to be 25.715 (cc mol^-1) and Beta= 8.0 x 10^-7 (atm)^-1

    2. Relevant equations
    mechanical work = -PdV
    dV=V.alpha.dT - V.beta.dP
    (maybe: dU = (Cp - PV.alpha)dT + V(P.beta - T.alpha)dP )

    3. The attempt at a solution
    I have tried and tried to come up with a solution but I just can't figure it out.
    See attached jpg.
    I would appreciate any thoughts on the question

    Attached Files:

    Last edited: Oct 2, 2015
  2. jcsd
  3. Oct 2, 2015 #2
    Your equation in the snapshot involving P's is correct. Just substitute numbers into it. Assume that the starting pressure is 1 atm.

  4. Oct 2, 2015 #3
    Thank you, can I use that equation for part (b) as well? It does not specify that it is mechanical work or reversible there.
  5. Oct 2, 2015 #4
    Same as (a).
  6. Oct 2, 2015 #5
    Oh wow, I've been really over thinking this question it seems! thanks again!
  7. Sep 15, 2016 #6
    can I have a solution to this problem?
  8. Sep 15, 2016 #7


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    Staff: Mentor

    Nope. That's contrary to Physics Forum rules (you should read the guidelines). If you wish to solve this problem and you can't glean enough hints from the current thread's content then you can show your own attempt and present what part you don't understand, or start a new, separate thread and show your attempt there.
  9. Oct 1, 2017 #8
    Is there any unit conversions that need to be done if my molar volume is given in cc/mol I thought I might need to convert to m^3. The beta value for silver given in my textbook is 9e7 atm^-1, and the molar volume is given as V^S= 10.27 cc/mol, but there is also a V^L=11.54 cc/mol, and I'm not sure of the difference. I know this is an old thread, but any help would be greatly appreciated. I came to the same equation, but cannot get a correct answer.
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