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!

Thermodynamics, Callen's problem, it doesn't make sense to me

  1. Mar 17, 2012 #1

    fluidistic

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Two moles of a particular single-component system are found to have a dependence of internal energy U on pressure and volume given by [itex]U=APV^2[/itex] (for N=2) where [itex]A=10cm ^{-3}[/itex]. Note that doubling the system doubles the volume, energy, and the mole number but leaves the pressure unaltered, and write the complete dependence of U on P, V and N for arbitrary mole number. Answer: [itex]U=BPV^2/N[/itex] where [itex]B=20 cm ^3[/itex].


    2. Relevant equations
    Not sure there are any. To be found.


    3. The attempt at a solution
    First I tried to understand the problem and makes sense of the given expression [itex]U=APV^2[/itex]. I checked out the units in SI units and instead of joule I reach [itex]Jm^2[/itex]. So that this expression doesn't make sense to me, I cannot even proceed further.
    The book is considered of one of the most cited apparently for thermodynamics so I'm sure the book is right and I'm missing something. I just don't know what I'm missing, any help is appreciated.
     
  2. jcsd
  3. Mar 17, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    How'd you get that?
     
  4. Mar 17, 2012 #3

    fluidistic

    User Avatar
    Gold Member

    By forgetting to divide the force by the area in the pressure expression. :smile:
     
  5. Mar 17, 2012 #4

    fluidistic

    User Avatar
    Gold Member

    I don't know how to reach the answer. Attempt:
    Let [itex]V_0[/itex] be the volume when [itex]n=2[/itex]. So that [itex]V(n)=\frac{V_0 n}{2}[/itex] (linear relation). P is constant. I do the same relation for [itex]U(n)[/itex]: [itex]U(n)=\frac{U_0 n}{2}[/itex].
    The relation they provide becomes [itex]U_0=APV_0^2[/itex]. Written with my terms I get [itex]U(n)=\frac{2APV^2(n)}{n}=\frac{APnV_0 ^2}{2}[/itex].
    Something is definitely wrong, I shouldn't be able to get a dependence of U only in terms of n.
    I've tried other things but reached nothing like the result.
     
  6. Mar 18, 2012 #5

    rude man

    User Avatar
    Homework Helper
    Gold Member

     
    Last edited: Mar 18, 2012
  7. Mar 18, 2012 #6

    fluidistic

    User Avatar
    Gold Member

    Thank you for your help. But I think that's what I've done in my last post. I don't see anything wrong with my answer.
    When n=2 I get the given relation. When I duplicate n, I get twice the same energy and the volume duplicates too; the pressure remaining the same just as it should. Yet my answer differs from the one given (worth apparently [itex]\frac{200PV^2}{An}[/itex] if I write it in terms of A rather than B.)
     
  8. Mar 18, 2012 #7

    rude man

    User Avatar
    Homework Helper
    Gold Member

     
    Last edited: Mar 18, 2012
  9. Mar 18, 2012 #8

    fluidistic

    User Avatar
    Gold Member

    Ok rude man, thanks for helping me. I am the OP in case you didn't notice :rofl:
    If my answer [itex]U(n)=\frac{2APV^2(n)}{n}[/itex] is right and the answer provided by the textbook is (written in terms of A instead of B) [itex]\frac{200PV^2}{An}[/itex] which is right by definition, how would one shows they are equivalent?
    To me, this would mean that 2A=200/A. Wait, that means A=10 (I didn't check out the units though) which is... RIGHT!
    Wow.
    Problem solved!
     
  10. Mar 19, 2012 #9

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yep, and my apologies for thinking you were not the OP! :blushing:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Thermodynamics, Callen's problem, it doesn't make sense to me
Loading...