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Basic thermo problem

  1. Jun 5, 2007 #1
    we know that $\Delta U$ $=$ $\Delta K_{cm}$ $+$ $\Delta I$
    where $\Delta K_{cm}$ is the kinetic energy from cm frame of refrence and the other part is the change in change in potential energy due to various conservative forces.

    also by first law of thermodynamics $\Delta U$ $=$ $W_{ext} + Q$
    can we define $W_{ext}$ and $Q$ in terms of the other variables kinetic energy and potential energy
     
  2. jcsd
  3. Jun 5, 2007 #2

    cepheid

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    I changed all of your dollar signs to [ tex ] [ /tex ] tags (don't put the spaces) so that the LaTeX would work and the post would be readable. Now let me see if it can be understood...
     
  4. Jun 5, 2007 #3
    work energy theorem

    [tex] W_{ext} = \Delta KE + \Delta U[/tex]

    first law of thermo stated that way is just a restatement of this
     
  5. Jun 6, 2007 #4
    but how do you define [tex]W[/tex] and [tex]Q[/tex] using these
     
  6. Jun 6, 2007 #5
    you don't, it's the same formula, Q=ke and w=w
     
  7. Jun 6, 2007 #6
    what is [tex]k,e,w[/tex]
     
  8. Jun 6, 2007 #7
    it's simple
    [tex]Q=KE[/tex] and [tex]W_{ext}=W[/tex]

    your statement of the first law of thermo is the work-energy theorem.
     
  9. Jun 6, 2007 #8
    i don't think that's true because consider an ideal gas let it undergo adiabatic compression although the heat gained is [tex]0[/tex] the temp. of body which is measured by change in kinetic energy changes
     
  10. Jun 6, 2007 #9
    I think the mistake i've made is the the U from the thermo equation equals KE from the work energy theorem and the Q from the thermo equation equals the U from the work energy theorem.

    I would love to confirm or deny though.
     
  11. Jun 6, 2007 #10
    actually the problem while doing that is we know as in the original [tex]U[/tex] equation the change in kinetic energy as seen from c.m is work done by ext forces ,work done by internal conservative forces,work done by internal non conservative forces on the particles minus the work done by external forces on c.m but what
    while change in internal energy is negative of work done by internal conservative forces .then while adding they cancel out then what's the use of defining them.

    also now consider a gas in a cylinder the usual example with a pison.the pressure P of the gas be say less than [tex]P_{ext}[/tex]then the net force on c.m due to these should be [tex]P_{ext}-P[/tex].P due to lower wall by newtons third law,and [tex]P_{ext}[/tex] due to external force above by piston
     
  12. Jun 8, 2007 #11
    someone please....
     
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