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Thermodynamic identity (non-quasistatic?)

  1. Nov 20, 2008 #1
    1. The problem statement, all variables and given/known data

    A cylinder contains one liter of air at room termperature and atmospheric pressure. At one end of the cylinder is a massless piston, whose surface area is 0.01 m^2. Suppose that you push the piston _very_ suddenly, exerting a force of 2 kN. The piston moves only one millimeter, before it is stopped by an immovable barrier of some sort.

    Use the thermodynamic identity to calculate the change in the entropy of the gas (once it has again reached equilibrium)

    2. Relevant equations

    [itex]dU = TdS - PdV[/itex]

    3. The attempt at a solution

    Very sudden processes can be assumed adiabatic. The pressure at the beginning and the end is atmospheric, therefore rearranging the thermodynamic identity gives

    [itex]dS = \frac{-P_\mathrm{atm}A d\ell + dQ}{T} = 0.003 \mathrm{J/K}[/itex].

    My concern here is that since the process is non-quasistatic, that the thermodynamic identity does not hold.

    Is it possible to apply the thermodynamic identity to processes which begin and end in equilibrium, but are non-quasistatic in between?
     
  2. jcsd
  3. Nov 22, 2008 #2
    Any thoughts on this one?
     
  4. Nov 22, 2008 #3
    All the variables in your problem are state variables - so you can, in fact, use the thermodynamic relation. Even if you do not know along what path the system changed state, but you can always relate the final and initial states.
     
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