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Relationship between Pressure of Gas and Mechanical Work

  1. Jan 2, 2008 #1
    Is there a mechanical work equivalent to a change in the pressure of a gas?
     
  2. jcsd
  3. Jan 2, 2008 #2
    Mechanical work is energy, pressure is force per unit area. How can they be equivalent? :smile:
     
  4. Jan 2, 2008 #3
    but if the pressure changes doesn't that imply a change in volume, which could do work or a change in temperature which means a change in energy?
     
  5. Jan 2, 2008 #4
    No it doesn't. If you have a box with fixed walls you can change it's pressure by heating or cooling it. The volume will remain the same and the work equals zero.
     
  6. Jan 2, 2008 #5
    Your right, thanks. But if the change in temperature did not account for the entire amount of work and a change in volume did occur, given the change in temperature and pressure I could find the change in volume, so would this have a mechanical work equivalent?
     
  7. Jan 2, 2008 #6
    It would, by that's the 1st law of thermodynamics

    [tex]Q=W+\Delta U[/tex]

    i.e. the heat given equals the work done + internal's energy change
     
  8. Jan 2, 2008 #7
    and internal energy change would be the change in pressure?
     
  9. Jan 2, 2008 #8
    A change in the pressure will cause a change in the amount of work that a change in volume accomplishes, but without a change in volume no work occurs. Remember that dW = -PdV
     
  10. Jan 2, 2008 #9
    No! The internal enery is given by

    [tex] dU=n\,C_V\,d\,T[/tex]

    i.e. it is proportional to the change of the temperature not to the pressure!
     
  11. Jan 2, 2008 #10
    so mcdT - nCdT = Mechanical Work ?
     
  12. Jan 2, 2008 #11
    Correct! 1st law of thermodynamics :smile:
     
  13. Jan 2, 2008 #12
    what if p changes for the equation W = pdV? would I use the initial p or the final p?
     
  14. Jan 2, 2008 #13
    If temperature is constant, you can use the ideal gas law to write the pressure as a function of the volume and integrate...
     
  15. Jan 2, 2008 #14
    The work done through a revisible process is given by

    [tex]d\,W=p\,d\,V\Rightarrow W=\int_{V_i}^{V_f}p\,d\,V[/tex]

    if you are familiar with integrals.
     
  16. Jan 2, 2008 #15
    Perhaps an example will help me. If the pressure and temperature of a gas increase, while the # of moles is the same, then the volume has decreased. How could I find the amount of mechanical work done.
     
  17. Jan 2, 2008 #16

    Mute

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    Homework Helper

    The volume doesn't need to decrease in that process. The increase in temperature could compensate for the increase of pressure in the ideal gas law, and so the volume could remain constant. It doesn't have to remain constant, I guess, but that would probably be complicated to calculate a simple form for the work, since the temperature changes with the volume too.

    If temperature happened to remain constant, along with the particle number, while pressure increased, then volume would have to decrease, and if the process is quasistatic, then [itex]dW = PdV[/itex], and so

    [tex] W = \int_{V_0}^{V_f}P dV = n k_B T \int_{V_0}^{V_f}\frac{dV}{V}[/tex]

    which you can integrate to find the work done. (Note that [itex]nk_B = NR[/itex] - I've just expressed the ideal gas law in terms of boltzmann's constant and particle number instead of the gas constant and number of moles).
     
    Last edited: Jan 2, 2008
  18. Jan 2, 2008 #17
    o haha I see I can sub with the one which is assumed constant during a particular section... ok I'm satisfied. Thanks everyone.
     
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