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Thermodynamics question

  1. Aug 5, 2005 #1
    Why in a reversible adiabatic process you can plot the exact process path in the P-V-T coordinates, but you can't do so under different conditions?
     
  2. jcsd
  3. Aug 5, 2005 #2

    Chi Meson

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    I'll take a stab at this although I'm not sure if I understand the question:

    Could this be referring to the fact that an isothermal process must be performed "quasi-statically"? That is, to get close to isothermal, you would have to progress very slowly to allow for the system to canstantly return to equilibrium. The real isothermal process would appear to be a "staircase" that is overlaid on a perfect isotherm.

    Stop me now if I'm on the wrong path. I could go on.
     
  4. Aug 6, 2005 #3

    Andrew Mason

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    I assume that "under different conditions" refers to a non-reversible adiabatic process. An example of such a process would be a sudden expansion of gas in an insulated container due to a sudden reduction of outside pressure. Such a process is generally thought to be adiabatic since there is no exchange of heat with the surroundings.

    When this occurs, however, the system is not in equilibrium. So PV=nRT does not apply until the gas reaches equilibrium, which is after expansion has stopped and the translational kinetic energy of the gas has been converted to heat of the gas. Consequently, we cannot provide precise values for P, V and T during the process.

    A reversible adiabatic process requires the system to be in equilibrium at all times during the process (an very slow expansion of gas in an insulated container). In such a process PV=nRT applies at all times so we can determine exactly what P, V and T will be at all times.

    AM
     
  5. Aug 6, 2005 #4
    I thought that PV=nRT is true no matter what for ideal gases. Why can't it be true if the system isn't in equilibrium?
     
  6. Aug 6, 2005 #5

    Andrew Mason

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    PV=nRT is a relationship between P, V and T in a gas that is in equilibrium. Suppose you have a can of compressed air and open it up in outer space. [itex]P_0V_0 =nRT_0[/itex] = the energy of the compressed gas, and the gas does no work in a free expansion and no heat is exchanged with the surroundings. Does P'V' (pressure and volume at time t', say) represent the energy of the expanding gas? If so, it must be the same as the original energy: [itex]P'V' = P_0V_0 =nRT_0 = nRT'[/itex]. This would mean that the temperature is constant. Is that true? How do you define the temperature of an expanding gas? What about the translational kinetic energy of the gas molecules? Where does that fit in to PV=nRT?

    AM
     
  7. Aug 6, 2005 #6
    Thank you! :)
     
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