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Problem of gas expansion

  1. Jun 4, 2014 #1

    kelvin490

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    In the case of gaseous expansion, if the pressure of gas is P1 and the external pressure is P2, suppose P2<P1, we know from textbook that the work W is the negative of P2 times the increase in volume, but why P2 instead of P1? We would get different result if we use P1, should the work done on external pressure equal that done by the gas?
     
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  3. Jun 4, 2014 #2

    CAF123

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    Consider a system separated from its surroundings by a well-defined boundary. The work done by the surroundings on the system is equal to W = -∫PdV if the process is a reversible one, meaning that the system and surroundings are in equilibrium and any pressure difference over the interface is zero. The expansion of the gas can be generated by an infinitesimal change in this pressure.

    The work done is interpreted as the area under the P-V diagram, where P is this common pressure of the system/surroundings. If P was not the same then the system does not evolve through a series of well-defined quasi-static states and above interpretation is not so meaningful.
     
  4. Jun 4, 2014 #3

    kelvin490

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    If the external pressure is not equal to the system's pressure, how to evaluate the work done?

    In addition, we know that in free expansion the work done by the system and the surrounding are zero. Is it due to the fact that W = -∫PdV where P is the external pressure?
     
  5. Jun 5, 2014 #4

    CAF123

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    In this case, the system and the surroundings are not in equilibrium and work done by the gas in expanding is W = ∫PextdV which is not the negative of the work done by the surroundings because of this pressure difference over the boundary. The surroundings applies a force F = PextA on the system so for the gas to expand it must do work against this force.

    Yes, the external pressure is zero (vacuum) so typically the gas expands into a container and there is nothing preventing it from doing so. So no force on the gas from the surroundings, no work done.

    Related links:
    http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node12.html
    http://www.chem.arizona.edu/~salzmanr/480a/480ants/pvwork/pvwork.html
     
  6. Jun 5, 2014 #5

    kelvin490

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    Does it mean that the work done by the gas and that done by the surrounding have equal magnitude W= ∫PextdV but just different sign (assume massless piston), and we don't use internal pressure just because the system is not in equilibrium and so cannot be described by state properties?

    What if the piston is not massless?
     
  7. Jun 5, 2014 #6

    CAF123

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    I think that would only true if they were constantly in equilibrium throughout the process. In which case there is a well-defined pressure and Wgas on surr = ∫PsurrdV = -Wsurr on gas = ∫PgasdV if Pgas = Psurr. Perhaps someone else could clarify this though.

    If the set up is vertical, then as the gas expands the piston gains gravitational potential energy. This energy will have to be supplied by the gas.

    If the set up is horizontal then GPE is constant, but e.g frictional forces may be at work, again additional energy needs to be supplied by the gas should the piston move.
     
  8. Jun 5, 2014 #7
    In my Physics Forums Blog, I focus specifically on these issues. It would be worth your while to read my Blog, which is not very long. In the Blog, I point out the following:


    • At the boundary, the pressure of the gas always matches the pressure of the surroundings.
    • The work done by the gas on the surroundings is always minus the work done by the surroundings on the gas.
    • For a irreversible process, the pressure of the gas within the cylinder typically varies with spatial location. So there is no one value to use for the gas pressure. However, to get the work, you only need to use the pressure at the boundary.
    • For a reversible process, the pressure of the gas within the cylinder does not vary with spatial location. So the pressure of the gas (which is uniform) can be used to calculate the work done.

    Hope this, together with my Blog (which has been given the Good Housekeeping seal of approval by Wannabenewton) helps.

    Chet
     
  9. Jun 5, 2014 #8

    kelvin490

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    Thanks both, your answers are very useful. Now I know that W= ∫PextdV because we can only be sure what the external pressure is, while the internal pressure varies spatially. By conservation of energy if we figure out the work received by surrounding we know the work done by the system.

    Now I am interested in the case that the piston consumed some of the energy during expansion. If the piston gains PE or have friction, that means the work done by the system equals the energy consumed by piston plus the work done on surrounding. If we use W= ∫PextdV would we underestimate the work done by the gas?
     
  10. Jun 5, 2014 #9
    This all depends on what you call the system. If the system is the gas, then the Pext is the force per unit area on the base of the piston, and the piston is regarded as part of the surroundings. If the system also includes the piston, then Pext is the force per unit area on the top of the piston, and the piston is regarded as part of the system and not part of the surroundings.

    I have been involved in some threads recently where we examined these same issues in detail. See if you can find them. If not, I will try to locate them.

    Chet
     
  11. Jun 5, 2014 #10
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