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I have a trouble in understanding the concept of the work.

In the closed system, Work done by system=Work on surroundings

Here is the part of the process that prove "Work on surroundings=pressure of surrounding X volume change"

source:http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node12.html

[From web.mit.edu]

I can't understand why "force on surr./area" equals pressure of surr. Well, I may have a problem in pressure

enter image description here

but it seems "force on surr." is taken by the system.(see the second picture) so it seems proper to change "pressure of surr." to "pressure of system"

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  • $\begingroup$ Think of it as a quasi equilibrium process so essentially the two pressures are equal. $\endgroup$
    – Farcher
    Mar 15, 2016 at 17:22
  • $\begingroup$ @Farcher but if it is not quasi equilibrium (for example piston fixed by notch and surrounding is in vacuum), it seems it doesn't assume quasi equilibrium $\endgroup$ Mar 15, 2016 at 17:54

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At the interface between the system and the surroundings, the pressure exerted by the system is exactly equal to the pressure exerted by the surroundings. That is, pressure is continuous at the interface. However, in an irreversible process (non-quasistatic), the pressure within the system will typically not be uniform spatially, so at other locations within the system, the pressure differs from the interface value. Therefore, we need to use the surroundings pressure (at the interface) to calculate the work, because that matches the system pressure at the location where work is being done (and this is the only pressure we may have access to). In addition to all that, in an irreversible process, the pressure within the system depends not only on the system volume (say using the ideal gas law), but also the rate of change of volume. This is because, with an irreversible change, viscous stresses contribute to the system pressure at the boundary. These stresses are not readily accounted for without performing a detailed transient gas dynamics analysis of the system. So, in the case of irreversible processes, we need to depend much more upon externally controlling the pressure imposed at the interface.

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  • $\begingroup$ Great answer. Very succinct and better than in some thermodynamics books. $\endgroup$
    – John Darby
    Nov 29, 2020 at 16:48

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