Q u asi - be slow in relation to the time needed for the...

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Q u asi-equilibrium processes: A system in thermodynamic equilibrium satisfies: a) mechanical equilibrium (no unbalanced forces) b) thermal equilibrium (no temperature differences) c) chemical equilibrium. For a finite, unbalanced force, the system can pass through non-equilibrium states. We wish to describe processes using thermodynamic coordinates, so we cannot treat situations in which such imbalances exist. An extremely useful idealization, however, is that only "infinitesimal" unbalanced forces exist, so that the process can be viewed as taking place in a series of "quasiequilibrium" states. (The term quasi can be taken to mean "as if"; you will see it used in a number of contexts such as quasi-one-dimensional, quasi-steady, etc.) For this to be true the process must
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Unformatted text preview: be slow in relation to the time needed for the system to come to equilibrium internally. For a gas 0-4 at conditions of interest to us, a given molecule can undergo roughly 10 10 molecular collisions per second, so that, if ten collisions are needed to come to equilibrium, the equilibration time is on the order of 10-9 seconds. This is generally much shorter than the time scales associated with the bulk properties of the flow (say the time needed for a fluid particle to move some significant fraction of the lighten of the device of interest). Over a large range of parameters, therefore, it is a very good approximation to view the thermodynamic processes as consisting of such a succession of equilibrium states....
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