Hello. I'm new here. I'd like to ask a question which has been puzzling me for quite some time. The illustration is attached below the text.

Imagine a tall, airtight container C_{1}having a certain amount of liquid (say water) at an air pressure of P_{1}. An airtight valve V_{1}separates C_{1}from another container C_{2}containing air at a pressure of P_{2}. C_{2}is at an elevation with respect to C_{1}. Pressure P_{1}is greater P_{2}. When this valve is opened, the water is forced up through the tube into C_{2}. The height through which it rises h, is given by P_{1}'-P_{2}'/ (rho g), where P_{1}' and P_{2}' are the final pressures in containers 1 and 2 respectively.

The time taken to reach equilibrium condition is t_{1}.

Now, supposing there is a friction loss 4flv^{2}/2gD, where f is Fanning friction factor, v is the varying velocity of the flow and D is the diameter of the pipe. As a result of this, there is a kinetic energy loss, and therefore the flow is slower. The time taken to reach equilibrium condition is now t_{2}.

In either case (with or without friction loss), the work done is the same- the same mass of water is moved through the same height, and the final pressures are also the same.

In the first case, only PV work is done and there is no change in internal energy (adiabatic). The process is also reversible- if the same amount of energy is given to a pump/blower operating between the two air chambers, the initial pressures P_{1}and P_{2}are reached in container C_{1}and C_{2}respectively.

So how is it that in spite of the irreversibility occurring in the second case, the work done is the same (though it takes a longer time to be done)?

Please enlighten me.

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