Irodov Problem 2.7

An example of the pumping process is described in detail in the figure. The pump comprises of two chambers - a lower chamber of volume V and an upper chamber of volume . The upper and lower chambers are separated by a valve (valve 1) that can be opened and closed. There is also another valve - valve 2, above valve 1, that can not only be opened and closed but also moved up and down along the upper chamber. The goal of the pump is to reduce the pressure in the volume V under the lower chamber.

Initially, in step I, valve 1 is open, valve 2 is in the lower position and closed. Say the gas in the lower chamber has a pressure and volume of P and V.

In step II, valve 2 is moved to the upwards position and so the gas expands and fills up the entire volume of comprising both the chambers. Due to this expansion the pressure of the gas reduces to




In step III, valve 1 is closed, and the chambers are isolated from each other.

In step IV, valve 2 is opened and the air from outside environment flows into the upper chamber.

In step V, valve 2 is moved to the downward position to the bottom of the upper chamber.

In step VI, valve 2 is closed and valve 1 is opened.

Now the valves of the pump are in the same situation as the beginning of step I, except that the pressure in the lower chamber has decreased to P'. Thus, by repeating steps I to VI again and again one can keep reducing the pressure in the lower chamber. At each cycle the pressure in the lower chamber reduced by a factor of ( as seen in Eqn (1)).

Suppose that after n cycles, the pressure is Pn. Then we have the recursive relation,







Hence to reduce the pressure by a factor of times we have,