Let us return to our previous analogy, of an air pump transferring air between two rigid vessels. Rather than using two isolated chambers, let us consider two chambers that are formed in a single rigid vessel, separated by an elastic membrane. (The initial volumes of the two chambers are the same as in the previous example). We again apply a constant force to the pump handle and begin to transfer air between the two chambers. Now, however, the increase in pressure in chamber B and the suction from chamber A, will cause the membrane to flex and this simultaneously increases the volume of chamber B, while reducing the volume of chamber A. The result of this action, is that the pressure increase in chamber B and the suction in chamber A, will be less than it was, for the rigid isolated chambers, when the same amount of air is transferred.
i.e. When the same amount of air is transferred to chamber B, as in the previous example (with a rigid vessel), the volume of the chamber will now also increase and so the air is less compressed. For chamber A, the same amount of air is removed, but the volume of the chamber has been reduced. Therefore there is less space for the remaining gas to expand into, and so the pressure does not drop as much as before (i.e. there is less suction). The pump will continue to transfer air until the same pressure limits as before are reached and then the pump handle will stop turning. Now, however, more air will have been transferred, because the flexing of the membrane changes the volume of the chambers.
