Air pump physics problem

by Yegor
Tags: pump
 P: 120 Air pump physics problem Each cycle corresponds to a T=const. process. (in another version of this problem the pumping process is very fast and a the transformation would correspond to an adiabatic process - all you have to do is to add the exponent $$\gamma$$ to the equations below). After the first cycle the air volume increases from V to V+dV and the pressure decreases at $$p_1$$. You have $$p_0\cdot V=p_1\cdot (V+dV)$$ After the first cycle, the pump takes out the air from dV, and you get again $$p_1=p_0 \frac{V}{V+dv}$$ In the second cycle, $$p_1\cdot V=p_2\cdot (V+dV)$$ and $$p_2=p_1 \frac{V}{V+dv}=p_0 \frac{V^2}{(V+dv)^2}$$ and so on..... $$p_n=p_{n-1} \frac{V}{V+dv}=p_0 \frac{V^n}{(V+dv)^n}$$