Question

A vessel of volume $V_0$ is evacuated by means of a piston air pump. One piston stroke captures volume $\Delta V=0.2V_0$. If the process is assumed to be isothermal, then find the minimum number of strokes after which pressure in the vessel becomes $\left(\frac{1}{1.728}\right)$ times the initial pressure.

Answer 1

Answer: 3

Let ${\rho }_1$ be the density after first stroke.
$V\rho =(V+\Delta V){\rho }_1$
$\Rightarrow {\rho }_1=\frac{V}{(V+\Delta V)}\rho$
After $n^{\text{th }}$ stroke,
${\rho }_n={\left(\frac{V}{V+\Delta V}\right)}^n\rho$
$\frac{{\rho }_n}{\rho }={\left(\frac{V}{V+\Delta V}\right)}^n$
$\because$ Pressure $\propto$ density
$\therefore \frac{P_n}{P}={\left(\frac{V}{V+\Delta V}\right)}^n$
$\frac{1}{1.728}={\left(\frac{V_0}{1.2V_0}\right)}^n$
$(1.2{)}^n=(1.2{)}^3$
$n=3$