Question

A vessel of volume $V_{0}$ is evacuated by means of a piston air pump. One piston stroke captures volume $\Delta V=0.2 \,V_{0}$. 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

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}}{1.2 V_{0}}\right)^{n} $
$(1.2)^{n}=(1.2)^{3} $
$n=3$