Question

The volume of a vessel is $V$. The volume of the barrel of pump is $v$. After $n$ strokes of the exhausted pump, what is the pressure of air inside the vessel if the original pressure inside the vessel is $P_0$?

Answer 1

$\frac{P_1{V}_1}{{\eta }_1{T}_1}=k$

At pressure, $P_0$, $x$ moles are present $V=$ constant , $T=$ constant.

After $dn$ states, moles of gas drawn = $\frac{x}{V}.vdn$

$\therefore \frac{P-dP}{(x-\frac{x}{V}.vdn)}=\frac{P}{x}$

$\frac{dP}{P}=\frac{1}{87}.cdt$

$\therefore \frac{dP}{P}=\frac{1}{V}.vdn$

Integrating we get;

${\int }_{P_0}^P\frac{dP}{P}=$${\int }_0^n\frac{1}{V}.vdn$

$\log \frac{P}{P_0}=\frac{nv}{V}$

$P=P_0{e}^{\frac{nv}{V}}$