Question

A $10cm$ column of air is trapped by a column of $Hg,8cm$ long, in a capillary tube horizontally fixed as shown below, at $1atm$ pressure. Calculate the length of air column where the tube is fixed at the same temperature (a) vertically with open end up (b) vertically with open end down. (c) at ${45}^{\circ }$ with the horizontal with open end up.

Answer 1

(a) $P_1{V}_2=P_2{V}_2$
or $P_1{l}_{1}a=P_2{l}_{2}a$
where, $a=area$ of cross-section of tube
$l_1,l_2=$ length of air column
$\because P_2=76+8=84cm$
$\therefore l_2=\frac{P_1{l}_1}{P_2}=\frac{76\times 10}{84}$
$=9.04cm$
(b) $P_1{l}_{1}a=P_2{l}_{2}a$
$P_2=76-8=68cm$
$\therefore l_2=\frac{P_1{l}_1}{P_2}=\frac{76\times 10}{68}=11.17cm$.
(c) When the tube is held at ${45}^{\circ }$ with open end up, the of $Hg$ is borne partially by the gas and partially by the Vertical height of $Hg$ is a measure of additional pressure of gas.
$\therefore l=\frac{8}{\sqrt{2}}$
Also $P_2{l}_{1}a=P_1{l}_{1}a$
$\therefore l_2=\frac{P_1{l}_1}{P_2}$
$=\frac{76\times 10}{76+\frac{8}{\sqrt{2}}}=9.3cm$.