Question

A metre long narrow bore held horizontally $\left(\text{and closed at one end}\right)$ contains a $76cm$ long mercury thread, which traps a $15cm$ column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer 1

Initially,

Length of the narrow bore, $L=1m=100cm$

Length of the mercury thread, $l=76cm$

Length of the air column trapped between mercury and the closed end, $x=15cm$

Air column at the top $=100-(76+15)=9cm$

Initial pressure of air trapped, $P_1=P_a=76cm$ of mercury

Taking cross section of the bore to be 𝐴,

Initial volume of air,

$V_1=A\times x=15Ac{m}^3$

Now, the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is:
$100–(76+15)=9cm$

Hence, the total length of the air column $=15+9=24cm$.

Let $hcm$ of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore, $y=(24+h)cm$

And, length of the mercury column remaining $=(76-h)cm$

Final pressure, $P_2=P_a-(76-h)=76-(76-h)=hcm$ of mercury

Final volume of air,

$V_2=A\times y=(24+h)Ac{m}^3$

Temperature remains constant throughout the process.
Therefore, applying $\text{Boyle’s law},$

$P_1{V}_1=P_2{V}_2$

$\Rightarrow 76\times 15A=h(24+h)A$

$\Rightarrow h^2+24h-1140=0$

Using quadratic formula,

$h=\frac{-24\pm \sqrt{{24}^2+4\times 1\times 1140}}{2}$

$=23.8cm\text{or}-47.8cm$

Height cannot be negative.
Hence, $23.8cm$ of mercury will flow out from the bore.
Remaining column of mercury $=76-23.8=52.2cm.$

The length of the air column will be $24+23.8=47.8cm.$