Question

A column of Hg of 100 mm in length is contained in the middle of a narrow tube 1 m long which is closed at both ends. Both the halves of the tube contained air at a pressure of 760 mm of Hg. By what distance will the column of Hg lie displaced if the tube is held vertical. Assume decrease in length of mercury column to be negligible, also take the process at constant temperature. (isothermal process).

Answer 1

Pressure volume work is measured in litre atmosphere (1 L. amt=101.3 J) and it happens when the volume of a system varies.

Assume x= Column of Hg that it will come closer by

P1=Air pressure at upper side of the tube

P2= Air pressure at lower side of the tube

$P1+100mm\ Hg\ =P2\ \ \ ...(i)$

$1cm=10mm\newline 100mm=10cm\newline 760mm=76cm$

The upper part equation is

$76*45 *a=P1*(45+x)*a\ \ \\newline Making\ P1 the\ subject \newline P1=(76*45)/(45+x) \ \ \ ...(ii)$

The lower part equation is

$76*45*a=P2*(45-x)*a\newline Making\ P2\ the \ subject\newline P2=(76*45)/(45-x)\ \ \ ...(iii)$

Substitute $(ii)\ and \ (iii)\ in(i)$

$\dfrac{76*45}{45+x}+10=P2(=\dfrac{76*45}{45-x})$

$\dfrac{76*45}{45+x}\ - \dfrac{76*45}{45-x}\ =-10$

$x^2\ +684x\ -45^2=0\newline solving\ using \ algebra\newline x=3cm$