Question

An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of ${60}^0$ ? Assume the temperature to remain constant.

Answer 1

Case 1 : Atmospheric pressure +Pressure due to mercury column

Case 2: Atmospheric pressure +Component of the pressure due to mercury column

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

$\Rightarrow (76\times {\rho }_{Hg}\times g+{\rho }_{Hg}\times g\times 20)\times A\times 43$

= $(76\times {\rho }_{Hg}\times g+{\rho }_{Hg}\times g\times 20\times cos{60}^0)A\times 1$

$\Rightarrow 96\times 43=86\times L$

$\Rightarrow L=\frac{96\times 43}{86}$ = 48 cm