An ideal gas is confined between a mercury column and the closed end of a narrow vertical tube of uniform cross-section. The upper end of the tube is open to the atmosphere, and the atmospheric pressure is equal to 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm , respectively. Determine the length of the air column after the tube is slowly tilted in a vertical plane

Step-by-step solution

Let A be the cross-sectional area of the tube. In the initial vertical position, the pressure of the trapped air is the sum of the atmospheric pressure and the pressure due to the mercury column. P₁ = P₀ + l = 76 + 20 = 96 cm of Hg Initial volume of the trapped air, V₁ = L₁ A = 43A When the tube is tilted through an angle of 60^ with the vertical, the effective vertical height of the mercury column becomes l 60^ . Final pressure of the trapped air, P₂ = P₀ + l 60^ P₂ = 76 + 20 1 2 = 86 cm of Hg Final volume of the