Question

Mercury is added to a cylindrical container to a depth d and then the rest of the cylinder is filled with water. If the cylinder is 0.4 m tall and the absolute (or total) pressure at the bottom is 1.1 atmospheres, determine the depth of the mercury. (Assume the density of mercury to be 1.36 104 kg/m3, and the ambient atmospheric pressure to be 1.013e5 Pa)

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of water = ₁ = 1000 kg/m³

Density of mercury = ₂ = 1.36 x 10⁴ kg/m³

Atmospheric pressure = P_atm = 1.013 x 10⁵ Pa

Absolute pressure at the bottom of the container = P = 1.1 atm = 1.1 x (1.013x10⁵) Pa = 1.1143 x 10⁵ Pa

Height of the cylinder = H = 0.4 m

Height of the water in the cylinder = H₁

Height of the mercury in the cylinder = H₂

H = H₁ + H₂

H₁ = H - H₂

P = P_atm + ₁gH₁ + ₂gH₂

P = P_atm + ₁g(H - H₂) + ₂gH₂

1.1143x10⁵ = 1.013x10⁵ + (1000)(9.81)(0.4 - H₂) + (1.36x10⁴)(9.81)H₂

1.013x10⁴ = 3924 - 9810H₂ + 133416H₂

143226H₂ = 6206

H₂ = 4.333 x 10^-2 m

Depth of the mercury = 4.333 x 10^-2 m