You come across an open container which is filled with two liquids. Since the two liquids have different density there is a distinct separation between them. Water fills the lower portion of the container to a depth of 0.209 m which has a density of 1.00 * 103 kg/m3. The fluid which is floating on top of the water is 0.300 m deep. If the absolute pressure on the bottom of the container is 1.049 * 105 Pa, what is the density of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s2 and atmospheric pressure is P0 = 1.013 * 105 Pa.

526.57 Pa

Explanation:

P (pressure at the bottom of the container) = 1.049 * 10^5 pa

Using the formula of pressure in an open liquid

Pw (pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²

Pw = 1000 * 9.81 * 0.209 = 2050.29 Pa

P (atmospheric pressure) = 1.013 * 10^5 Pa

Pl (pressure due to the liquid) = ρ (density of the liquid) * h (depth of the liquid) * g

Subtract each of the pressure from the absolute pressure at the bottom

P (bottom) - atmospheric pressure

(1.049 * 10^5) - (1.013 * 10^5) = 0.036 * 10^5 = 3600 Pa

subtract pressure due to water from the remainder

3600 - 2050.29 = 1549.71 Pa

1549.71 = ρ (density of the liquid) * h (depth of the liquid) * g

ρ (density of the liquid) = 1549.71 / (h * g) = 1549.71 / (0.3 * 9.81) = 526.57 Pa