An open industrial mixing vat contains a liquid chemical with a specific gravity of 1.25. … — Armed Services Vocational Aptitude Battery (ASVAB)

An open industrial mixing vat contains a liquid chemical with a specific gravity of $1.25$ . The liquid in the vat reaches a depth of $4\text{ meters}$ . Assuming the density of fresh water is $1,000\text{ kg/m}^3$ and the acceleration due to gravity is $10\text{ m/s}^2$ , what is the hydrostatic pressure exerted by the liquid at the bottom of the vat? (Enter your answer in kilopascals)

Answer: 50 kPa

Answer

The hydrostatic pressure at the bottom of a fluid column is determined by the formula

P = \rho g h

. First, the specific gravity of

1.25

indicates the fluid is

1.25

times as dense as water, giving a density of

1,250\text{ kg/m}^3

. Multiplying this density by the acceleration due to gravity (

10\text{ m/s}^2

) and the depth (

4\text{ meters}

) yields

50,000\text{ Pascals}

. Since

1,000\text{ Pa}

equals

1\text{ kPa}

, this is equivalent to

50\text{ kPa}

Step-by-Step Solution

Determine the density of the fluid.

Density = 1,250 kg/m³

Specific gravity is the ratio of the fluid's density to the density of water. Multiplying 1.25 by 1,000 kg/m³ gives the actual density.

Calculate the hydrostatic pressure in Pascals.

Pressure = 50,000 Pa

Hydrostatic pressure is the product of density, gravity, and depth (1,250 × 10 × 4 = 50,000).

Convert the pressure to kilopascals (kPa).

50 kPa

The question asks for the answer in kilopascals, so divide the Pascal value by 1,000.

Key Concept

Hydrostatic Pressure

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