A 1.0-m-diameter vat of liquid is 2.0 m deep. The pressure at the bottom of the vat is 1.3 atm. What is the mass of the liquid in the vat?

The mass of the liquid = 10538 kg

Explanation:

The pressure in a liquid is

P = ρgh ... Equation 1

ρ = P/gh ... Equation 2

Where P = pressure, ρ = density, g = acceleration due to gravity, h = height.

Given: p = 1.3 atm, h = 2.0 m, g = 9.81 m/s²

If, 1 atm = 1.013*10⁵ N/m²

Then, P = 1.3*1.013*10⁵ N/m² = 1.3169*10⁵ N/m²

Substituting in equation 2,

ρ = 1.3169*10⁵ / (9.81*2)

ρ = 1.3169*10⁵/19.62

ρ = 6712.03 kg/m³.

But Density,

ρ = m/v

m = ρ * v ... Equation 3

Where m = mass of the liquid, v = volume of the liquid in the vat

v = πd²h/4, where d = diameter = 1.0 m, h = 2.0 m.

v = 3.14 (1) ²*2/4

v = 1.57 m³ also, ρ = 6712.03 kg/m³.

Substituting into equation 3

m = 1.57*6712.03

m = 10537.887 kg

m ≈ 10538 kg.

Thus the mass of the liquid = 10538 kg