Question

A 2.0-m-diameter vat of liquid is 2.7 m deep. The pressure at the bottom of the vat is 1.5 atm .What is the mass of the liquid in the vat?

Answer 1

here,

P = p * g * h

If the 1.50 atm is gauge pressure,

(1.50 atm)(101300 Pa/atm) = p * (9.81 m/s^2)(2.70 m)

p = 4629 kg/m^3 (This is uncommonly dense for a liquid, so p is likely absolute pressure, unless this is a contrived problem with unrealistic numbers...)

If the 1.50 atm is absolute pressure,
(1.5 - 1 atm)(101300 Pa/atm) = p * (9.81 m/s^2)(2.70 m)

p = 1068 kg/m^3

p = m/V

p = m / [pi * r^2 * h]

1068 kg/m^3 = m / [pi * (2.00 m / 2)^2 * (2.70 m)]

m = 9059 kg

the mass of the liquid is 9059 kg