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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?

Respuesta :

Answer:

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

The mass of the liquid in the vat is 2435.54 kg

Suppose we consider [tex]\mathbf{\rho}[/tex] to be the density of the liquid;

Then the pressure situated at the bottom with a depth (h) of the vat can now be:

[tex]\mathbf{P = P_o + \rho gh }[/tex]

where;

  • [tex]\mathbf{ P_o }[/tex] = atmospheric pressure.

[tex]\mathbf{P -P_o = \rho gh }[/tex]

[tex]\mathbf{ \rho = \dfrac{P -P_o }{gh }}[/tex]

We know that:

[tex]\mathbf{density (\rho) = \dfrac{mass}{volume}}[/tex]

The mass of the liquid in the vat can be determined by the formula:

[tex]\mathbf{m = \rho V}[/tex]

where;

[tex]\mathbf{ \rho = \dfrac{P -P_o }{gh }}[/tex]

Then;

[tex]\mathbf{m = \Big( \dfrac{P -P_o }{gh } \Big)\times Area (A) \times h}[/tex]

[tex]\mathbf{m = \Big( \dfrac{P -P_o }{g } \Big)\times Area (A) }[/tex]

[tex]\mathbf{m = \Big( \dfrac{P -P_o }{g } \Big)\times \pi \Big(\dfrac{d}{2}\Big )^2}[/tex]

replacing the values, we have:

[tex]\mathbf{m = \Big( \dfrac{1.3 \ atm - 1.0 \ atm }{9.8 \ m/s^2 }\Big)\times \pi \Big(\dfrac{1.0}{2}\Big )^2}[/tex]

[tex]\mathbf{m = \Big( \dfrac{1.3 \ atm - 1.0 \ atm }{9.8 \ m/s^2 } \Big) \times \Big ( \dfrac{1.013 \times 10^5 \ Pa}{1.0 \ atm } \Big) \times \pi \Big(\dfrac{1.0}{2}\Big )^2}[/tex]

[tex]\mathbf{m =(3101.020408 \times 0.7853981634) \ kg}[/tex]

[tex]\mathbf{m \simeq 2435.54 \ kg}[/tex]

Therefore, we can conclude that the mass of the liquid in the vat is 2435.54 kg

Learn more about the relation of density here:

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