Question

Consider a container filled with coconut oil having a density of 0.903 g/ml, What is the gauge pressure (Pa) 6.35 cm below the surface? 1ml = 1cm^3, 1000g = 1kg

Answer 1

The gauge pressure is 562.5 Pa.

Given data:

The density of oil is ρ=0.903 g/ml.

The distance below the surface is h=6.35 cm.

The density in kg/cm³ will be,

`\begin{gathered} \rho=0.903\text{ g/ml }*\frac{1\text{kg}}{1000\text{ g}}*\frac{1\text{ml}}{1\text{ }cm^3}*\frac{10^6\text{ }cm^3}{1m^3} \\ \rho=(903kg)/(m^3) \end{gathered}`

The gauge pressure can be calculated as,

`\begin{gathered} p=\rho gh \\ p=((903kg)/(m^3))((9.81m)/(s^2))(6.35cm*\frac{1\text{ m}}{100cm}) \\ p=(562.5kg)/(ms^2)*\frac{1\text{ Pa}}{(kg)/(ms^2)} \\ p=562.5\text{ Pa} \end{gathered}`

Thus, the gauge pressure is 562.5 Pa.