Question

What is the absolute pressure at a point on a submarine, at a depth of 30 m below the surface of the sea, when the atmospheric pressure is 101 k Pa ? Take the density of sea water as 1030 kg m^-3 and the gravitational acceleration as 9.8 ms^-2

Answer 1

The absolute pressure is 404 k Pa.

Explanation:

Pressure, P, under water, is given by `P = rho*g*h` where `rho` (the Greek letter rho) is density, in this case of the water. Therefore
`P = rho*g*h = 1030 kg*m^-3*9.8 m*s^-2*30 m`
`P = 3.03*10^5 kg*m^2*m^-3*s^-2`

The next conversion will seem strange, but see if it does not work out. I will cancel one of the m's in the numerator against one of the m's in the denominator giving this expression.
`P = 3.03*10^5 kg*m*m^-2*s^-2`
The next step is to group the units in a strategic way.
`P = 3.03*10^5 kg*m*s^-2*m^-2`

We hope to have the units of pressure, which are `N*m^-2`. The Newton is equivalent to the combination kgms^-2 so we can change the expression to
`P = 3.03*10^5 N*m^-2`
And since `10^3*N*m^-2` is equivalent to the unit kPa,
`P = 303* kPa`

That is the pressure due to the water above the submarine. The air above the water must be added to that to yield absolute pressure.
`P_"abs" = 303*k Pa + 101 k Pa = 404 k Pa`

I have asked that someone check my answer. Many steps. I suggest that even if someone does say "looks good" you go thru my math step by step.

I hope this helps,
Steve