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Question: Find the absolute pressure at a depth of 1000 m in an ocean. Given, the density of seawater is \(1.0...

Find the absolute pressure at a depth of 1000 m in an ocean. Given, the density of seawater is 1.03×103kgm31.03 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}} and g=10ms1g = 10{\text{m}}{{\text{s}}^{ - 1}} .
A) 104 atm
B) 100 atm
C) 108 atm
D) 110 atm

Explanation

Solution

The absolute pressure refers to the pressure that is measured from zero or relative to zero. It is the definite value of pressure. The absolute pressure will be the sum of the pressure of water at the given depth known as gauge pressure and the atmospheric pressure.

Formulas used:
The absolute pressure at a point is given by, P=Patm+PgP = {P_{atm}} + {P_g} where Patm{P_{atm}} is the atmospheric pressure and Pg{P_g} is the gauge pressure at that point.
The gauge pressure of a substance at a height hh from its surface is given by, Pg=ρgh{P_g} = \rho gh where ρ\rho is the density of the substance and gg is the acceleration due to gravity.

Complete step by step answer.
Step 1: List the parameters known from the question.
The density of seawater is given as ρ=1.03×103kgm3\rho = 1.03 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}} .
The acceleration due to gravity is g=10ms1g = 10{\text{m}}{{\text{s}}^{ - 1}} .
The height from the surface of the ocean at which the absolute pressure is to be determined is h=1000mh = 1000{\text{m}} .
Step 2: Express the relation for the gauge pressure of water at the given depth.
The gauge pressure of seawater is given by, Pg=ρgh{P_g} = \rho gh ---------- (1) where ρ\rho is the density of the seawater, gg is the acceleration due to gravity and hh is the depth.
Substituting values for ρ=1.03×103kgm3\rho = 1.03 \times {10^3}{\text{kg}}{{\text{m}}^{ - 3}}, g=10ms1g = 10{\text{m}}{{\text{s}}^{ - 1}} and h=1000mh = 1000{\text{m}} in equation (1) we get, Pg=1.03×103×10×1000=103×105Pa{P_g} = 1.03 \times {10^3} \times 10 \times 1000 = 103 \times {10^5}{\text{Pa}}
Thus the gauge pressure of seawater is Pg=103×105Pa{P_g} = 103 \times {10^5}{\text{Pa}} .
Step 3: Express the relation for the absolute pressure at the given depth of the ocean.
The absolute pressure at a depth of 1000 m point in the ocean is given by, P=Patm+PgP = {P_{atm}} + {P_g} ------ (2)
where Patm{P_{atm}} is the atmospheric pressure and Pg{P_g} is the gauge pressure of water.
Substituting for Pg=103×105Pa{P_g} = 103 \times {10^5}{\text{Pa}} and Patm=1×105Pa{P_{atm}} = 1 \times {10^5}{\text{Pa}} in equation (2) we get, P=(103+1)×105=104×105PaP = \left( {103 + 1} \right) \times {10^5} = 104 \times {10^5}{\text{Pa}}
Thus the absolute pressure at a depth of 1000 m in the ocean is P=104atmP = 104{\text{atm}} .

Hence the correct option is A.

Note: The atmospheric pressure refers to the pressure of the air above the surface of the ocean. The gauge pressure refers to the pressure which is measured relative to the atmospheric pressure i.e., for gauge pressure, the base value of the observations will be the atmospheric pressure. The conversion of the unit from Pa to atm is given as 1atm=105Pa1{\text{atm}} = {10^5}{\text{Pa}} .