Answer:
A, 96.71 kPa
B, 138.4 kPa
Explanation:
The absolute pressure is the sum of the atmospheric and gauge pressure this can be expressed as;
[tex]P_{ab} = P_{atm} +P_{gauge}[/tex] ............................1
Part A
The local atmospheric pressure can be calculated by making the atmospheric pressure in equation 1 the subject formula;
[tex]P_{atm} = P_{ab}-P_{gauge}[/tex] ...........................2
the gauge pressure can be obtained with the expression;
[tex]P_{gauge}[/tex] = ρgh ......................3
where ρ is the density of water = 1000 kg/[tex]m^{3}[/tex];
g is the acceleration due to gravity = 9.81 m/[tex]s^{2}[/tex];
h is the depth = 9 m
Therefore,
[tex]P_{gauge}[/tex] = 1000 kg/[tex]m^{3}[/tex] x 9.81 m/[tex]s^{2}[/tex] x 9 m
[tex]P_{gauge}[/tex] = 88290 Pa
Now we take our value for the gauge pressure and substitute in equation 2;
Given [tex]P_{gauge}[/tex] = 88290 Pa
[tex]P_{ab}[/tex] = 185 kPa = 185000 Pa (1000 Pa = 1 kPa)
[tex]P_{atm} = 185000 Pa - 88290 Pa[/tex]
[tex]P_{atm} = 96710 Pa = 96.71 kPa[/tex]
The local atmospheric pressure is 96.71 kPa
Part B
To obtain the gauge pressure at a depth of 5 m in the liquid, we have to calculate the density of the liquid;
Specific gravity(s.p) = [tex]\frac{Density of the liquid}{Density of water}[/tex]
making the density of the liquid the subject formula we have;
The density of the liquid = s.p x Density of water
= 0.85 x 1000 kg/[tex]m^{3}[/tex] = 850 kg/[tex]m^{3}[/tex]
the density of the liquid is substituted into the equation shown below;
[tex]P_{gauge}[/tex] = ρ[tex]_{l}[/tex] x g x h .
where ρ[tex]_{l}[/tex] is the density of the liquid
h is the depth = 5 m
g is the acceleration due to gravity.
[tex]P_{gauge}[/tex] = 850 kg/[tex]m^{3}[/tex] x 9.81 m/[tex]s^{2}[/tex] x 5
[tex]P_{gauge}[/tex] = 41692.5 Pa
The absolute temperature is calculated using equation 1
since the liquid is at the same location as the water, the atmospheric temperature is the same;
[tex]P_{ab} = P_{atm} +P_{gauge}[/tex] ............................1
[tex]P_{ab} =96710 Pa +41692.5 Pa[/tex]
[tex]P_{ab}[/tex] = 138402.5 = 138.4 kPa
Therefore the absolute pressure is 138.4 kPa
