Answer:
<em>A, 96.71 kPa</em>
<em>B, 138.4 kPa</em>
Explanation:
The absolute pressure is the sum of the atmospheric and gauge pressure this can be expressed as;
............................1
Part A
The local atmospheric pressure can be calculated by making the atmospheric pressure in equation 1 the subject formula;
...........................2
the gauge pressure can be obtained with the expression;
= ρgh ......................3
where ρ is the density of water = 1000 kg/;
g is the acceleration due to gravity = 9.81 m/;
h is the depth = 9 m
Therefore,
= 1000 kg/ x 9.81 m/ x 9 m
= 88290 Pa
Now we take our value for the gauge pressure and substitute in equation 2;
Given = 88290 Pa
= 185 kPa = 185000 Pa (1000 Pa = 1 kPa)
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) =
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/ = 850 kg/
the density of the liquid is substituted into the equation shown below;
= ρ x g x h .
where ρ is the density of the liquid
h is the depth = 5 m
g is the acceleration due to gravity.
= 850 kg/ x 9.81 m/ x 5
= 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;
............................1
= 138402.5 = 138.4 kPa
Therefore the absolute pressure is 138.4 kPa