A 2.0-m-diameter vat of liquid is 3.0m deep. The pressure at the bottom of the vat is 1.4atm. what is the mass of the liquid in the vat?
ive been trying to use P=(density)*g*h and finding the velocity of the liquid in the vat with V=pi r^2 *h and with the density and the velocity found from those two formulas use M=density* V to find the M. i got .45 but that didnt work so then i changed the pressure to pascals and used that instead and got 45451 (two sig figs: 45000) and that was wrong too.
could you please explain?? Thank you so much :)
ive been trying to use P=(density)*g*h and finding the velocity of the liquid in the vat with V=pi r^2 *h and with the density and the velocity found from those two formulas use M=density* V to find the M. i got .45 but that didnt work so then i changed the pressure to pascals and used that instead and got 45451 (two sig figs: 45000) and that was wrong too.
could you please explain?? Thank you so much :)
-
Fluid volume = 9.426m^3.
(101325 x 1.4) = 141,855N/m^2.
(141,855/3)/9.8 = mass of 4,825kg/m^3.
Total mass = (4,825 x 9.426) = 45,480.45kg.
If the pressure were absolute,
(141,855 - 101,325) = 40,530N/m^2.
(40,530/3)/9.8 = 1,378.57kg/m^3.
Total mass = (1,378.57 x 9.426) = 12,994.4kg.
(101325 x 1.4) = 141,855N/m^2.
(141,855/3)/9.8 = mass of 4,825kg/m^3.
Total mass = (4,825 x 9.426) = 45,480.45kg.
If the pressure were absolute,
(141,855 - 101,325) = 40,530N/m^2.
(40,530/3)/9.8 = 1,378.57kg/m^3.
Total mass = (1,378.57 x 9.426) = 12,994.4kg.
-
P(bottom) = P(surface) + P(depth)
=>P(depth) = 0.4 atm
=>0.4 x 101325 = hdg
=>0.4 x 101325 = 3 x d x 9.8
=>d = 1378.57 kg/m^3
By M = V x d
=>M = (πr^2h) x d
=>M = 3.14 x (1)^2 x 3 x 1378.57
=>M = 12986.14 kg
=>P(depth) = 0.4 atm
=>0.4 x 101325 = hdg
=>0.4 x 101325 = 3 x d x 9.8
=>d = 1378.57 kg/m^3
By M = V x d
=>M = (πr^2h) x d
=>M = 3.14 x (1)^2 x 3 x 1378.57
=>M = 12986.14 kg
-
First, using density is not profitable because it is not provided in the question.
Second did you note that the diameter was 2 not the radius?
Is the pressure an absolute 1.4 atm or is it 1.4 above atmospheric ( 2.4 in total) you will get two different answers depending on which interpretation of the question is used.
I will use 1.4 above atmospheric.
The pressure of the atmosphere is 101 kN / m^2
The pressure of the liquid is the weight ( a force) per square metre.
so the WEIGHT of liquid per square metre is 1.4 * 101 * 10 ^ 3 N
The total weight of liquid is then this number by the area of the vat
= (1/4 *pi()* 2^2) * 1.4 * 101 * 10 ^ 3 N
And as weight = mg
m = weight / g
which is 4.5 * 10^4 kg
or 45 tonne.
Using your method.
P = (density) * g h
(density) = pressure /(gh)
mass = density * volume
= density * area * height
= pressure / (gh) * pi() * r^2 * h
= 1.4 *101 * 10 ^ 3 / 9.8 * pi() * 1 ^2
Which is the same value as I calculated and you also calculated.
So I suspect that the answer they are seeking might be using absolute pressure not relative pressure ( as I mentioned at the beginning)
Second did you note that the diameter was 2 not the radius?
Is the pressure an absolute 1.4 atm or is it 1.4 above atmospheric ( 2.4 in total) you will get two different answers depending on which interpretation of the question is used.
I will use 1.4 above atmospheric.
The pressure of the atmosphere is 101 kN / m^2
The pressure of the liquid is the weight ( a force) per square metre.
so the WEIGHT of liquid per square metre is 1.4 * 101 * 10 ^ 3 N
The total weight of liquid is then this number by the area of the vat
= (1/4 *pi()* 2^2) * 1.4 * 101 * 10 ^ 3 N
And as weight = mg
m = weight / g
which is 4.5 * 10^4 kg
or 45 tonne.
Using your method.
P = (density) * g h
(density) = pressure /(gh)
mass = density * volume
= density * area * height
= pressure / (gh) * pi() * r^2 * h
= 1.4 *101 * 10 ^ 3 / 9.8 * pi() * 1 ^2
Which is the same value as I calculated and you also calculated.
So I suspect that the answer they are seeking might be using absolute pressure not relative pressure ( as I mentioned at the beginning)