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The air pressure above the liquid in an open container is 1.12 atm. The depth of an air bubble in the liquid is 35.2 cm (measured from the middle of the air bubble to the top of the liquid). The liquid's density is 831 kg/m^3. Determine the air pressure in the bubble suspended in the liquid (in Pa). Acceleration of gravity is 9.8 m/s^2.

My ans. thus far that are WRONG:
127167 Pa
114300 Pa
116.353 Pa

2006-12-01 03:34:26 · 3 answers · asked by Mariska 2 in Science & Mathematics Physics

3 answers

At the surface the pressure is 101325 * 1.12 = 113484 Pa
The pressure added due to the depth is ρgh = 831*9.8*.352 = 2866.6 Pa

Total absolute P = 113484 + 2866.6 = 116350.6 Pa

Looks like your 3rd one with the decimal pt bumped.........

firefly: It's not water. Read the ?

2006-12-01 03:57:41 · answer #1 · answered by Steve 7 · 0 0

The equation required to solve the problem is:

Total pressure is equal to air pressure above the open container plus the liquid pressure over the bubble which is 35.2 cm from the surface of the container.

Total pressure = 1.12 atmosphere + density of liquid times H(height of bubble from the surface)* Gravity .

1.12* 101000 pascal(one atmospheric pressure)+ 831Kg/m^2* 0.352* 9.8m/second^2 = 113120 + 292.512*9.8 =
113120+(2866.6 Kg/meter*3*m/second^2)

113120+ 2866 Kg/.meter^2/second^2 =115986newton/m^2

=115986 newton/m^2
= 115986 Pascal
The total pressure on the bubble is = 115986 Pascal

2006-12-01 04:13:59 · answer #2 · answered by lonelyspirit 5 · 0 0

OK, I et 1.12*101325 pascals for the pressure just due to the air.
For the additional due to water, I use the fact that 10 meters is 1 atmosphere, so 35.2 cm = .0352 atmospheres

so the total pressure should be (1.12+.0352)*101325
which is 117051

2006-12-01 03:47:08 · answer #3 · answered by firefly 6 · 0 0

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