A 1.5 m long glass tube, closed at one end, is weighted and lowered to the bottom of a fresh-water lake. When the tube is recovered, an indicator mark shows that water rose to within 0.38 m of the closed end. Determine the depth of the lake. Assume constant temperature.
2007-07-13 05:49:29 · 3 answers · asked by makk 3 in Science & Mathematics ➔ Physics
vivek's answer would be correct if the tube were not closed at the top!
When the tube is at the bottom of the lake, the water pressure is so great that it compresses the air into a smaller volume. At the top of the lake, the column of air is 1.5 m long; but at the bottom of the lake, the air has been compressed so that it's only 0.38m long.
This means the air's volume shrank by a factor of 1.5/0.38 = 3.95
This means the air's pressure _increased_ by the same factor (assuming the temperature didn't change much). The Ideal Gas Law says so.
So, if the air's pressure at the top of the lake was 1 atm., then its pressure at the bottom of the lake is 3.95 atm.
And since the air's pressure must have been balancing the water's pressure, that means the water's pressure at the bottom of the lake was 3.95 atm also.
Now, use the formula for figuring out the change in pressure of water as a function of depth. Use that to determine how deep you have to go to change from 1 atm to 3.95 atm.
2007-07-13 06:21:59 · answer #1 · answered by RickB 7 · 2⤊ 0⤋
P stands for pressure, S for section, V for volume. We have
P* V = K constant.
P * S * 1.5 = P' * S * (1.5 - 0.38)
We obtain P' = P * 1.5 / 1.12 = 1.34 atm.
But every atm = 10 m water depth, so lake depth = 13.4 meters.
2007-07-13 06:48:28 · answer #2 · answered by Jano 5 · 0⤊ 0⤋
1.12 m
2007-07-13 05:56:21 · answer #3 · answered by Anonymous · 0⤊ 1⤋