Question

A particular cylindrical bucket has a height of 50.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is empty, aside from containing air. The bucket is then inverted so that its open end is down and, being careful not to lose any of the air trapped inside, the bucket is lowered below the surface of a fresh-water lake so the water-air interface in the bucket is 40.0 m below the surface of the lake. To keep the calculations simple, use g = 10 m/s $^2$ , atmospheric pressure = 1 x 10 $^5$ Pa, and the density of water as 1000 kg/m $^3$ .
If the temperature is the same at the surface of the lake and at a depth of 40.0 m below the surface , what is the height of the cylinder of air in the bucket when the bucket is at a depth of 40.0 m below the surface of the lake?
_______ cm

A particular cylindrical bucket has a height of 50.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is empty, aside from containing air. The bucket is then inverted so that its open end is down and, being careful not to lose any of the air trapped inside, the bucket is lowered below the surface of a fresh-water lake so the water-air interface in the bucket is 40.0 m below the surface of the lake. To keep the calculations simple, use g = 10 m/s $^2$ , atmospheric pressure = 1 x 10 $^5$ Pa, and the density of water as 1000 kg/m $^3$ .
If, instead, the temperature changes from 300K at the surface of the lake to 275K at a depth of 40.0 m below the surface of the lake, what is the height of the cylinder of air in the bucket?
_______ cm

Answer 1

Temperature is same at surface and also at 40 m depth

we have P_i V_i = P_f V_f ........(1)

where P_i = initial pressure at surface of lake = 1$\times$10⁵ Pa

V_i = initial volume = $\pi$ $\times$15²$\times$ 50 cm³

P_f = pressure at depth of 40 m = 1$\times$10⁵ + $\rho$ gh = 1$\times$10⁵ + 1000$\times$10$\times$40 = 5 $\times$ 10⁵ Pa

Hence from eqn.(1), we get V_f = ( P_i / P_f ) V_i = (1/5) $\times$ $\pi$$\times$15²$\times$ 50 = $\pi$ $\times$15²$\times$ 10 cm³

Water height inside bucket = 10 cm

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Temperature at surface of lake = 300 K ; Temperature at 40 m depth = 275 K

we have, ( P_i V_i ) / T_i = ( P_f V_f ) / T_f   .....................(2)

V_f = ( P_i / P_f ) $\times$ ( T_f / T_i ) $\times$ V_i = (1/5) $\times$ (275/300) $\times$ $\pi$$\times$15²$\times$ 50

height of water in the bucket, when it is at 40 m deep inside lake = (1/5) $\times$ (275/300) $\times$ 50 = 9.17 cm