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A particular cylindrical bucket has a height of 36.0 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 20.0 m below the surface of the lake. To keep the calculations simple, use g = 10 m/s2, atmospheric pressure = 1. 105 Pa, and the density of water as 1000 kg/m3. a. If the temperature is the same at the surface of the lake and at a depth of 20.0 m below the surface, what is the height of the cylinder of air in the bucket when the bucket is at a depth of 20.0 m below the surface of the lake? b. If, instead, the temperature changes from 300 K at the surface of the lake to 275 K at a depth of 20.0 m below the surface of the lake, what is the height of the cylinder of air in the bucket?

Respuesta :

MechEngineer

Answer:

a. 0.000002 m

b. 0.00000182 m

Explanation:

36 cm = 0.36 m

15 cm = 0.15 m

a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

[tex] P = \ro g h = 1000 * 10 * 20 = 200000 Pa[/tex]

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

[tex] P_1V_1 = P_2V_2[/tex]

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

[tex]V_1 = Ah[/tex]

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:

[tex]P_1Ah_1 = P_2Ah_2[/tex]

[tex]h_2 = h_1\frac{P_1}{P_2} = 0.36\frac{1.105}{200000} = 0.000002 m[/tex]

b) If the temperatures changes, we can still reuse the ideal gas equation above:

[tex]\frac{P_1Ah_1}{T_1} = \frac{P_2Ah_2}{T_2}[/tex]

[tex]h_2 = h_1\frac{P_1T_2}{P_2T_1} = 0.36\frac{1.105 * 275}{200000*300} =0.00000182 m[/tex]

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