diving bell 3.0 m3 50 m 1.025 g/cm3

diving bell 3.0 m3 50 m 1.025 g/cm3

A diving bell has an air space of 3.0 m3 when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g/cm3 and assume that the temperature is the same as on the surface.

P_h = P_0 + ρgh

> P_h: 깊이(수심) h에서 압력

> P_0: 대기압

> ρ: 액체의 밀도

> g: 중력가속도

> h: 깊이(수심)

( 참고 https://ywpop.tistory.com/21288 )

( 참고 https://ywpop.tistory.com/20902 )

(1.025 g/cm3) (1 kg / 1000 g) (100^3 cm3 / 1 m3)

= (1.025) (1 / 1000) (100^3)

= 1025 kg/m3

정수압, P = ρgh

= (1025 kg/m3) (9.8 m/s2) (50 m)

= (1025) (9.8) (50)

= 502250 kg/m•s2

= 502250 Pa

1 atm = 101325 Pa 이므로,

(502250 Pa) (1 atm / 101325 Pa)

= (502250) (1 / 101325)

= 4.9568 atm

P_h = P_0 + ρgh

= 1 + 4.9568

= 5.9568 atm

보일의 법칙, P1V1 = P2V2

( 참고 https://ywpop.tistory.com/1977 )

V2 = P1V1 / P2

= (1 atm) (3.0 m3) / (5.9568 atm)

= 0.5036 m3

답: 0.50 m3

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