A diver ascends quickly to the surface from the bottom of a lake of depth H. During this period, he neither exhales nor inhales air. Assuming constant temperature, what would be the fractional increase in volume of his lungs? The atmospheric pressure is 7H metre of water.
Asked by Mansi
17th September 2018,
12:24 AM
Answered by Expert
Answer:
Since, Diver neither exhale nor inhale.
so, mole is constant.
here temperature is also constant.
so, Boyle's law can applied, as-
P1V1 = P2V2
Here, P1 = (7H + H) meter of water
= 8H meter of water
P2 = 1 atm = 7H meter of water.
so, from Boyle's law,
8H V1 = 7H V2
V2/V1 = 8/7
(V2-V1)/V1 = 1/7 (ans)
Answered by Expert
4th October 2018,
8:52 PM
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