A diver ascends quickly to the surface from the bottom of a lake of depth H metre. During this period, he neither exhales nor inhales air. Assuming constant temperature, what would be the fractional i

Question

A diver ascends quickly to the surface from the bottom of a lake of depth H metre. 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.

Prashant Kotian · Accepted Answer

Correct answer: D. By Boyle's Law: P₁V₁ = P₂V₂. P₁ = 7H + H = 8H, P₂ = 7H. So V₂/V₁ = 8/7. Fractional increase = (V₂ - V₁)/V₁ = 1/7.

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