A balloon contains 14.0 L of air at 760 torr. What will be the volume of the balloon when it is taken to a depth of 10 ft. in a swimming pool? Assume that the temperature of the air and water are equal. (density : Hg = 13.6 g/ml.)

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11.0

11.3

10.8

answer is D.

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Detailed Solution

$\begin{array}{l} P_{1} V_{1} = P_{2} V_{2}, P_{2} = P_{0} + hdg \\ P_{2} = 101325 + 3.048 \times 1000 \times 9.8 = 131125.4 pa \\ (∵ 1 atm = 101325 pa, 1 ft = 0.3048 m) \\ V_{2} = \frac{101325 \times 14}{131125.4} = 10 .8 L \end{array}$

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