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: $H g =$ $13.6$ g/mL.)

47.5

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42.5

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44.5

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50

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Solution

The correct option is A $47.5$
Given, $P_{1} = 760$ torr, $V_{1} = 14 L$
When immersed, $P_{2} = S g h$
density, $S = 1 g / {c m}^{3} = 1000 k g / m^{3}$ (water)
$h = 10 f t = 3.05 m$
$g = 9.8 m / 8^{2}$
$∴ P_{2} = 1000 \times 3.05 \times 9.8$
$= 29870.4 P a$
$= 224.05$ torr
Since temp. is constant
$P_{1} V_{1} = P_{2} V_{2}$
$V_{0} = \frac{P_{1} V_{1}}{P_{2}} = 47.5 L$

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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.)

The correct option is A 47.5Given, P1=760torr, V1=14LWhen immersed, P2=Sghdensity, S=1g/cm3=1000kg/m3 (water)h=10ft=3.05mg=9.8m/82∴P2=1000×3.05×9.8 =29870.4Pa =224.05torrSince temp. is constantP1V1=P2V2V0=P1V1P2=47.5L

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: $H g =$ $13.6$ g/mL.)