An air bubble of volume ' $V_{1}$ ' is at the bottom of a lake of depth ' $d$ ', where the temperature is $T_{1} .$ The bubble rises to the surface, which is at a temperature of $T_{2}$ . Assuming that the bubble’s air is at the same temperature as that of the surrounding water, the volume of the bubble at the surface is (Let $P_{0}$ is the atmospheric pressure and $P_{1}$ is the pressure at $T_{1}$ )

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$\frac{n R T_{2}}{P_{1}}$

$\frac{T_{2}}{T_{1}} (\frac{P_{0} + ρ g d}{P_{0}}) V_{1}$

$\frac{n R T_{1}}{P_{1}}$

$\frac{T_{1}}{T_{2}} (\frac{ρ g d}{P_{0}}) V_{1}$

answer is B.

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Detailed Solution

From ideal gas equation :

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$

at bottom

$\frac{(P_{0} + ρ g d) V_{0}}{T_{1}} = \frac{P V_{2}}{T_{2}}$

$V_{2} = \frac{T_{2}}{T_{1}} (\frac{P_{0} + ρ g d}{P_{0}}) (V_{0})$

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