An air bubble of volume $1\,cm ^3$ rises from the bottom of a lake $40\,m$ deep to the surface at a temperature of $12^{\circ}\,C$. The atmospheric pressure is $1 \times 10^5 Pa$, the density of water is $1000\,kg / m ^3$ and $g =10\,m / s ^2$. There is no difference of the temperature of water at the depth of $40\,m$ and on the surface. The volume of air bubble when it reaches the surface will be $..........\,cm^{3}$
JEE MAIN 2023, Medium
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$P = P _0+\rho gh =10^5\,Pa +10^3 \times 10 \times 40=5 \times 10^5\,Pa$
At $T$ is constant
$PV = P _0 V _0$
$\Rightarrow 5 \times 10^5\,Pa \times 1\,cm ^3=10^5 Pa \times V _0 \Rightarrow V _0=5\,cm ^3$
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