At what depth in an ocean will a bubble of air have one fourth the volume it will have on reaching the surface? (Atmospheric pressure $= 76 c m$ of $H g$ and density of $H g = 13.6 g c m^{- 3}$ )

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Solution

Since number of molecules will remain same and we can assume temperature to be constant.
Let at the surface pressure be $P$ and volume be V.
when the volume become $V / 4$ , let at that point pressure be $P_{1}$
So $P_{1} V_{1} = P_{2} V_{2}$
$P V = P_{1} \frac{V}{4}$
so $P_{1} = 4 P$
Also $P_{1} = P + ρ g H$
So $ρ_{w a t e r} g H = 3 P$
Also at $P = ρ_{m e r c u r e y} g H = 13.6 \times g \times 76$
Putting the value of P in above equation.
$ρ_{w a t e r} g h = 3 \times 13.6 \times g \times 76$
$h = 3100.8 c m$

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At what depth in an ocean will a bubble of air have one fourth the volume it will have on reaching the surface? (Atmospheric pressure =76 cm of Hg and density of Hg=13.6 gcm−3)

At what depth in an ocean will a bubble of air have one fourth the volume it will have on reaching the surface? (Atmospheric pressure $= 76 c m$ of $H g$ and density of $H g = 13.6 g c m^{- 3}$ )