Question

An air bubble of radius 2.0 mm is formed at the bottom of a 3.3 m deep river. Calculate the radius of the bubble as it comes to the surface. Atmospheric pressure $=1.0\times {10}^5$ Pa and density of water $=1000kg{m}^{-3}$

• A. 2.2mm
• B. 1.8mm
• C. 2.8mm
• D. 2mm

Answer 1

The correct option is A 2.2mm

The pressure at a depth h = 3.3 m under water will be -

$P=P_0+\rho gh$ (symbols have their usual meaning)

$=105+1000\times 10\times 3.3(P_0={10}^{5}Pa;{\rho }_{water}=1000kg{m}^{-3};g=10m{s}^{-2})$

$=1.33\times {10}^{5}Pa.$

For simplicity we have to assume that gases inside the bubble don't get dissolved in the water as the bubble rises up. So, the number of moles stays fixed. It is also safe to assume constant temperature.

Volume of the bubble at the bottom of the river = $[\left(\frac{4}{3}\right)\times \pi \times (2.0{)}^3]m{m}^3$.

Let the radius of the bubble when it reaches the top be rmm , and volume $\left(\frac{4}{3}\right)\pi r^{3}m{m}^3$ denoted by V.

Since PV = nRT, if n and T are fixed, the product PV will be a constant.

$\therefore (1.33\times {10}^5)\times (\frac{4}{3}\times \pi \times {2.0}^3)=(1\times {10}^5)\times (\frac{4}{3}\times \pi \times r^3)$

$\Rightarrow r^3=1.33\times 2.03m{m}^3$

$\Rightarrow r=2.199mm\approx 2.2mm.$