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 × 105 Pa and density of water = 1000 kg m−3.

Answer 1

Radius of the bubble as it comes to surface = 2.2 mm

Explanation:

Depth of river = 3.3 m

Air Bubble of Radius 2.0 mm = 2 * 10⁻³

Atmospheric Pressure = 1.0 × 10⁵

Volume of the air bubble at the bottom of the river:

V₁ = 4/3 π r³ = 4/3 π (2 * 10⁻³)³

h = 3.3 m

Pressure at the surface = pressure at bottom + pgh

P₁ = Po + pgh = (1.0 × 10⁵) + (1000 * 9.8 * 3.3)

P₁ = 1.32 * 10⁵ Pa

P₂ = 1.0 * 10⁵

According to Boyle's law, we know that:

P₁V₁ = P₂V₂

V₂ = P₁V₁ / P₂

= 1.32 * 10⁵ * 4/3 π (2 * 10⁻³)³ /  1.0 × 10⁵

If R is the new radius of the buble, then new volume:

4/3 π R³ = 1.32 * 10⁵ * 4/3 π (2 * 10⁻³)³ /  1.0 × 10⁵

R³ =  1.32 * (2 * 10⁻³)³  

R = cube root of [1.32 * (2 * 10⁻³)³ ]

 = 2.2 * 10⁻³ m

 = 2.2 mm

Answer 2

The radius of the bubble when it comes to the surface is $2.19 \times 10^{-3} \mathrm{m}$

Explanation:

Given data in the question

$V_{1}=\frac{4}{3} \pi\left(2.0 \times 10^{-3}\right)^{3}$

h = 3.3 m

$P_{1}=P_{0}+\rho g h$

$\begin{aligned}&P_{1}=1.0 \times 10^{5}+1000 \times 9.8 \times 3.3\\&P_{1}=1.0 \times 10^{5}+10 \times 98 \times 33\end{aligned}$

$\begin{aligned}&P_{1}=1.0 \times 10^{5}+980 \times 33\\&P_{1}=1.0 \times 10^{5}+32340\end{aligned}$

$\begin{aligned}&P_{1}=1.32 \times 10^{5} \mathrm{Pa}\\&\mathrm{P}_{2}=1.0 \times 10^{5} \mathrm{Pa}\end{aligned}$

Since temperature remains the same, applying Boyle’s law we get

$\begin{aligned}&\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{P}_{2} \mathrm{V}_{2}\\&V_{2}=\frac{P_{1} V_{1}}{P_{2}}\end{aligned}$

$V_{2}=\frac{1.32 \times 10^{5} \times \frac{4}{8} \pi\left(2.0 \times 10^{-8}\right)^{3}}{1.0 \times 10^{5}}$

Let $R_2$ be the new radius. Then,

$\frac{4}{3} \pi R^{3}_{2}=\frac{1.32 \times 10^{5} \times \frac{4}{3} \pi\left(2.0 \times 10^{-8}\right)^{3}}{1.0 \times 10^{5}}$

$R^{3}_{2}=\frac{1.32 \times 10^{5} \times\left(2.0 \times 10^{-3}\right)^{3}}{1.0 \times 10^{5}}$

$R_{2}=\sqrt[2]{\frac{1.32 \times 10^{5} \times\left(2.0 \times 10^{-3}\right)^{3}}{1.0 \times 10^{5}}}$

$R_{2}=\sqrt[3]{\frac{1.32 \times 10^{5} \times 8 \times 10^{-9}}{1.0 \times 10^{5}}}$

$R_{2}=\sqrt[3]{\frac{10.56 \times 10^{-4}}{1.0 \times 10^{5}}}$

$R_{2}=\sqrt[3]{\frac{10.56 \times 10^{-5} \times 10^{-4}}{1.0}}$

$R_{2}=\sqrt[3]{\frac{10.56 \times 10^{-9}}{1.0}}$

$R_{2}=\sqrt[3]{10.56 \times 10^{-9}}$

$R_{2}=2.19 \times 10^{-3} \mathrm{m}$

Therefore the radius of the air bubble when it comes to the surface is $2.19 \times 10^{-3} \mathrm{m}$ when the radius of the bubble formed at the bottom of a 3.3 m deep river is 2.0 mm