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Expert-verified Found in: Page 578 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Question: An air bubble of volume 20 cm3 is at the bottom of a lake 40 m deep, where the temperature is 4. 0 0C . The bubble rises to the surface, which is at a temperature of . Take the temperature of 20 0C the bubble’s air to be the same as that of the surrounding water. Just as the bubble reaches the surface, what is its volume?

The volume of the bubble as it reaches the surface is $1.02×{10}^{2}{\text{cm}}^{3}$.

See the step by step solution

## Step 1: Given

1. Volume of the air bubble at the bottom of the lake,${V}_{1}=20{\text{cm}}^{3}=20×{10}^{-6}{\text{m}}^{3}.$
2. The depth of the lake,h = 40 m
3. Temperature of the air bubble at the bottom of the lake,${T}_{1}=4.0°\text{C}=277\text{K}.$

Temperature of the air bubble at the surface of the lake,${T}_{0}=20°\text{C}=293\text{K}.$

## Step 2: Determine the concept

Find the number of moles using the gas law. Using this value of number of moles, find the volume of the bubble as it reaches the surface by applying the gas law to the air bubble at the surface.

Formula is as follow:

${p}_{i}{v}_{i}=nR{T}_{i}$

Here, p is pressure, v is volume, T is temperature, R is universal gas constant and n is number of moles.

## Step 3: Determine the volume of the bubble as it reaches the surface

According to the gas law,

pV = n RT

Pressure at the bottom of the lake is given by,

,${p}_{1}={p}_{0}+\rho gh$

Where, ${p}_{0}$is the atmospheric pressure and $\rho$ is the density of water.

Thus,

$n=\frac{\left({p}_{0}+\rho gh\right){V}_{1}}{R{T}_{1}}\phantom{\rule{0ex}{0ex}}n=\frac{\left(1.013×{10}^{5}+998\left(9.8\right)\left(40\right)\right)\left(20×{10}^{-6}\right)}{8.314×277}\phantom{\rule{0ex}{0ex}}n=4.277×{10}^{-3}\text{moles}$

Consider the bubble reaches the surface the gas, the law becomes:

${p}_{0}{V}_{0}=nR{T}_{0}\phantom{\rule{0ex}{0ex}}{V}_{0}=\frac{nR{T}_{0}}{{P}_{0}}$

Substitute the values and solve as:

${V}_{0}=\frac{4.277×{10}^{-3}\left(8.314\right)\left(293\right)}{1.013×{10}^{5}}\phantom{\rule{0ex}{0ex}}{V}_{0}=1.02×{10}^{-4}{\text{m}}^{3}\phantom{\rule{0ex}{0ex}}{V}_{0}=1.02×{10}^{2}{\text{cm}}^{3}$

Hence, the volume of the bubble as it reaches the surface is$1.02×{10}^{2}{\text{cm}}^{3}$ .

Therefore, the volume of the bubble as it reaches the surface can be found using the gas law. ### Want to see more solutions like these? 