Question

The volume of an air bubble is doubled as it rises from the bottom of a lake to its surface. The atmospheric pressure is 75 cm of mercury and the ratio of the density of mercury to that of lake water is 40/3. The depth of the lake is:

• A. 15 m
• B. 10 m
• C. 20 m
• D. 25 m

Hint:

Boyle’s law helps relate the pressure and volume of a gas at constant temperature. The pressure and volume are inversely proportional.

Answer 1

The Correct Option is C

The problem involves the relationship between pressure and volume for an air bubble as it rises to the surface of a lake. We can use Boyle's Law, which states that for a constant temperature, the product of pressure and volume remains constant, i.e., \( P_1V_1 = P_2V_2 \). Given:

\( V_2 = 2V_1 \) 

Let \( P_1 \) be the pressure at the bottom of the lake and \( P_2 \) be the pressure at the surface.

Atmospheric pressure \( P_2 = 75 \) cm Hg.

Density of mercury \( \rho_{Hg} = 40 \times \) density of water \( \rho_w \).

The pressure at the bottom is the sum of atmospheric pressure and the pressure due to water column:

\( P_1 = P_2 + h \rho_w g \)

Using Boyle’s Law:

\( P_1V_1 = P_2 \cdot 2V_1 \)

Therefore:

\( P_1 = 2P_2 \)

Substitute in the pressure equation:

\( 2P_2 = P_2 + h \rho_w g \)

\( P_2 = h \rho_w g \)

Convert \( P_2 \) from cm of Hg to pascals:

\( P_2 = 75 \) cm Hg \( = 75 \) cm \( \times 1333 \) Pascal/cm (since 1 cm Hg = 1333 Pa)

\( P_2 = 99975 \) Pascal

Using the conversion of mercury to water density:

\( \rho_w = \rho_{Hg} / (40/3) \)

\( \rho_w = 13600 \) kg/m\(^3\) / (40/3)

Calculate pressure in water as \( h \cdot \rho_w \cdot g = P_2 \):

\( \rho_w g = 99975 \) Pascal

Finally, solve for \( h \):

\( h = 99975 \) Pascal \(/ (1000 \times 9.81)\)

\( h \approx 20 \) meters

Thus, the depth of the lake is 20 meters.

Answer 2

Treat the bubble's isothermal rise using Boyle's law, \( P_1V_1=P_2V_2 \). Since the volume doubles by the time the bubble reaches the surface, \( V_2=2V_1 \), so \( P_1=2P_2 \), where \( P_2 \) is atmospheric pressure and \( P_1 \) is the absolute pressure at the bottom.

The absolute pressure at depth \( h \) is the atmospheric pressure plus the water column above it: \( P_1=P_2+\rho_w g h \). Combining this with \( P_1=2P_2 \) gives:

\[ 2P_2 = P_2 + \rho_w g h \quad\Rightarrow\quad P_2 = \rho_w g h \]

Now use the fact that the atmospheric pressure itself is quoted as a 75 cm column of mercury, i.e. \( P_2=\rho_{Hg}\,g\,(0.75\,\text{m}) \). Setting the two expressions for \( P_2 \) equal and cancelling \( g \):

\[ \rho_w h = \rho_{Hg}(0.75) \quad\Rightarrow\quad h = \frac{\rho_{Hg}}{\rho_w}\times0.75\ \text{m} \]

Using the given ratio \( \rho_{Hg}/\rho_w = 40/3 \):

\[ h = \frac{40}{3}\times0.75 = 10\ \text{m} \]

Checking each option:

1. Option A, 15 m: would require the density ratio times 0.75 to equal 15, i.e. a ratio of 20, not 40/3.
2. Option B, 10 m: matches the ratio calculation above exactly.
3. Option C, 20 m: would need the ratio to be roughly 26.7, well off from the given 40/3.
4. Option D, 25 m: would need a density ratio of about 33, also inconsistent with 40/3.

The correct answer is 10 m.