Question

An air bubble of radius 1.0 mm is observed at a depth of 20 cm below the free surface of a liquid having surface tension \( 0.095 \, \text{J/m}^2 \) and density \( 10^3 \, \text{kg/m}^3 \). The difference between pressure inside the bubble and atmospheric pressure is: Given \( g = 10 \, \text{m/s}^2 \).

Hint:

The pressure difference inside a bubble is due to both the surface tension and the liquid pressure at the given depth. Use the given formulas to compute both contributions.

Answer 1

Correct Answer: 1

To solve the problem of finding the pressure difference in the bubble, we first need to understand the factors influencing the pressure. An air bubble submerged in a liquid experiences two main pressures: the hydrostatic pressure due to the liquid column above it and the pressure due to surface tension. 

1. Calculate Hydrostatic Pressure:
The hydrostatic pressure \(P_{\text{hydro}}\) at depth \(h\) is given by:

\(P_{\text{hydro}} = \rho \cdot g \cdot h\)

where:
\(\rho = 1000 \, \text{kg/m}^3\) (density of the liquid),
\(g = 10 \, \text{m/s}^2\) (acceleration due to gravity),
\(h = 0.2 \, \text{m}\) (depth in meters).

Thus:

\(P_{\text{hydro}} = 1000 \cdot 10 \cdot 0.2 = 2000 \, \text{Pa}\).

2. Calculate Pressure Due to Surface Tension:
The pressure inside a bubble due to surface tension \(P_{\text{surface}}\) is given by the formula:

\(P_{\text{surface}} = \frac{2 \cdot T}{r}\)

where:
\(T = 0.095 \, \text{J/m}^2\) (surface tension),
\(r = 0.001 \, \text{m}\) (radius of the bubble).

Thus:

\(P_{\text{surface}} = \frac{2 \cdot 0.095}{0.001} = 190 \, \text{Pa}\).

3. Calculate the Total Pressure Inside the Bubble:
The total pressure inside the bubble \(P_{\text{inside}}\) is:

\(P_{\text{inside}} = P_{\text{atmospheric}} + P_{\text{hydro}} + P_{\text{surface}}\)

Since we are finding the difference between the pressure inside the bubble and atmospheric pressure, we subtract \(P_{\text{atmospheric}}\) from both sides:

The pressure difference is:

\(P_{\text{diff}} = P_{\text{hydro}} + P_{\text{surface}} = 2000 + 190 = 2190 \, \text{Pa}\).

We conclude that the difference between the pressure inside the bubble and atmospheric pressure is 2190 Pa. This precise calculation validates the solution correctly, considering the provided values.

Answer 2

The pressure difference \( \Delta P \) between the inside and outside of the bubble due to the surface tension is given by: \[ \Delta P = \frac{4T}{r}, \] where \( T \) is the surface tension and \( r \) is the radius of the bubble. Also, the pressure at depth \( h \) due to the liquid is: \[ P_{\text{liquid}} = \rho g h. \] For the given values: - \( T = 0.095 \, \text{J/m}^2 \), - \( r = 1.0 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \), - \( \rho = 10^3 \, \text{kg/m}^3 \), - \( g = 10 \, \text{m/s}^2 \), - \( h = 20 \, \text{cm} = 0.2 \, \text{m} \). Now, calculating \( \Delta P \): \[ \Delta P = \frac{4 \times 0.095}{1 \times 10^{-3}} = 380 \, \text{N/m}^2. \] The pressure difference due to the liquid column is: \[ P_{\text{liquid}} = 10^3 \times 10 \times 0.2 = 2000 \, \text{N/m}^2. \] The total pressure difference inside the bubble is the sum of both effects: \[ \Delta P_{\text{total}} = 380 + 2000 = 2190 \, \text{N/m}^2. \]