Question

The excess pressure inside a small air bubble of radius 0.05 mm in water of surface tension $70\,{\text{dyne}} - {\text{c}}{{\text{m}}^{ - 1}}$ is _ _ _ _ _Pa:
A. $28.2$
B. $8 \times {10^2}$
C. $5600$
D. $2800$

Answer 1

Convert the surface tension from dyne/cm to N/m. Use the formula for excess pressure inside the bubble and calculate the excess pressure in Pa. Convert the radius of curvature of the air bubble from mm to m.

Formula used:
Excess pressure inside the bubble, $P = \dfrac{{4T}}{R}$
Here, T is the surface tension of the bubble and R is the radius of the bubble.

Complete step by step answer:
We have given the radius of small air bubble $R = 0.05\,{\text{mm}} = 0.05 \times {10^{ - 3}}\,{\text{m}}$ and surface tension $T = 70\,{\text{dyne}} - {\text{c}}{{\text{m}}^{ - 1}}$. We have to convert the surface tension from dyne/cm to N/m. we know that ${10^5}\,{\text{dyne}} = 1\,{\text{N}}$.

Let’s convert the surface tension from dyne/cm to N/m as follows,
${\text{T}} = \left( {{\text{70}}\,\dfrac{{{\text{dyne}}}}{{{\text{cm}}}}} \right)\left( {\dfrac{{{\text{1 N}}}}{{{\text{1}}{{\text{0}}^{\text{5}}}\,{\text{dyne}}}}} \right)\left( {\dfrac{{{{10}^2}\,{\text{cm}}}}{{1\,{\text{m}}}}} \right)$
$\Rightarrow {\text{T}} = 70 \times {10^{ - 3}}\,{\text{N/m}}$

We know the expression for excess pressure inside the bubble,
$P = \dfrac{{4T}}{R}$
Here, T is the surface tension of the bubble and R is the radius of the bubble.

Substituting ${\text{T}} = 70 \times {10^{ - 3}}\,{\text{N/m}}$ and $R = 0.05 \times {10^{ - 3}}\,{\text{m}}$ in the above equation, we get,
$P = \dfrac{{4\left( {70 \times {{10}^{ - 3}}} \right)}}{{0.05 \times {{10}^{ - 3}}}}$
$\therefore P = 5600\,{\text{Pa}}$
Therefore, the excess pressure inside the small air bubble is 5600 Pa.

Hence, the correct answer is option C.

Additional information:
The molecules residing on the outer surface of the bubble experience a net inward force. Therefore, the pressure inside the bubble becomes greater than the pressure outside the bubble (atmospheric pressure). This excess pressure is expressed as,
$P = {P_i} - {P_0}$, where, ${P_i}$ is the pressure inside the bubble and ${P_0}$ is the pressure outside the bubble.

Note: Students should always check whether all the units are in the S.I system or not. If not, you should convert it into S.I units. Note that the dyne is the C.G.S unit of the force and that of pressure is ${\text{dyne/c}}{{\text{m}}^2}$. The excess pressure inside the bubble does not work in displacing the area of the bubble but the work is stored in the form of potential energy.