Question 223: Properties of Matter

Question

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

Accepted Answer

To find the difference between the pressure inside the air bubble and the atmospheric pressure, we use the fact that at depth $h$ in a liquid of density $ho$, the external (hydrostatic) pressure exceeds atmospheric by $ho g h$. Additionally, for a spherical bubble in a liquid, the internal pressure exceeds the external pressure by $\frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the bubble.
Hence, the internal pressure of the bubble is
$ P_	ext{inside} = P_	ext{atm} + ho g h + \frac{2T}{r}. $
Therefore, the difference between the bubble's internal pressure and the atmospheric pressure is
$ \Delta P = P_	ext{inside} - P_	ext{atm} = ho g h + \frac{2T}{r}. $
Substituting the given values:
$ho = 10^3 \,	ext{kg/m}^3$
$g = 10 \,	ext{m/s}^2$
$h = 0.20 \,	ext{m}$
$T = 0.095 \,	ext{J/m}^2$
$r = 1.0 	imes 10^{-3} \,	ext{m}$
we compute each term:
Hydrostatic term
:
$    ho g h = (10^3 \,	ext{kg/m}^3) \cdot (10 \,	ext{m/s}^2) \cdot (0.20 \,	ext{m})             = 2000 \,	ext{N/m}^2.    $
Surface tension term
:
$    \frac{2T}{r} = \frac{2 	imes 0.095 \,	ext{J/m}^2}{1.0 	imes 10^{-3} \,	ext{m}}                 = \frac{2 	imes 0.095}{10^{-3}}                 = 2 	imes 95                 = 190 \,	ext{N/m}^2.    $
Hence,
$ \Delta P = 2000 \,	ext{N/m}^2 + 190 \,	ext{N/m}^2           = 2190 \,	ext{N/m}^2. $
Thus, the required pressure difference between the inside of the bubble and the atmospheric pressure is
$ \boxed{2190 \,	ext{N/m}^2.} $

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