What is the excess pressure inside a bubble of soap solution of radius $5.00 \;mm$, given that the surface tension of soap solution at the temperature ($20\,^{\circ} C$) is $2.50 \times 10^{-2}\; N m ^{-1}$ ? If an air bubble of the same dimension were formed at depth of $40.0 \;cm$ inside a container containing the soap solution (of relative density $1.20$), what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \;Pa$ ).
What is the excess pressure inside a bubble of soap solution of radius $5.00 \;mm$, given that the surface tension of soap solution at the temperature ($20\,^{\circ} C$) is $2.50 \times 10^{-2}\; N m ^{-1}$ ? If an air bubble of the same dimension were formed at depth of $40.0 \;cm$ inside a container containing the soap solution (of relative density $1.20$), what would be the pressure inside the bubble? ($1$ atmospheric pressure is $1.01 \times 10^{5} \;Pa$ ).
Excess pressure inside the soap bubble is $20 Pa$
Pressure inside the air bubble is $1.06 \times 10^{5} Pa$
Soap bubble is of radius, $r=5.00 mm =5 \times 10^{-3} m$
Surface tension of the soap solution, $S=2$ $\times 10^{-2} Nm ^{-1}$
Relative density of the soap solution $=1.20$
$.$ Density of the soap solution, $\rho=1.2 \times 10^{3} kg / m ^{3}$
Air bubble formed at a depth, $h=40 cm =0.4 m$
Radius of the air bubble, $r=5 mm =5 \times 10^{-3} m$
$1$ atmospheric pressure $=1.01 \times 10^{5} Pa$
Acceleration due to gravity, $g=9.8 m / s ^{2}$
Hence, the excess pressure inside the soap bubble is given by the relation
$P=\frac{4 S}{r}$
$=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$
$=20 Pa$
Therefore, the excess pressure inside the soap bubble is $20 Pa$.
The excess pressure inside the air bubble is given by the relation:
$P^{\prime}=\frac{2 S}{r}$
$=\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}$
$=10 Pa$
Therefore, the excess pressure inside the air bubble is $10 Pa$.
At a depth of $0.4 m ,$ the total pressure inside the air bubble
$=$ Atmospheric pressure $+h \rho g +P$
$=1.01 \times 10^{5}+0.4 \times 1.2 \times 10^{3} \times 9.8+10$
$=1.057 \times 10^{5} Pa$
$=1.06 \times 10^{5} Pa$
Therefore, the pressure inside the air bubble is $1.06 \times 10^{5} \;Pa$
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