
An air bubble which is lying just below the surface of water. The surface tension of water is given as $7\times {{10}^{-3}}N{{m}^{-1}}$ and the atmospheric pressure is $1.013\times {{10}^{5}}N{{m}^{-2}}$. If the radius of bubble is $1mm$, then the pressure inside the bubble will be given as,
$\begin{align}
& A.1.0270\times {{10}^{5}}Pa \\
& B.1.0160\times {{10}^{5}}Pa \\
& C.1.0144\times {{10}^{5}}Pa \\
& D.1.0131\times {{10}^{5}}Pa \\
\end{align}$
Answer
509.7k+ views
Hint: Since it is said that the bubble is just below the surface of water, we can suppose that the pressure outside the bubble will be similar as atmospheric pressure. Therefore the pressure inside the bubble will be greater than that of the outside.
Formula used:
\[{{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R}\]
Where \[T\] be the surface tension of water and \[R\] be the universal gas constant.
Complete answer:
As the bubble is said to be just beneath the surface of the water, we can assume safely that the pressure outside the bubble is similar to the atmospheric pressure. Hence the pressure inside the bubble will be greater than the pressure outside the bubble,
So that we can write it in the form of an equation,
\[{{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R}\]
Where \[T\] be the surface tension of water and \[R\] be the radius of bubbles.
It is given in the question that,
The surface tension of the water is,
$T=7\times {{10}^{-3}}N{{m}^{-1}}$
And the radius of the bubble is given as,
$R=1mm$
And also the outside pressure is,
${{P}_{out}}=1.013\times {{10}^{5}}N{{m}^{-2}}$
Substituting this in the equation will give,
\[\begin{align}
& {{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R} \\
& {{P}_{in}}={{P}_{out}}+\dfrac{2T}{R} \\
& {{P}_{in}}=1.013\times {{10}^{5}}+\dfrac{2\times 7\times {{10}^{-3}}N{{m}^{-1}}}{1\times {{10}^{-3}}} \\
& {{P}_{in}}=1.013\times {{10}^{5}}+\dfrac{140\times {{10}^{-3}}}{0.0001} \\
\end{align}\]
Therefore the value of inside pressure will be,
\[{{P}_{in}}=1.0144\times {{10}^{5}}N{{m}^{-2}}\]
Therefore the correct answer is option C.
Note: The attraction of the molecules in a surface is known as the surface tension. It will contract the soap bubble into the possible smallest area. This will make the air inside the soap bubble presses against this simultaneously. That is the reason why the bubble is forming into a sphere with large contents and small circumference.
Formula used:
\[{{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R}\]
Where \[T\] be the surface tension of water and \[R\] be the universal gas constant.
Complete answer:
As the bubble is said to be just beneath the surface of the water, we can assume safely that the pressure outside the bubble is similar to the atmospheric pressure. Hence the pressure inside the bubble will be greater than the pressure outside the bubble,
So that we can write it in the form of an equation,
\[{{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R}\]
Where \[T\] be the surface tension of water and \[R\] be the radius of bubbles.
It is given in the question that,
The surface tension of the water is,
$T=7\times {{10}^{-3}}N{{m}^{-1}}$
And the radius of the bubble is given as,
$R=1mm$
And also the outside pressure is,
${{P}_{out}}=1.013\times {{10}^{5}}N{{m}^{-2}}$
Substituting this in the equation will give,
\[\begin{align}
& {{P}_{in}}-{{P}_{out}}=\dfrac{2T}{R} \\
& {{P}_{in}}={{P}_{out}}+\dfrac{2T}{R} \\
& {{P}_{in}}=1.013\times {{10}^{5}}+\dfrac{2\times 7\times {{10}^{-3}}N{{m}^{-1}}}{1\times {{10}^{-3}}} \\
& {{P}_{in}}=1.013\times {{10}^{5}}+\dfrac{140\times {{10}^{-3}}}{0.0001} \\
\end{align}\]
Therefore the value of inside pressure will be,
\[{{P}_{in}}=1.0144\times {{10}^{5}}N{{m}^{-2}}\]
Therefore the correct answer is option C.
Note: The attraction of the molecules in a surface is known as the surface tension. It will contract the soap bubble into the possible smallest area. This will make the air inside the soap bubble presses against this simultaneously. That is the reason why the bubble is forming into a sphere with large contents and small circumference.
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