What is the excess pressure inside the bubble of soap solution of radius 3mm, given that surface tension of soap solution is 25 dyne/cm? Also find the pressure inside the bubble if atmospheric pressure is
$1.013\times {{10}^{6}}dyne/c{{m}^{2}}$
Answer
606.3k+ views
Hint: A soap bubble is basically a result of equilibrium between the pressure inside the soap film and that of the pressure outside. There is an excess pressure inside the bubble because of the surface tension of the soap film. Hence using the expression of the excess pressure inside the soap bubble, we can determine the excess pressure as well as the pressure inside.
Formula used:
${{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R}$
Complete answer:
Let us say there is a soap bubble of radius ‘R’ and having surface tension ‘S’ .The pressure inside is directed outwards whereas the external or the atmospheric pressure along with the surface tension is directed inwards. Hence the pressure inside the bubble is always greater than that of the outside. Hence the excess pressure ${{P}_{E}}$ i.e. the difference between the internal pressure (${{P}_{i}}$ )and the external pressure (${{P}_{o}}$ ) is given by,
${{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R}.....(1)$
Hence the excess pressure for the above soap bubble is given by,
$\begin{align}
& {{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R} \\
& {{P}_{E}}=\dfrac{4(25\text{ }dyne/cm)}{0.3cm}=333.33dyne/c{{m}^{2}} \\
\end{align}$
The atmospheric pressure is given as $1.013\times {{10}^{6}}dyne/c{{m}^{2}}$. Therefore using equation 1 the internal pressure is,
$\begin{align}
& {{P}_{E}}={{P}_{i}}-{{P}_{o}} \\
& 333.33dyne/c{{m}^{2}}={{P}_{i}}-1.013\times {{10}^{6}}dyne/c{{m}^{2}} \\
& \Rightarrow {{P}_{i}}=(333.33+1.013\times {{10}^{6}})dyne/c{{m}^{2}} \\
& \Rightarrow {{P}_{i}}=1.0133\times {{10}^{6}}dyne/c{{m}^{2}} \\
\end{align}$
Hence the internal pressure of the soap bubble is $1.0133\times {{10}^{6}}dyne/c{{m}^{2}}$
Note:
The excess pressure inside the soap bubble is twice as that of the air bubble in water. This is due to the fact that the soap bubble has an appreciable thickness and is double layered. Therefore the pressure across the two layers as we go inside increases twice.
Formula used:
${{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R}$
Complete answer:
Let us say there is a soap bubble of radius ‘R’ and having surface tension ‘S’ .The pressure inside is directed outwards whereas the external or the atmospheric pressure along with the surface tension is directed inwards. Hence the pressure inside the bubble is always greater than that of the outside. Hence the excess pressure ${{P}_{E}}$ i.e. the difference between the internal pressure (${{P}_{i}}$ )and the external pressure (${{P}_{o}}$ ) is given by,
${{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R}.....(1)$
Hence the excess pressure for the above soap bubble is given by,
$\begin{align}
& {{P}_{E}}={{P}_{i}}-{{P}_{o}}=\dfrac{4S}{R} \\
& {{P}_{E}}=\dfrac{4(25\text{ }dyne/cm)}{0.3cm}=333.33dyne/c{{m}^{2}} \\
\end{align}$
The atmospheric pressure is given as $1.013\times {{10}^{6}}dyne/c{{m}^{2}}$. Therefore using equation 1 the internal pressure is,
$\begin{align}
& {{P}_{E}}={{P}_{i}}-{{P}_{o}} \\
& 333.33dyne/c{{m}^{2}}={{P}_{i}}-1.013\times {{10}^{6}}dyne/c{{m}^{2}} \\
& \Rightarrow {{P}_{i}}=(333.33+1.013\times {{10}^{6}})dyne/c{{m}^{2}} \\
& \Rightarrow {{P}_{i}}=1.0133\times {{10}^{6}}dyne/c{{m}^{2}} \\
\end{align}$
Hence the internal pressure of the soap bubble is $1.0133\times {{10}^{6}}dyne/c{{m}^{2}}$
Note:
The excess pressure inside the soap bubble is twice as that of the air bubble in water. This is due to the fact that the soap bubble has an appreciable thickness and is double layered. Therefore the pressure across the two layers as we go inside increases twice.
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