Question

Two spherical soap bubbles coalesce. If V is the consequent change in volume of the contained air and S the change in total surface area, show that
$3PV+4ST=0$, where T is the surface tension of the soap bubble and P is atmospheric pressure.

Answer 1

We are given two soap bubbles which are coalesced together. The change in volume and surface area is given to us. Since the temperature change is not mentioned we can assume this as an isothermal condition and apply Boyle’s law. By using the equation for pressure inside a soap bubble and applying the known terms in the Boyle’s law equation, we will get the solution.
Formula used:
${{P}_{i}}=P+\dfrac{4T}{r}$
$V=\dfrac{4}{3}\pi {{r}^{3}}$
$PV=\text{constant}$

Complete answer:
In the question it is said that two soap bubbles join together.
Let A and B be the bubbles that coalesce and let C be the new bubble formed.
Then ‘a’, ‘b’, ‘c’ will be the radius of the soap bubbles ‘A’, ‘B’, ‘C’, respectively.
Then we know that ${{P}_{a}},{{P}_{b}},{{P}_{c}}$ will be the pressure inside the bubbles A, B, C and ${{V}_{a}},{{V}_{b}},{{V}_{c}}$ will be the volume of the bubbles A, B, C respectively.
We know that the equation for the pressure inside a soap bubble is given as,
${{P}_{i}}=P+\dfrac{4T}{r}$, were ‘${{P}_{i}}$’ is the inside pressure, ‘P’ is the atmospheric pressure, ‘T’ is the surface tension and ‘r’ is the radius.
Then the pressure inside the bubble A, B and C will be,
${{P}_{a}}=P+\dfrac{4T}{a}$
${{P}_{b}}=P+\dfrac{4T}{b}$
${{P}_{c}}=P+\dfrac{4T}{c}$
We know that volume of a sphere is given by the equation,
$V=\dfrac{4}{3}\pi {{r}^{3}}$
Therefore volume of the soap bubbles A, B, C will be,
${{V}_{a}}=\dfrac{4}{3}\pi {{a}^{3}}$
${{V}_{b}}=\dfrac{4}{3}\pi {{b}^{3}}$
${{V}_{c}}=\dfrac{4}{3}\pi {{c}^{3}}$
In the question nothing is said about the temperature. Hence we can assume that the temperature is constant.
For an isothermal situation we know that,
$PV=\text{constant}$
${{P}_{a}}{{V}_{a}}+{{P}_{b}}{{V}_{b}}={{P}_{c}}{{V}_{c}}$
By substituting for ${{P}_{a}},{{P}_{b}},{{P}_{c}}$ and ${{V}_{a}},{{V}_{b}},{{V}_{c}}$ in the above equation, we get
$\Rightarrow \left( P+\dfrac{4T}{a} \right)\left( \dfrac{4}{3}\pi {{a}^{3}} \right)+\left( P+\dfrac{4T}{b} \right)\left( \dfrac{4}{3}\pi {{b}^{3}} \right)=\left( P+\dfrac{4T}{c} \right)\left( \dfrac{4}{3}\pi {{c}^{3}} \right)$
By solving this we get,
$\Rightarrow \left( P+\dfrac{4T}{a} \right)\left( \dfrac{4}{3}\pi {{a}^{3}} \right)+\left( P+\dfrac{4T}{b} \right)\left( \dfrac{4}{3}\pi {{b}^{3}} \right)-\left( P+\dfrac{4T}{c} \right)\left( \dfrac{4}{3}\pi {{c}^{3}} \right)=0$
$\Rightarrow P\dfrac{4}{3}\pi {{a}^{3}}+\dfrac{4T}{a}\left( \dfrac{4}{3}\pi {{a}^{3}} \right)+P\dfrac{4}{3}\pi {{b}^{3}}+\dfrac{4T}{b}\left( \dfrac{4}{3}\pi {{b}^{3}} \right)-P\dfrac{4}{3}\pi {{c}^{3}}-\dfrac{4T}{c}\left( \dfrac{4}{3}\pi {{c}^{3}} \right)=0$
$\Rightarrow \left( P\dfrac{4}{3}\pi a+P\dfrac{4}{3}\pi {{b}^{3}}-P\dfrac{4}{3}\pi {{c}^{3}} \right)+\dfrac{4T}{3a}\times 4\pi {{a}^{3}}+\dfrac{4T}{3b}\times 4\pi {{b}^{3}}-\dfrac{4T}{3c}\times 4\pi {{c}^{3}}=0$
$\Rightarrow P\left( \dfrac{4}{3}\pi {{a}^{3}}+\dfrac{4}{3}\pi {{b}^{3}}-\dfrac{4}{3}\pi {{c}^{3}} \right)+\dfrac{4T}{3}\left( 4\pi {{a}^{2}}+4\pi {{b}^{2}}-4\pi {{c}^{2}} \right)=0$
Here we know that $\dfrac{4}{3}\pi {{a}^{3}}+\dfrac{4}{3}\pi {{b}^{3}}$ is the volume of the bubbles before getting coalesced and $\dfrac{4}{3}\pi {{c}^{3}}$ is the volume of the bubble formed.
Hence the difference between them is the consequent change in volume, which is given as ‘V’ in the question.
Similarly we know that $4\pi {{a}^{2}}+4\pi {{b}^{2}}$ is the total surface area before getting coalesced and $4\pi {{c}^{2}}$ is the surface area of the bubble formed.
Therefore their difference is the change in surface area which is given as ‘S’ in the question.
By substituting ‘V’ and ‘S’ in the equation, we get
$\Rightarrow PV+\dfrac{4T}{3}S=0$
By solving this we get,
$\Rightarrow 3PV+4TS=0$
Hence the proof.

Note:
According to Boyle’s law at constant temperature the pressure of a given quantity of gas will be inversely proportional to the volume of that gas.
This statement is mathematically stated as,
$P\propto \dfrac{1}{V}$
$\Rightarrow PV=\text{constant}$
If ${{P}_{1}},{{V}_{1}}$ is the pressure and volume of a certain quantity of gas respectively and ${{P}_{2}},{{V}_{2}}$ is the pressure and volume of another quantity of gas, then according to Boyle’s law we can say that
${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$