# What is this equation of volumes and pressures of the soap bubble?

Initially the uncharged soap bubble exists with the radius $$r$$ and the surface tension $$T$$

Nextly the bubble has been held the potential $$V$$at the surface of it and the radius of the bubble became $$R~~$$

$$p_{0}:=\text{atmospheric pressure}$$

$$p:=\text{inner pressure of the bubble before the given of the potential}$$

The below equation is held.

$$p-p_{0}=\frac{4T}{r}$$

$$p':=\text{inner pressure of the bubble after the given of the potential}$$

$$p'=+p_{0}+\frac{4T}{r}-\frac{\epsilon_{0}E^{2}}{2}$$

$$=+p_{0}+\frac{4T}{r}-\frac{\epsilon_{0}}{2}\left(\frac{V}{R}\right)^{2}~~~~~~~~\text{(rightmost term represents the electrostatic energy density)}$$

$$v:=\text{volume of the bubble before the given of the potential}$$

$$v':=\text{volume of the bubble after the given of the potential}$$

What I can't get currently is the below equation.

$$pv=p'v'$$

Why this equation can be held?

or $$pV = \text{constant}$$
Now, in this case there is no mention of temperature change, and the mass remains constant, and we can assume air to be roughly an ideal gas, so that $$p_1V_1=p_2V_2$$