In figure (a) a gas bubble in liquid is shown and in figure (b) this bubble is shown spherical and in figure (c ) a soap bubble in air is shown .
Suppose , bubble formed by this way radius of bubble is r and inside pressure is `P_(i)` and outside pressure is `P_(0)`.
Let us obtain `P_(i)-P_(0)` for bubble in liquid or drop of liquid .
The pressure inside the drop is `P_(i)` and outside is `P_(0).P_(i)` is slightly higher than `P_(0)`.
When bubble increase by `Deltar`, new radius becomes `r+Deltar`.
Area of this surface `A_(2)=4pi(r+Deltar)^(2)`
`A_(2)=4pi(r^(2)+2r*Deltar+Deltar^(2))`
`Deltar^(2)` is very small comparedto others terms
`thereforeA_(2)=4pi(r^(2)+2rDeltar)` ....(2)
The change inarea of surface of drop
`DeltaA=A_(2)-A_(1)`
`DeltaA=4pi(r^(2)+2rDeltar)-4pir^(2)`
`=4pi(r^(2)+2rDeltar-2r^(2))`
=`4pi(2rDeltar)`
`Deltar=8pirDeltar` ...(3)
The work done for increase area `DeltaA` of free surface of drop.
`W=S_(la)(DeltaA)`
`W=S_(la)(8pirDeltar)` ....(4)
In the process of blowing bubble the force due to pressure difference `(P_(i)-P_(0)` on the inside surface of area `4pir^(2)` is `(P_(i)-P_(0))4pir^(2)` . Under the effect of this force , surface moved by `Deltar` displacement , hence work done on the surface,
`W=("force")xx("Displacement")`
`W=("pressure difference")xx("area")xx("displacement")`
`W=(P_(i)-P_(0))(4pir^(2))(Deltar)`...5
Comparing equation (4) and (5),
`(P_(i)-P_(0))(4pir^(2)Deltar)=S_(la)(8pirDeltar)`
`thereforeP_(i)-P_(0)=(2S_(la))/(r)`.....(6)
Bubble in liquid and drop of liquid has one free surface . For them equation of pressure difference ontain according to equation (6).
Bubble in air have two free surface . For them formula for pressure difference obtained as below.
`P_(i)-P_(0)=(4S_(la))/(r)` ....(7)
From this it can be said that for drop or bubble
`P_(i)-P_(0)prop(1)/(r)`
