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| 1. |
Derive an expression for excess of pressure (pressure difference) inside the drop and bubble. |
Answer» Solution :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)`. Letus obtain `P_(i)-P_(0)` for bubble in liquid or drop of liquid . The pressure insidethe drop is `P_(i)` and outsideis `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 surfaceof 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)` |
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