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define excess pressure. obtain an expression for excess pressure inside a liquid drop of radius R.assume surface tension of liquid as T.

Asked by Deepshikha deepsi 7th March 2013, 6:51 AM
Answered by Expert
  Consider the figure below 
    we know that as the drop/bubble remains in an equilibrium condition work done = increase in potential energy If we consider the atmospheric pressure to be 'Ps' and the internal pressure to be 'Pi'
then the net pressure would be written as  
Pnet = Pi - Pa  
Now, as the radius of the drop has increased, work has been done by the net internal pressure.
So, this work done will be   W = force x displacement or W = (pressure x area) x displacement  
and in this case W = (pressure x area) x increase in drop radius  
so, the work done will be W = (P net x 4πr 2 ) x dr ........(1)
but, the increase potential energy will be twice dU = 16πr.dr x S ...........(2)   t
his is because in case of a bubble there are two free surfaces, the outside surface and the inside surface.
The potential energy will thus change accordingly.  
so, as (1) equals (2), we have (Pnet x 4?r2) x dr = 16πr.dr x S
or   the net pressure inside a bubble will be:
Pnet = 4S / r [twice than the that inside the drop]   which is the required expression
Answered by Expert 7th March 2013, 11:13 AM
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