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Asked by Prashant DIGHE | 20 Oct, 2019, 07:29: AM
answered-by-expert Expert Answer
If Po is the pressure outside soap bubble, then pressure P1 inside soap bubble  is given by
 
P1 = [ Po + ( 4T/R) ] .................(1)
 
where T is surface tension and R is radius of bubble.
 
Volume of bubble V1 = (4/3)πR3 ................(2)
 
If pressure outside the bubble is reduced n-fold and diameter increaded r-fold,
then changed pressure P2 and changed volume V2 are given by,
 
P2 = [ ( Po /n) + ( 4T/ (r R) ) ] .........................(3)
 
V2 = (4/3)π r3 R3 .....................................(4)
 
From Boyl's law, we have, P1V1 = P2V2 ................(5)
 
using all the above equations, we get ,
 
[ Po + ( 4T/R) ] (4/3)πR3 = [ ( Po /n) + ( 4T/R) ] (4/3)π r3 R3 ....................(6)
 
After simplification,  T = (1/4) Po R [ 1 - (r3 / n ) ] [ 1 / (r2 - 1) ] 
 
In terms of diameter D, we get , T = (1/8) Po D [ 1 - (r3 / n ) ] [ 1 / (r2 - 1) ]
Answered by Thiyagarajan K | 20 Oct, 2019, 15:53: PM

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