If the radius and surface tension of a spherical soap bubble be ‘ R ‘ & ‘ T ‘ respectively , show that the charge required to double its radius would be, 8 π R [ ε 0 R [7 PR + 12 T]] 1/2 . (where P is the atmospheric pressure and process isothermal)
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
by conservation of mole:
P A × V = P’ A × (8V) { ∴ doubling the radius, volume gets 8 times}
⇒ P’ A = 
Now for radius R:
P A – P 0 = 
& 
= 
{ ∴ pressure by charge density σ is given by 
}
Now 
we get 
= – 
⇒ σ 2 = 
⇒ σ = 
⇒ Q = 4 π (2R) 2 
= 8 π R 
= 8 π R 
Hence proved.
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