Question

Drive laplaee's low for sherical membranes of bubble due to surface tenion

Answer 1

Answer:

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Explanation:

Consider a spherical liquid drop and let the outside pressure be P0 and inside pressure be Pi, such that the excess pressure is  Pi−Po.  

Let the radius of the drop increase from Δr, where Δr is very small, so that the pressure, inside the drop remains almost constant. 

Initial surface area (A1)=4πr2

Final surface area (A2)=4πr2

=4π(r+Δr)2

=4π(r2+2rΔr+Δr2)

=4πr2+8πrΔr+4πΔr2

As Δr is very small Δr2 is neglected (i.e. 4πΔr2=0)

Increase in surface area (dA)=A2=−A1=4πr2+8πrΔr−4πr2

In crease in surface area (dA)=8πrΔr

Work done to increase the surface area 8πrΔris extra energy.

∴dW=TdA

∴dW=T×8πrΔr  .....(1)

This work done is equal to the product of the force and the distance Δr

dF=(Pi−Po)4πr2×r2

The increase in the radius of the bubble is Δr. 

dW=dFΔr=