Derive Laplace's law for spherical membrane of
bubble due to surface tension.
Answers
Consider a spherical liquid drop and let the outside pressure be P
0
and inside pressure be P
i
, such that the excess pressure is P
i
−P
o
.
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 (A
1
)=4πr
2
Final surface area (A
2
)=4πr
2
=4π(r+Δr)
2
=4π(r
2
+2rΔr+Δr
2
)
=4πr
2
+8πrΔr+4πΔr
2
As Δr is very small Δr
2
is neglected (i.e. 4πΔr
2
=0)
Increase in surface area (dA)=A
2
=−A
1
=4πr
2
+8πrΔr−4πr
2
In crease in surface area (dA)=8πrΔr
Work done to increase the surface area 8πrΔr is extra energy.
∴dW=TdA
∴dW=T×8πrΔr .....(1)
This work done is equal to the product of the force and the distance Δr
dF=(P
i
−P
o
)4πr
2
×r
2
The increase in the radius of the bubble is Δr.
dW=dFΔr=(P
i
−P
o
)4πr
2
×Δr
2
...(2)
Comparing equation (1) and (2), we get
(P
i
−P
o
)4πr
2
×Δr=T×8πrΔr
∴(P
i
−P
o
)=2T/r