Explain the term axcess pressure in liquid and find its expression
Answers
The work done by the excess pressure is given by
dW = Force x Displacement
= (Excess pressure x Area) x (increase in radius)
=[(Pi - Po) × 4πr²] × Δr .......(1)
Let the initial surface area of the drop be A1=4πr²
The final surface area of the drop is A2=4π(r+Δr)²
A2 = 4π(r² + 2rΔr + Δr²)
= 4πr² + 8πrΔr +4πΔr²
As Δr is very small, Δr² can be neglected, i.e. 4πΔr² ≡ 0
Therefore, A2 = 4πr² + 8πrΔr
Therefore, increase in the surface area of the drop dA = A2 - A1
= 4πr² + 8πrΔr - 4πr²
= 8πrΔr
Work done to increase the surface area by 8πrΔr is the extra surface energy.
dW = T x dA
where T is the surface energy
dW = T x 8πrΔr .........(2)
Comparing (1) & (2), we get
[(Pi - Po) × 4πr²] × Δr = T × 8πrΔr
Pi - Po = 2T/r
The above equation gives the excess pressure inside a liquid drop. i hope you understand....