Derive expression for excess pressure inside a liquid drop
Answers
Answer:The net upward force on the top hemisphere of the liquid is just the pressure difference times the area of the equatorial circle :
F
upward
=(P
1
′
−P
0
)πr
2
where F
upward
= upward force on the liquid due to air-currents;
P
1
′
= Final pressure; P
0
= Initial pressure; r= radius of the drop, assuming it to be spherical.
The surface tension force downward around circle is equal to the surface tension of the circumference, since only one surface contributed to the force :
Thus, F
downward
=T(2πr) where, T= surface tension.
This gives, P
1
−P
0
=
r
2T
which represents the excess pressure inside the liquid drop.
Explanation:
Answer:
Let the pressure outside the liquid drop be Po and the inside pressure be Pi.
The excess pressure inside the drop = (Pi − Po).
Let T be the surface tension of the liquid.
Let the radius of the drop increase from r to (r + Δr) due to excess pressure.
The work done by the excess pressure is given by
dW = Force x Displacement
= (Excess pressure x Area) x (Increase in radius)
= [(Pi - Po) x 4πr2] x 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 × d A
where T is the surface energy
dW = T × 8πrΔr..........(2)
Comparing (1) and (2), we get
[(Pi - Po) x 4πr2] x Δr = T x 8πrΔ
Pi - Po = 2T/r
The above equation gives the excess pressure inside a liquid drop.