Question

derive laplace's law for spherical membrane of bubble due to surface tension​

Answer 1

Answer:

$\huge\mathfrak\orange{answer :-⤵}$

$\bigstar\underline{\underline{\sf Derivation :}}$

Consider a spherical liquid drop and let the outside pressure be P0 and inside pressure be P i , 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) = $4\pi {r}^{2}$

Final surface area (A2) = $4\pi {r}^{2}$

$= 4\pi( {r +Δr)}^{2} </p><p></p><p>$

$= 4\pi( {r}{2} + 2rΔr + {Δr)}{2} </p><p></p><p>$

${4\pi \: r}^{2} + 8\pi \: rΔr + {4\pi \:Δr}{2}$

As Δr is very small Δr^2 is neglected (i.e = 0)

Increase in surface area (dA) = A2 =

- A1 = ${4\pi \: r}^{2} + 8\pi \: rΔr - {4\pi \: r}^{2}$

In crease in surface area (dA)=8πrΔr

Work done to increase the surface area 8πrΔ

${4\pi \: r}^{2} \times Δr$

r 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 = (pi- p0)

${4\pi \: r}^{2} \times {r}^{2}$

The increase in the radius of the bubble is Δr.

dW = dFΔr = (pi - p0)

${4\pi \: r}^{2} \times {Δr}^{2} - - - - - - 2$

Comparing equation (1) and (2), we get

(pi - p0)

${4\pi \: r}^{2} \times Δr$

$= t \times 8\pi \: rΔr$

$(pi - p0) =$2T/r.

Explanation:

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Answer 2

Explanation:

Expression for excess pressure inside a drop :

delta r is very small

$excess \: pressure \: = \ p_i - p_{o}$

$4\pi {r}^{2}$