Question

(a) Show from first principles, that when the surface area of a system, subject to a surface tension ¶, changes by a small amount dA, the work done by the system is (-¶dA)
(b). A bubble rose from the bottom of a lake to near its surface. The
bubble was initially 1 cm in diameter, at a pressure of 2 bar. The final pressure
experienced by the bubble was 1 bar. The initial as well as the final shape of the bubble was spherical, and its volume varied inversely to the pressure. Determine
the net work done by the bubble. For water, ¶ = 0.073 N/m. Which additional
assumptions did you make? Please comment on your answer.

Answer 1

solution

(a)consider a fluid of tension T then displaced dx by a constant force F very slowly then

work done can be given as

$dW=Fdx$

and

F can be given as

F= - 2T.L

so

$dW=-2Tldx$

and

Ldx=dA

so

$dW=-2TdA$

(b)pressure

$p_1=2bar\quad p_2=1bar$

radius of bubble(r₁)=1 cm

Pv= constant and considering it isotherm process

$p_1V_1=p_2V_2$

$2\times \frac{4}{3}\pi r_1^3=1\times\frac{4}{3}\pi r_2^3\\r_2=2^{\frac{1}{3}}cm=1.25cm$

work done can be

$W=T\Delta A$

$W=0.073\times4\pi(1.25^2-1^2)\times10^{-4}$

$W=0.51\times10^{-4}J$