Question

The difference between the mean square velocity of air molecules inside and outside the soap bubble of radius ' $r$ ' and surface tension ' $T$ ' is found to be $\left(\frac{ aT }{ r \rho}\right)$, where ' $a$ ' is a constant with value $=$

Answer 1

For soap bubble the excess pressure inside is,
$\Delta P =\frac{4 T }{ r }$
From kinetic theory of gases;
$P =\frac{1}{3} \rho c ^{2}$
$\therefore \quad\left( c ^{2}\right)_{\text {in }}-\left( c ^{2}\right)_{\text {out }}=\frac{3}{\rho}\left( P _{\text {in }}- P _{\text {out }}\right) =\frac{3}{\rho} \Delta P$
$=\frac{3}{\rho}\left(\frac{4 T }{ r }\right)$
$\therefore \left(\frac{ aT }{ r \rho}\right)=\frac{12 T }{ r \rho}$
$\therefore a =12$