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

Answer: 12

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