Question

A spherical bubble inside water has radius $R$. Take the pressure inside the bubble and the water pressure to be $p_{0}$. The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R-a)$. For $a < < R$ the magnitude of the work done in the process is given by $\left(4 \pi p_{0} R a^{2}\right) X$, where $X$ is a constant and $\gamma=C_{ p } / C_{ V }=41 / 30$. The value of $X$ is ______.

Answer 1

In adiabatic process
$dp =-\frac{\gamma p }{ V } dV $
$=-\frac{\gamma p _{0}}{ V }\left(-4 \pi R ^{2} a \right)$
Work done in the process $=-( dp )_{ avg } d V =-\frac{ dp }{2} dV$
$=-\frac{\gamma p _{0}}{2 V }\left(4 \pi R ^{2} a \right)\left(-4 \pi R ^{2} a \right)$
$=\left(4 \pi p _{0} Ra ^{2}\right) \frac{3}{2} \times \frac{41}{30}$
$=(2.05)\left(4 \pi p _{0} Ra ^{2}\right)$
$\Rightarrow x =2.05$