Question

A spherical bubble inside water has a 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 $a\ll R$ 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

Workdone in reducing the radius of bubble by a very small amount ${}^{\prime }{a}^{\prime }$, [given that $a<<R$]
$W=(\Delta p{)}_{avg}\times 4\pi R^{2}a$
Here $\Delta V=4\pi R^{2}a$, is the change in volume during the pressure
$W=\left|\frac{dp}{2}.4\pi R^{2}a\right|...\left(i\right)$
{for small change $(\Delta p{)}_{avg}\simeq \text{arithmetic mean}=\frac{dp}{2}$
and $\Delta V\simeq dV$
For the adiabatic compression, we know the slope of process is given by
$\frac{dp}{dV}=-\gamma \frac{p}{V}$
$\Rightarrow dp=-\gamma \frac{p}{V}dV=-\frac{\gamma p_o}{V}4\pi R^{2}a$

Therefore from Eq.$\left(i\right)$ the workdone,
$W=\frac{\gamma p_o}{2V}4\pi R^{2}a\times 4\pi R^{2}a$

$W=\frac{\gamma p_0}{2\times \left(\frac{4\pi R^3}{3}\right)}\times 4\pi R^{2}a\times 4\pi R^{2}a$
$\Rightarrow W=\left(4\pi p_{0}R{a}^2\right)\frac{3\gamma }{2}$

$\Rightarrow x=\frac{3}{2}\gamma =\frac{123}{60}\simeq 2.05$