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

Answer: 2.05

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 $=-(dp{)}_{avg}dV=-\frac{dp}{2}dV$
$=-\frac{\gamma p_0}{2V}\left(4\pi R^{2}a\right)\left(-4\pi R^{2}a\right)$
$=\left(4\pi p_{0}R{a}^2\right)\frac{3}{2}\times \frac{41}{30}$
$=(2.05)\left(4\pi p_{0}R{a}^2\right)$
$\Rightarrow x=2.05$