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 $\left(4\pi p_0{\left(Ra\right)}^2\right)X,$ , where $X$ is a constant and $\gamma =\frac{C_P}{C_v}=\frac{41}{30}.$ The value of $X$ is _______ correct to nearest integer.

Answer 1

Answer: 2

$W=(\Delta P\times 4\pi R^{2}a=\left|\frac{dP}{2}\cdot 4\pi R^{2}a\right|$
{for small change $(\Delta P<P>$ Arithmetic mean]
$\Rightarrow P{V}^{\gamma }=c\Rightarrow dP=-\gamma \frac{P}{V}dV=-\frac{\gamma P_0}{V}4\pi R^{2}a$
$=\frac{\gamma P_0}{2V}\times 4\pi R^{2}a\times 4\pi R^{2}a$
$=\frac{\gamma P_0}{2\times 4\pi R^3}4\pi R^{2}a\times 4\pi R^{2}a$
$=\left(4\pi RP\times a^2\right)\frac{3\gamma }{2}$
$\therefore X\simeq 2.05\approx 2$