A spherical bubble inside water has radius R. Take the pressure i… - Vidyadip

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 $a \ll R$ 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. . . . . .

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Answer

$W =(\Delta P )_{ arg } \times 4 \pi R ^2 a$

$\simeq\left|\frac{ dP }{2} \cdot 4 \pi R ^2 a \right|$

$\text { \{for small change } \left.(\Delta P )_{ avg }< P >\text { arithmetic mean }\right\}$

$= PV = c \Rightarrow dP =-\gamma \frac{ P }{ V } dV =-\frac{\gamma P _0}{ V } 4 \pi R ^2 a$

$=\frac{\gamma P _0}{2 V } \times 4 \pi R ^2 a \times 4 \pi R ^2 a$

$=\frac{\gamma P _0}{2 \times 4 \pi R ^3} 4 \pi R ^2 a \times 4 \pi R ^2 a$

$=\left(4 pRP \times a ^2\right) \frac{3 \gamma}{2}$

$\therefore x =2.05$

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