A spherical bubble inside water has radius R. Take the pressure inside the bubble and the water pressure to be 𝑝₀. 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 (4𝜋𝑝₀𝑅²𝑎)X, where X is a constant and $γ = C_{p} / C_{V} = 41 / 30$ . The value of X is________.

Difficult

Solution

$\begin{array}{l} P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \\ \Rightarrow P_{0} {(\frac{4}{3} π R^{3})}^{γ} = P_{2} {(\frac{4}{3} π (R - a)^{3})}^{γ} \\ \Rightarrow P_{0} {(\frac{R}{R - a})}^{3 γ} = P_{2} \end{array}$
$Work done in adiabatic process$
$\begin{array}{l} W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} \\ = \frac{P_{0} \frac{4}{3} π R^{3} - P_{0} {(\frac{R}{R - a})}^{3 γ} \frac{4}{3} π (R - a)^{3}}{(\frac{41}{30} - 1)} [h e r e 3 γ = \frac{41}{10}] \\ \begin{array}{l} = \frac{P_{0} \frac{4}{3} π}{\frac{11}{30}} [R^{3} - R^{41 / 10} (R - a)^{3 - \frac{41}{10}}] \\ = \frac{30 P_{0}}{11} \frac{4}{3} π [R^{3} - R^{41 / 10} (R - a)^{- \frac{11}{10}}] \\ \begin{array}{l} = \frac{10 P_{0} 4 π}{11} [R^{3} - R^{41 / 10} R^{- 11 / 10} {(1 - \frac{a}{R})}^{- 11 / 10}] \\ = \frac{10 P_{0} 4 π}{11} [R^{3} - R^{3} {(1 - \frac{a}{R})}^{- 11 / 10}] \\ \begin{array}{l} = \frac{10 P_{0} 4 π}{11} R^{3} [1 - 1 - \frac{11 a}{10 R}] \\ = - (4 π P_{0} R^{2} a) \end{array} \end{array} \end{array} \end{array}$
Magnitude of work done $W = (4 π P_{0} R^{2} a)$
So, $X = 1$

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