A spherical bubble inside water has radius . Take the pressure inside the bubble and the water pressure to be . The bubble now gets compressed radially in an adiabatic manner so that its radius becomes . For , the magnitude of the work done in the process

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 $(4 π p_{0} R a^{2}) X$ where $X$ is a constant and $γ = \frac{C_{p}}{C_{v}} = \frac{41}{30} .$ The value of $X$ is _______. [2020]

Smart Achievers · Accepted Answer

(2.05)

$Here, we will neglect surface tension.$

$As P_{inside} = P_{outside} = P_{0}$

$Here, Δ k = 0$

$So, W_{all forces} = 0$

$W_{ext} + W_{w} + W_{g} = 0, W_{w} \to work done by water$

$W_{ext} = - (W_{w} + W_{g}), W_{g} \to work done by gas$

$Now, W_{w} = P_{0} [\frac{4}{3} π R^{3} - \frac{4}{3} π {(R - a)}^{3}]$

$= \frac{4}{3} P_{0} π [R - (R - a)] [R^{2} + R (R - a) + {(R - a)}^{2}]$

$= \frac{4}{3} P_{0} π (a) [R^{2} + R^{2} + R^{2} - 3 R a + a^{2}]$

$= 4 P_{0} π [R^{2} a - R a^{2}]$

$and,$ $W_{g} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = P_{0} \frac{4}{3} π R^{3} - P_{0} {(\frac{R}{R - a})}^{41 / 10} \frac{4}{3} π {(R - a)}^{3}$ $[∵ P V^{γ} = constant]$

$= \frac{30}{11} P_{0} \times \frac{4}{3} π R^{3} [1 - {(\frac{R}{R - a})}^{41 / 10} {(\frac{R - a}{R})}^{3}]$

$= \frac{40}{11} P_{0} π R^{3} [1 - {(1 - \frac{a}{R})}^{- 11 / 10}]$

$= \frac{40}{11} P_{0} π R^{3} [1 - 1 - \frac{11 a}{10 R} - \frac{(- \frac{11}{10}) (- \frac{21}{10})}{2} \frac{a^{2}}{R^{2}}]$

$= - 4 P_{0} π R^{2} a - 4.2 P_{0} π R a^{2}$

$So, W_{w} + W_{g} = - 4 P_{0} π R_{0}^{2} [1 + 1.05] = - 4 P_{0} π R_{0}^{2} \times 2.05$

$∴$ $W_{ext} = 4 P_{0} π R a^{2} \times 2.05$

$So, X = 5$

Solution

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