Q.

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=Cp/CV=41/30. The value of X is________.

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answer is 1.

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Detailed Solution

P1V1γ=P2V2γ⇒P043πR3γ=P243π(R−a)3γ⇒P0RR−a3γ=P2Work done in adiabatic process W=P1V1−P2V2γ−1=P043πR3−P0RR−a3γ43π(R−a)34130−1     here 3γ=4110=P043π1130R3−R41/10(R−a)3−4110=30P01143πR3−R41/10(R−a)−1110=10P04π11R3−R41/10R−11/101−aR−11/10=10P04π11R3−R31−aR−11/10=10P04π11R31−1-11a10R=-4πP0R2aMagnitude of work done W=4πP0R2aSo, X=1
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