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 $$(4$$𝜋𝑝$$0$$𝑅$$2$$𝑎$$)X$$, where X is a constant and γ$$=Cp/CV=41/30$$. The value of X $$is________$$.

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 $$(4$$𝜋𝑝$$0$$𝑅$$2$$𝑎$$)X$$, where X is a constant and γ$$=Cp/CV=41/30$$. The value of X $$is________$$.

Infinity Learn · Accepted Answer

$$P1V1$$γ$$=P2V2$$γ⇒$$P043$$$$π$$$$R3$$γ$$=P243$$π(R$$−$$a)3$$γ⇒$$P0RR$$−$$a3$$γ$$=P2Work$$ done in adiabatic process $$W=P1V1$$−$$P2V2$$γ−$$1=P043$$$$π$$$$R3$$−$$P0RR$$−$$a3$$γ$$43$$π$$(R$$−$$a)34130$$−$$1$$     here $$3$$γ$$=4110=P043$$$$π$$$$1130R3$$−$$R41/10(R$$−$$a)3$$−$$4110=30P01143$$$$π$$$$R3$$−$$R41/10(R$$−$$a)−$$1110=10P04$$$$π$$$$11R3$$−$$R41/10R$$−$$11/101$$−aR−$$11/10=10P04$$$$π$$$$11R3$$−$$R31$$−aR−$$11/10=10P04$$$$π$$$$11R31$$−$$1-11a10R=-4$$$$π$$$$P0R2aMagnitude$$ of work done $$W=4$$$$π$$$$P0R2aSo$$, $$X=1$$

Want to Fund your own JEE / NEET / Foundation preparation ??

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!