A container filled with air under pressure P_0 contains a soap bubble of radius R. The air pressure has been reduced to half isothermally and the new radius of the bubble becomes 5R/4. If the surface tension of the soap water solution is S, P_0 is found to be nS/R SI unit,where n is...

Hannah Richmond

Hannah Richmond

Answered question

2023-01-27

A container filled with air under pressure P_0 contains a soap bubble of radius R. The air pressure has been reduced to half isothermally and the new radius of the bubble becomes 5R/4. If the surface tension of the soap water solution is S, P_0 is found to be nS/R SI unit,where n is...

Answer & Explanation

Hazel Lam

Hazel Lam

Beginner2023-01-28Added 9 answers

P 1 = ( P 0 + 4 S R ) P 2 = ( P 0 + 4 S 5 R 4 ) P 1 V 1 = P 2 V 2 ( P 0 + 4 S R ) ( 4 3 π R 3 ) = ( P 0 2 + 4 S 5 R 4 ) [ 4 3 π ( 5 R 4 ) 3 ] ( P 0 + 4 S R ) = ( P 0 2 + 16 S 5 R ) ( 125 64 ) P 0 × 3 128 = 25 S 4 R 4 S R = 9 S 4 R P 0 = 96 S R

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