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A spherical balloon of radius 3 cm contain He  gas has a pressure of 48\times 10^{-3}  bar.  At the same temp. the pressure of spherical balloon of radius 12 cm containing the same amount of gas will be _____ \times 10^{-6}  bar.
 

Answers (1)

best_answer

We have:

r_{1} = 3cm

P_{1} =48\times 10^{-3} bar

Thus, according to Boyle's law, we have:

P_{1}V_{1} = P_{2}V_{2}

On putting the values we get:

\\\mathrm{(48\: x\: 10^{-3}bar)(4/3\pi (3)^{3})\: =\: P_{2}(4/3\pi (12)^{3})}\\\\\mathrm{P_{2}\: =\: \frac{48\: x\: 10^{-3}\: x\: (3)^{3}}{(12)^{3}}}

\mathrm{P_{2}\: =\: 750\: x\: 10^{-6}\: bar}

Thus, 750 is the answer.

 

Posted by

Kuldeep Maurya

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