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  • A bubble at air is underwater at Temperature and pressure 1 bar.<br /

A bubble at air is underwater at Temperature 10^{\circ} \mathrm{C} and pressure 1 bar.

If the bubble rise to the surface where the Temperature \mathrm{20^{\circ} \mathrm{C}} and Pressure is \mathrm{56\, \mathrm{ar}}

what will happen to the volume of the bubble? volume will become greater by a factor?

Option: 1

volume will become greater by a factor at 1-6


Option: 2

volume will become greater by a factor at 2.07


Option: 3

volume will become greater by a factor at 3.07


Option: 4

volume will be become greater by factor 5.09


Answers (1)

best_answer

\mathrm{\left.\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}} \text { [By ideal was equation }\right]}

or
\mathrm{\frac{1 \times v_{1}}{283} =\frac{5 \times v_{2}}{293}}

\mathrm{v_{1}=\frac{5 \times 283 v_{2}}{293} \text { or } v_{2} =\frac{v_{1} \times 293}{283 \times5} }

\mathrm{v_{2} =\frac{293 \mathrm{v_{1}}}{141.5} }
\mathrm{v_{2} =2.07 v_{1}}

i.e volume of bubble will be almost 2.07 time to instal volume of bubble.

Posted by

Anam Khan

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