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A bubble of air is underwater at temperature $15^{\circ} \mathrm{C}$ and the pressure $1.5 \mathrm{~bar}$. If the bubble rises to the surface where the temperature is $25^{\circ} \mathrm{C}$ and the pressure is $1.0 \mathrm{~bar}$, what will happen to the volume of the bubble?
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The correct answer is:
Volume will become greater by a factor of 1.6
$\because \quad \frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
(By ideal gas equation)
or $\frac{1.5 \times V_1}{288}=\frac{1 \times V_2}{298}$
$\therefore \quad V_2=1.55 V_1$
i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.
(By ideal gas equation)
or $\frac{1.5 \times V_1}{288}=\frac{1 \times V_2}{298}$
$\therefore \quad V_2=1.55 V_1$
i.e., volume of bubble will be almost 1.6 times to initial volume of bubble.
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