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A bubble of air is underwater at temperature $15^\circ C$ and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is $25^\circ C$ and the pressure is 1.0 bar, what will happen to the volume of the bubble?
A
Volume will become smaller by a factor of 0.70
B
Volume will become greater by a factor of 2.5
C
Volume will become greater by a factor of 1.6
D
Volume will become greater by a factor of 1.1
Detailed Solution
The amount of air in the bubble is fixed, so the combined gas equation applies: $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$
$P_1 = 1.5$ bar, $T_1 = 15 + 273 = 288$ K; $P_2 = 1.0$ bar, $T_2 = 25 + 273 = 298$ K
$\frac{1.5\times V_1}{288} = \frac{1.0\times V_2}{298}$
$V_2 = \frac{1.5\times298}{288}V_1$
$V_2 = 1.55V_1 \approx 1.6V_1$
So the volume becomes greater by a factor of 1.6.
$P_1 = 1.5$ bar, $T_1 = 15 + 273 = 288$ K; $P_2 = 1.0$ bar, $T_2 = 25 + 273 = 298$ K
$\frac{1.5\times V_1}{288} = \frac{1.0\times V_2}{298}$
$V_2 = \frac{1.5\times298}{288}V_1$
$V_2 = 1.55V_1 \approx 1.6V_1$
So the volume becomes greater by a factor of 1.6.