Question #bd0ea

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Question #bd0ea

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Answer 1 · 2016-01-15T02:17:16

The volume of the bubble at the surface will be 600 mL.
Actually this would be the volume just under the surface, as the bubble would dissipate into the atmosphere at the surface.

Explanation:

Use the combined gas law, with the equation $\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$ . For gas laws, the temperature needs to be in Kelvins.

Given/Known
$P_{1} = 2.4 atm$ $P_1="2.4 atm"$
$V_{1} = 250 mL$ $V_1="250 mL"$
$T_{1} = 15^{o} C + 273.15 K=285 K$ $T_1="15"^"o""C"+"273.15 K=285 K"$
$P_{2} = 1.0 atm$ $P_2="1.0 atm"$
$T_{2} = 27^{o} C + 273.15 K=300 K$ $T_2="27"^"o""C"+"273.15 K=300 K"$

Unknown
$V_{2}$ $V_2$

Solution
Rearrange the equation to isolate $P_{2}$ $P_2$ and solve.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1=(P_2V_2)/T_2$

$V_{2} = \frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}$ $V_2=(P_1V_1T_2)/(T_1P_2)$

$V_{2} = \frac{2.4 atm \times 250 mL \times 300 K}{285 K \times 1.0 atm} = 600 mL$ $V_2=(2.4"atm"xx250"mL"xx300"K")/(285"K"xx1.0"atm")="600 mL"$ (rounded to one significant figure)

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Question #bd0ea

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