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A scuba diver is 40 m40 \text{~m} below the surface of a lake, where the temperature is 5.0C5.0^{\circ} \text{C}. He releases an air bubble that has a volume of 15 cm315 \text{~cm}^3. The bubble rises to the surface, where the temperature is 25C25^{\circ} \text{C}. Assume that the air in the bubble is always in thermal equilibrium with the surrounding water, and assume that there is no exchange of molecules between the bubble and the surrounding water. What is the volume of the bubble right before it breaks the surface? Hint: Remember that the pressure also changes.

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Answered 2 years ago
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In this problem, following values are given:

Volume of the bubble, V=15 cm3=15106 m3Atmosphere pressure, Pa=1 atm=1.103105 PaT1=50C=278 KT2=250C=298 K\begin{align*} \text{Volume of the bubble},\ V&=15\text{~cm}^3\\ &=15\cdot10^{-6}\text{~m}^3\\ \text{Atmosphere pressure},\ P_a&=1\text{~atm}\\ &=1.103\cdot 10^5\text{~Pa}\\ T_1=5^0\text{C}&=278\text{~K}\\ T_2=25^0\text{C}&=298\text{~K} \end{align*}

We need to find the volume of the bubble.

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