Question 171: Thermodynamics & KTG

Question

An air bubble of volume $2.9 \mathrm{~cm}^3$ rises from the bottom of a swimming pool of 5 m deep. At the bottom of the pool water temperature is $17^{\circ} \mathrm{C}$. The volume of the bubble when it reaches the surface, where the water temperature is $27^{\circ} \mathrm{C}$, is $\_\_\_\_$ $\mathrm{cm}^3$.
( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$, density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$, and 1 atm pressure is $10^5 \mathrm{~Pa}$ )

Accepted Answer

Initial state (at the bottom)
Volume $=\mathrm{V}_1=2.9 \mathrm{~cm}^3$
Temperature $=\mathrm{T}_1=17^{\circ} \mathrm{C}=17+273=290 \mathrm{~K}$
Pressure $=\mathrm{P}_1$
Total pressure at the bottom is atmospheric pressure $\left(\mathrm{P}_0ight)$ plus the hydrostatic pressure $(\mathrm{h} ho \mathrm{g})$.
$$ P_1=10^5+10^3 	imes 10 	imes 5=1.5 	imes 10^5 \mathrm{~Pa} $$
Final state (at the surface)
Temperature $=\mathrm{T}_2=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$
Pressure $=P_2=P_0$ (Only atmospheric pressure acts at the surface.)
$$ \mathrm{P}_2=10^5 \mathrm{~Pa} $$
According to the Ideal Gas Law :
$$ \frac{\mathrm{P}_1 \mathrm{~V}_1}{\mathrm{~T}_1}=\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2} $$
$$\Rightarrow $$ $\frac{\left(1.5 	imes 10^5ight) 	imes 2.9}{290}=\frac{\left(10^5ight) 	imes V_2}{300}$
$$\Rightarrow $$ $\frac{1.5 	imes 2.9}{290}=\frac{V_2}{300}$
$$\Rightarrow $$ $V_2=\frac{1.5 	imes 2.9 	imes 300}{290}$
$$\Rightarrow $$ $$ \mathrm{V}_2=4.5 \mathrm{~cm}^3 $$
So, as the bubble rises, the external pressure decreases significantly while the temperature increases slightly. Both factors contribute to the expansion of the air bubble, resulting in a final volume of $4.5 \mathrm{~cm}^3$.
Therefore, the correct option is (D).

Answer

2.0

Answer

4.2

Answer

3.0

Answer

4.5

Thermodynamics & KTG

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