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An air bubble of volume $1 \mathrm{~cm}^3$ rises from the bottom of a lake $40 \mathrm{~m}$ deep at a temperature of $12^{\circ} \mathrm{C}$. To what volume does it grow when it reaches the surface, which is at a temperature of $35^{\circ} \mathrm{C}$ ? Given $1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}$.
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Verified Answer
Given, $V_1=1.0 \mathrm{~cm}^3=1 \times 10^{-6} \mathrm{~m}^3$,
$$
\begin{aligned}
T_1 &=12^{\circ} \mathrm{C}=285 \mathrm{~K}, P_1=1 \mathrm{~atm}+h_1 \rho g \\
&=1.01 \times 10^5+40 \times 10^3 \times 9.8=493000 \mathrm{~Pa}
\end{aligned}
$$
Let, $V_2=$ volume of bubble at the surface of the lake
$$
\begin{aligned}
&\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
&\therefore V_2=\frac{P_1 V_1 T_2}{T_1 P_2}\left[\text { Where, } T_2=35^{\circ} \mathrm{C}=308 \mathrm{~K},\right. \\
&\left.\quad P_2=1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}\right] \\
&=\frac{(493000) \times\left(1.0 \times 10^{-6}\right) \times 308}{285 \times 1.01 \times 10^5} \\
&=5.275 \times 10^{-6} \mathrm{~m}^3
\end{aligned}
$$
$$
\begin{aligned}
T_1 &=12^{\circ} \mathrm{C}=285 \mathrm{~K}, P_1=1 \mathrm{~atm}+h_1 \rho g \\
&=1.01 \times 10^5+40 \times 10^3 \times 9.8=493000 \mathrm{~Pa}
\end{aligned}
$$
Let, $V_2=$ volume of bubble at the surface of the lake
$$
\begin{aligned}
&\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2} \\
&\therefore V_2=\frac{P_1 V_1 T_2}{T_1 P_2}\left[\text { Where, } T_2=35^{\circ} \mathrm{C}=308 \mathrm{~K},\right. \\
&\left.\quad P_2=1 \mathrm{~atm}=1.01 \times 10^5 \mathrm{~Pa}\right] \\
&=\frac{(493000) \times\left(1.0 \times 10^{-6}\right) \times 308}{285 \times 1.01 \times 10^5} \\
&=5.275 \times 10^{-6} \mathrm{~m}^3
\end{aligned}
$$
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