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A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
When the gas bubble is at height $y$ from the bottom, its temperature is
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A small spherical monoatomic ideal gas bubble $\left(\gamma=\frac{5}{3}\right)$ is trapped inside a liquid of density $\rho_1$ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains $n$ moles of gas. The temperature of the gas when the bubble is at the bottom is $T_0$, the height of the liquid is $H$ and the atmospheric pressure is $p_0$ (Neglect surface tension).
Question:
When the gas bubble is at height $y$ from the bottom, its temperature is
Solution:
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Verified Answer
The correct answer is:
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$
$T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}$
As there is no exchange of heat. Therefore, process is adiabatic. Applying,
$$
\begin{array}{rlrl}
& T p^{\frac{1-\gamma}{\gamma}}=\text { constant } \\
\therefore \quad & T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \\
\text { or } & T_2 & =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}
\end{array}
$$
Substituting the values we have,
$$
\begin{aligned}
T_2 & =T_0\left[\frac{p_0+\rho \lg (H-y)^T}{p_0+\rho \lg H}\right]^{\frac{5 / 3-1}{5 / 3}} \\
& =T_0\left[\frac{p_0+\rho \lg (H-y)}{p_0+\rho \lg H}\right]^{2 / 5}
\end{aligned}
$$
$\therefore$ correct option is (b).
$$
\begin{array}{rlrl}
& T p^{\frac{1-\gamma}{\gamma}}=\text { constant } \\
\therefore \quad & T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \\
\text { or } & T_2 & =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}
\end{array}
$$
Substituting the values we have,
$$
\begin{aligned}
T_2 & =T_0\left[\frac{p_0+\rho \lg (H-y)^T}{p_0+\rho \lg H}\right]^{\frac{5 / 3-1}{5 / 3}} \\
& =T_0\left[\frac{p_0+\rho \lg (H-y)}{p_0+\rho \lg H}\right]^{2 / 5}
\end{aligned}
$$
$\therefore$ correct option is (b).
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