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JEE Advanced · Physics · 11. Properties of Fluids

Paragraph:
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

  1. A \(T_0\left(\frac{p_0+\rho \lg H}{p_0+\rho \lg y}\right)^{\frac{2}{5}}\)
  2. B \(T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}\)
  3. C \(T_0\left(\frac{p_0+\rho l g H}{p_0+\rho \lg y}\right)^{\frac{3}{5}}\)
  4. D \(T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{3}{5}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(T_0\left(\frac{p_0+\rho l g(H-y)}{p_0+\rho l g H}\right)^{\frac{2}{5}}\)

Step-by-step Solution

Detailed explanation

As there is no exchange of heat. Therefore, process is adiabatic. Applying,
\(T p^{\frac{1-\gamma}{\gamma}}=\text { constant } \)
\( \therefore T_2 p_2^{\frac{1-\gamma}{\gamma}}=T_1 p_1^{\frac{1-\gamma}{\gamma}} \)
\( \text {or } T_2 =T_1\left(\frac{p_1}{p_2}\right)^{\frac{1-\gamma}{\gamma}}=T_1\left(\frac{p_2}{p_1}\right)^{\frac{\gamma-1}{\gamma}}\)
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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