A small spherical monoatomic ideal gas bubble | vedclass

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(B,B,B) $1.$ As the bubble moves upwards, it experiences the force of gravity (weight) and the buoyant force (which is the resultant of the pressure f

Vedclass · Accepted Answer

Vedclass · Answer

(B,B,B) $1.$ As the bubble moves upwards, it experiences the force of gravity (weight) and the buoyant force (which is the resultant of the pressure force exerted by the liquid). Since the problem does not specify that the liquid is moving or that the bubble is moving at a constant terminal velocity, the primary forces acting on the bubble are gravity and the buoyant force (pressure force). Thus, option $(B)$ is correct.$2.$ The process is adiabatic as there is no heat exchange. For an adiabatic process, $P^{1-\gamma} T^{\gamma} = 	ext{constant}$.At the bottom, $P_1 = P_0 + ho_{\ell} gH$ and $T_1 = T_0$.At height $y$ from the bottom, the depth is $(H-y)$, so $P_2 = P_0 + ho_{\ell} g(H-y)$.Using $P_1^{1-\gamma} T_1^{\gamma} = P_2^{1-\gamma} T_2^{\gamma}$, we get $T_2 = T_1 \left(\frac{P_1}{P_2}ight)^{\frac{1-\gamma}{\gamma}} = T_0 \left(\frac{P_0 + ho_{\ell} gH}{P_0 + ho_{\ell} g(H-y)}ight)^{\frac{1-5/3}{5/3}} = T_0 \left(\frac{P_0 + ho_{\ell} gH}{P_0 + ho_{\ell} g(H-y)}ight)^{-2/5} = T_0 \left(\frac{P_0 + ho_{\ell} g(H-y)}{P_0 + ho_{\ell} gH}ight)^{2/5}$. Thus, option $(B)$ is correct.$3.$ Buoyant force $F_B = ho_{\ell} V_2 g$. From the ideal gas law, $V_2 = \frac{nRT_2}{P_2}$.Substituting $T_2 = T_0 \left(\frac{P_2}{P_1}ight)^{\frac{\gamma-1}{\gamma}} = T_0 \left(\frac{P_2}{P_1}ight)^{2/5}$, we get $V_2 = \frac{nR T_0}{P_2} \left(\frac{P_2}{P_1}ight)^{2/5} = \frac{nRT_0}{P_1^{2/5} P_2^{3/5}}$.Thus, $F_B = ho_{\ell} g \frac{nRT_0}{P_1^{2/5} P_2^{3/5}} = \frac{ho_{\ell} nRgT_0}{(P_0 + ho_{\ell} gH)^{2/5} [P_0 + ho_{\ell} g(H-y)]^{3/5}}$. Thus, option $(B)$ is correct.

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