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IIT JEE Advanced 2008 Paper 1

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A small spherical monoatomic ideal gas bubble (γ = 5/3) is trapped inside a liquid of density

\rho_{l}

(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 liquidis H and the atmospheric pressure is

P_{0}

(neglect surface tension).

Section: Physics

Question : 16 of 69

Marks: +1, -0

When the gas bubble is at a height y from the bottom, its temperature is

[JEE Adv 2008 P1]

$T_0 \left( \frac{P_0 + \rho_{lg} h}{P_0 + \rho_{lg} y} \right)^{2/5}$
$T_0 \left[ \frac{P_0 + \rho_{lg} (H-y)}{P_0 + \rho_g H} \right]^{2/5}$
$T_0 \left( \frac{P_0 + \rho_{lg} h}{P_0 + \rho_{lg} y} \right)^{3/5}$
$T_0 \left[ \frac{P_0 + \rho_{lg} (H-y)}{P_0 + \rho_g H} \right]^{3/5}$

Validate

Solution:

Applying ideal gas law, pV = nRT, at the bottom, we have

(\rho_0 + \rho g h) V_0

n R T_0

At a height y, we have

[\rho_0 + \rho_g (H-y)] V

= nRT

Since no heat exchange occurs with the liquid, the process is adiabatic:

{PV}^\gamma

= Constant

(P_0 + \rho_{lg} H) V_0^\gamma

[P_0 + \rho_{lg} (H-y)] V^\gamma

where

V =

\left[ \frac{P_0 + \rho_{lg} H}{P_0 + \rho_{lg} (H-y)} \right]^{1/\gamma} V_0

Also ,

T V^{\gamma-1}

= Constant

Therefore,

T =

\left[ \frac{P_0 + \rho_{lg} (H-y)}{P_0 + \rho_{lg} H} \right]^{(\gamma-1)/\gamma} T_0

Substituting r = 5 / 3, we get

T =

\left[ \frac{P_0 + \rho_{lg} (H-y)}{P_0 + \rho_{lg} H} \right]^{2/5} T_0

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