JEE Advanced 2022 — Paper 2 — PHY-II solution

JEE Advanced2022Paper 2PHY-II

A bubble has surface tension $S$ . The ideal gas inside the bubble has ratio of specific heats $γ = \frac{5}{3}$ . The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is $P_{a 1}$ , the radius of the bubble is found to be $r_{1}$ and the temperature of the enclosed gas is $T_{1}$ . When the atmospheric pressure is $P_{a 2}$ , the radius of the bubble and the temperature of the enclosed gas are $r_{2}$ and $T_{2}$ , respectively. Which of the following statement(s) is(are) correct?

A
If the surface of the bubble is a perfect heat insulator, then $(\frac{r _{1}}{r _{2}})^{5} = \frac{P _{a 2} + \frac{2 S}{r _{2}}}{P _{a 1} + \frac{2 S}{r _{1}}}$ .
B
If the surface of the bubble is a perfect heat insulator, then the total internal energy of the bubble including its surface energy does not change with the external atmospheric pressure.
C
If the surface of the bubble is a perfect heat conductor and the change in atmospheric temperature is negligible, then $(\frac{r _{1}}{r _{2}})^{3} = \frac{P _{a 2} + \frac{4 S}{r _{2}}}{P _{a 1} + \frac{4 S}{r _{1}}}$ .
D
If the surface of the bubble is a perfect heat insulator, then $(\frac{T _{2}}{T _{1}})^{5/2} = \frac{P _{a 2} + \frac{4 S}{r _{2}}}{P _{a 1} + \frac{4 S}{r _{1}}}$ .

Solution

For a soap bubble (two surfaces), the gauge pressure inside is $\frac{4 S}{r}$ , so the internal pressure is $P_{in} = P_{a} + \frac{4 S}{r}$ . (Note: the option (A) with $2 S / r$ assumes a single liquid surface, not a soap bubble.)

Adiabatic case: $P V^{γ} =$ const with $γ = 5/3$ and $V \propto r^{3}$ gives $(P_{a} + \frac{4 S}{r}) r^{5} =$ const. Hence $(\frac{r _{1}}{r _{2}})^{5} = \frac{P _{a 2} + 4 S / r _{2}}{P _{a 1} + 4 S / r _{1}}$ . Option (A) uses $2 S / r$ and is therefore incorrect.

Also $T V^{γ - 1} =$ const, so $\frac{T _{2}}{T _{1}} = (\frac{V _{1}}{V _{2}})^{2/3} = (\frac{r _{1}}{r _{2}})^{2}$ . Combining with the adiabatic result yields $(\frac{T _{2}}{T _{1}})^{5/2} = \frac{P _{a 2} + 4 S / r _{2}}{P _{a 1} + 4 S / r _{1}}$ . (D) correct.

Isothermal case (perfect conductor): $P V =$ const gives $(P_{a} + \frac{4 S}{r}) r^{3} =$ const, i.e. $(\frac{r _{1}}{r _{2}})^{3} = \frac{P _{a 2} + 4 S / r _{2}}{P _{a 1} + 4 S / r _{1}}$ . (C) correct.

Option (B): In an adiabatic process the gas does work on the surroundings, so the sum of internal energy and surface energy is not conserved. Incorrect.

$∴$ Correct options: (C), (D).

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