Question

Two soap bubbles of radius 5 cm and 12 cm at same temperature in vacuum coalesce to form a single bubble. The surface tension of the soap solution is 0.1 N/m. The system does not exchange heat with surroundings. The gas inside is monoatomic. What is the change in temperature (in K) of the system ?

Answer 1

The question is unsolvable for a specific numerical value in Kelvin as the initial temperature ($T_0$) is not provided.

Answer 2

The work done by surface tension during bubble coalescence is converted into internal energy of the monoatomic gas. This energy increase causes a temperature rise, calculated using the ideal gas law and the formula for internal energy of a monoatomic gas. The change in temperature is found to be proportional to the initial temperature, which is not provided, making the question unsolvable for a specific value in Kelvin. The change in temperature is given by $\Delta T = T_0 \frac{r_1^2 + r_2^2 - r_3^2}{r_1^2 + r_2^2}$, where $r_1=5$ cm, $r_2=12$ cm, and $r_3=(r_1^3+r_2^3)^{1/3}$.