Answer to Question #213170 in Physics for Veer Patel

Question

An air bubble of 19.4 cm3 volume is at the bottom of a lake 41.5 m deep. The bubble rises to the surface, which is at a temperature of 22.6°C, where its volume becomes 104 cm3 ,before finally bursting out. Assuming the temperature of the bubble to be the same as that of the surrounding water, find the temperature at the bottom of the lake.

Answer 1

According to the combined gas law:

$$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2},\\\space\\ T_1=T_2\frac{P_1V_1}{P_2V_2}=T_2\frac{(\rho gh+P_\text{atm})V_1}{P_\text{atm}V_2},\\\space\\ T_1=(22.6+273)\frac{(1000·9.8·41.5+101325)19.4}{101325·104}=276.5\text{ K},$$

or 3.5°C.