Question

An air bubble rises from the bottom of a well where the temperature is 25°C to the surface where the temperature is 27°C. Find the percent increase in the volume of the bubble if the depth of the well is 5 m. Atmospheric pressure is 101,528 Pascals.    

Answer 1

$T_1 = 25 +273 = 298 \;K \\ T_2 = 27 + 273 = 300 \;K \\ p_2 = p_{atm}=101528 \;Pa = 101528 \;N/m^2$

Pressure at bottom due to depth of fluid

$p = ρgh \\ p = 10^3 \times 9.81 \times 5 = 49050 \;N/m^2$

Total pressure at bottom

$p_1 = p_{atm} + p \\ = 101528 +49050 \\ = 150578 \;N/m^2 \\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\ \frac{150578 \times V_1}{298} = \frac{101528 \times V_2}{300} \\ V_2 = \frac{150578}{101528} \times \frac{300}{298} \times V_1 \\ V_2 = 1.4929V_1 \\ = \frac{1.4929V_1-V_1}{V_1} \times 100 \\ = 49.295 \; \%$