Question

An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?

Answer 1

Volume of bubble at the bottom ( V1) = 1 cm³
= 10^-6 m³
Pressure at the bottom ( P1) = 1 atm + dhg
= 1.013 × 10^5 + 40 × 10^3 × 9.8
= 493.3 × 10³ Pa

Temperature at the bottom ( T1) = 12°C
= 12 + 273.15 K = 285.15 K

Pressure of the bubble at the surface ( P2) = 1 atm = 1.013 × 10^5 N/m²
Temperature ( T2) = 35°C
= 35 + 273.15
= 308.15 K

Now, use gas equation,
P1V1/T1 = P2V2/T2
V2 = P1V1.T2/P2
= 493.3 × 10³ × 10^-6 × 308.15 /1.013 × 10^5
= 5.263 × 10^-6 m³

Answer 2

Hope it will help........◆◆◆◆◆◆◆

;)