An air bubble with a diameter of 1.9 cm rises from the bottom of the lake, where the water temperature is 6 °C, to the water surface, where the temperature reaches 23 °C. As the bubble rises, its volume increases 2.5 times.

Calculate the depth of the lake, if the atmospheric pressure is normal - p0 = 1atm, the density of the lake water is ρ = 1000kgm3 and the acceleration of free fall is g = 9.8ms2!

Thank you very much!!

Pressure x Volume is proportional to Temperature

PV = kTP1V1 = kT1 and P2V2 = kT2

Divide and the k's cancel

(P1V1)/(P2V2) = T1/T2

V1/V2 = 1/2.5

T1 = 279 Kelvin

T2 = 296 Kelvin

P1 = 1 atm = 10^5 N/m^2

This gives

P1= (279/296) x 10^5 x 2.5 =2.35x10^5 N/m^2P1 = P2 + (roh)gh

(2.35x10^5) N/m^2 = (1x10^5 N/m^2) + (1000 kg/m^3)(9.8 N/kg)(h)

h = 14 mThank you very much, now I understood!

You very welcome