Question

An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at the top 16.0°C, what is the volume of the bubble just before it reaches the surface?

Answer 1

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top  t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

so

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$  Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³