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?

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 × 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 = 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 = + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 ×  Pa

so from gas law

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

so put here value and find v1

V1 = 5.71 cm³

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

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .