An air bubble at the bottom of a lake 40.5 m deep has a volume of 1.00 cm3. part a if the temperature at the bottom is 2.3 ∘c and at the top 28.1 ∘c, what is the radius of the bubble just before it reaches the surface? express your answer to two significant figures and include the appropriate units.

5.4 cm³

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure, V is volume, n is number of moles, R is ideal gas constant, and T is absolute temperature.

Since n and R are constant, we can say:

PV / T = constant

At the bottom of the lake, the pressure is:

P = ρgh + Patm

P = (1000 kg/m³) (9.8 m/s²) (40.5 m) + 101,325 Pa

P = 498,225 Pa

And the temperature is:

T = 2.3 + 273.15 K

T = 275.45 K

At the top of the lake, the pressure is:

P = Patm

P = 101,325 Pa

And the temperature is:

T = 28.1 + 273.15 K

T = 301.25 K

Therefore:

PV / T = PV / T

(498225 Pa) (1.00 cm³) / (275.45 K) = (101325 Pa) V / (301.25 K)

V = 5.4 cm³

the ima (higher than 1m) will be greater than the ima with the height of 1m.