Question

a

34. (III) An air bubble at the bottom of a lake 41.0 m deep has a volume of 1.00 cm. If the temperature at the bottom is 5.5°C and at the top 18.5°C, what is the radius of the bubble just before it reaches the surface?

Answer 1

34. (III) Pressure at the bottom of a lake which will be given as -

P₁ = P_atm + P_w

P₁ = P_atm + _$\rho$w g h

where, _$\rho$w = density of water = 1000 kg/m³

g = acceleration due to gravity = 9.8 m/s²

h = depth of a lake = 41 m

P_atm = atmospheric pressure = 1.013 x 10⁵ Pa

then, we get

P₁ = [(1.013 x 10⁵ Pa) + (1000 kg/m³) (9.8 m/s²) (41 m)]

P₁ = 5.031 x 10⁵ Pa

Using a combined gas law & we get

P₁ V₁ / T₁ = P₂ V₂ / T₂

V₂ = P₁ V₁ T₂ / P₂ T₁

where, P₂ = pressure at the surface of a lake = 1.013 x 10⁵ Pa

V₁ = volume of an air bubble = 1 cm³

T₁ = temperature at the bottom = 5 ⁰C = 278.65 K

T₂ = temperature at the top = 18.5 ⁰C = 291.65 K

then, we get

V₂ = [(5.031 x 10⁵ Pa) (1 cm³) (291.65 K)] / [(1.013 x 10⁵ Pa) (278.65 K)]

V₂ = 5.198 cm³

Radius of the bubble just before it reaches the surface which will be given by -

V₂ = (4/3) $\pi$ r₂³

(5.198 cm³) = (4/3) (3.14) r₂³

r₂³ = [(5.198 cm³) / (4.186)]

r₂³ = 1.24176 cm³

r₂ = 1.07 cm

56. (II) Using a Gay-lussac's law & we get

P₁ / T₁ = P₂ / T₂

where, P₁ = air pressure = 0.80 atm

P₂ = atmospheric pressure = 1 atm

T₁ = temperature at which water boils = ?

T₂ = boiling temperature of water = 100 ⁰C = 373 K

then, we get

T₁ = [(0.80 atm) (373 K)] / (1 atm)

T₁ = 298.4 K