Question

An air bubble with a radius of 2.5 cm at the bottom of a lake where the temperature is 5.4 0C and the pressure is 3.2atm rises to the surface, where the temperature is 25.0 0C and the pressure is 1.0 atm. Calculate the radius of the bubble when it reaches the surface. Assume ideal gas behavior.

Answer 1

Answer:

3.8 cm

Explanation:

Given data

• Initial pressure (P₁): 3.2 atm

• Initial volume (V₁): 65 cm³

Volume of a sphere: V = 4/3 × π × r³ = 4/3 × π × (2.5 cm)³ = 65 cm³

• Initial temperature (T₁): 5.4°C + 273.15 = 278.6 K

• Final pressure (P₂): 1.0 atm

• Final volume (V₂): ?

• Final temperature (T₂): 25.0°C + 273.15 = 298.2 K

We can find the final volume using the combined gas law.

$\frac{P_{1}\times V_{1}}{T_{1}} =\frac{P_{2}\times V_{2}}{T_{2}}\\\frac{3.2atm\times 65cm^{3} }{278.6K} =\frac{1.0atm\times V_{2}}{298.2K}\\V_{2}=223cm^{3}$

The final radius of the bubble is:

V = 4/3 × π × r³

223 cm³ = 4/3 × π × r³

r = 3.8 cm