A scuba diver creates a spherical bubble with a radius of 3.0 cm at a depth of 30.0 m where the total pressure (including atmospheric pressure) is 4.00 atm. Part A What is the radius of the bubble when it reaches the surface of the water? (Assume atmospheric pressure to be 1.00 atm and the temperature to be 298 K.)

Radius of the bubble reaches the surface of the water = 4.68 cm

Explanation:

We know Ideal gas equation PV = nRT

Since amount of gas and temperature remains constant

so p₁v₁ = p₂v₂

Volume of Spherical bubble with a radius of 3.0 cm

⇒ v₁=

⇒ v₁ = x 3.14 x

Pressure at the depth p₁ = 4 atm

Volume of bubble when it reaches the surface of water =  

Pressure p₂ = 1 atm

      p₁v₁ = p₂v₂

⇒  4 x   x 3.14 x =  1 x  

⇒ r = 4.68 cm

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."