Question

Divers often inflate heavy duty balloons attached to salvage items on the sea floor. If a balloon is filled to a volume of 1.20 L at a pressure of 6.25 atm, what is the volume of the balloon when it reaches the surface?

Answer 1

The volume of the balloon when it reaches the surface is 7.50 L.

Step-by-step explanation:

To solve this problem, we can use the combined gas law

P1V1/T1 = P2V2/T2

Given:

• V1 = 1.20 L (initial volume of the balloon)
• P1 = 6.25 atm (initial pressure of the balloon)
• T1 = ? (initial temperature of the balloon)
• P2 = 1 atm (pressure at the surface)
• V2 = ? (final volume of the balloon at the surface)
• T2 = ? (final temperature of the balloon at the surface)

We need to find V2, the volume of the balloon at the surface. To do this, we can rearrange the combined gas law equation:

V2 = (P1V1T2) / (P2T1)

Since the temperature is constant, T1 = T2. We can substitute this into the equation:

V2 = (P1V1) / P2

Substituting the given values:

V2 = (6.25 atm * 1.20 L) / 1 atm

V2 = 7.50 L

Therefore, the volume of the balloon when it reaches the surface is 7.50 L.

Answer 2

7.50 L

Step-by-step explanation:

The balloon has a volume of 1.20 L (V₁) when the pressure at the sea floor is 6.25 atm (P₁). When it reaches the surface, the pressure is that of the atmosphere, that is, 1.00 atm (P₂). If we consider the gas to behave as an ideal gas and the temperature to be constant, we can calculate the final volume (V₂) using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 6.25 atm × 1.20 L / 1.00 atm

V₂ = 7.50 L