Question

A sample of an ideal gas at 1.00 atm and a volume of 1.91 L was placed in a weighted balloon and dropped into the ocean. As the sample descended, the water pressure compressed the balloon and reduced its volume. When the pressure had increased to 40.0 atm , what was the volume of the sample

Answer 1

The new volume is 47.8 mL

Step-by-step explanation:

Step 1: Data given

The initial pressure of the gas = 1.00 atm

The initial volume of the gas = 1.91 L

The pressure increased to 40.0 atm

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒with P1 = the initial pressure of the gas = 1.00 atm

⇒with V2 = the initial volume = 1.91 L

⇒with P2 = the increased pressure = 40.0 atm

⇒with V2 = the new volume = TO BE DETERMINED

1.00 atm * 1.91 L = 40.0 atm * V2

V2 = (1.91 atm* L / 40.0 atm)

V2 = 0.0478 L = 47.8 mL

The new volume is 47.8 mL

When the pressure increases 40x, the volume will decrease 40x

Answer 2

0.0478 L

Step-by-step explanation:

Given data

• Initial volume (V₁): 1.91 L

• Initial pressure (P₁): 1.00 atm

• Final volume (V₂): ?

• Final pressure (P₂): 40.0 atm

We can find the final volume of an ideal gas using Boyle´s law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁ / P₂

V₂ = 1.00 atm × 1.91 L / 40.0 atm

V₂ = 0.0478 L

The final volume is 0.0478 L = 47.8 mL.