Question

A sample of an ideal gas at 1.00 atm and a volume of 1.32 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 90.0 atm , what was the volume of the sample

Answer 1

Answer:

The volume of the new sample is 14.7 mL

Explanation:

Step 1: data given

Initial pressure = 1.00 atm

Initial volume = 1.32 L

Final pressure = 90.0 atm

Step 2: Calculate the new volume

P1 * V1 = P2 * V2

⇒with P1 = the initial pressure = 1.00 atm

⇒with V1 = the initial volume = 1.32 L

⇒with P2 = the final pressure = 90.0 atm

⇒with V2 = the final volume = TO BE DETERMINED

V2 = (P1*V1)/P2

V2 = (1.00 atm *1.32L)/90.0 atm

V2 = 0.0147 L = 14.7 mL

The volume of the new sample is 14.7 mL

Answer 2

Answer:

When the pressure increases to 90.0 atm , the volume of the sample is 0.01467L

Explanation:

To answer the question, we note that

P₁ = 1.00 atm

V₁ = 1.32 L

P₂ = 90 atm.

According to Boyle's law, at constant temperature, the volume of gas is inversely proportional to its pressure

That is P₁V₁ = P₂V₂

Solving the above equation for V₂ we have

$V_2 = \frac{P_1P_2}{P_2}$  that is V₂ = $\frac{1atm*1.32L}{90atm}$ = $\frac{11}{750}L$ or 0.01467L