Question

lion 10 of 15 > A sample of an ideal gas at 1.00 atm and a volume of 1.08 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? Assume that the temperature was held constant. VES

Answer 1

Initial

pressure P₁= 1atm

volume V₁ =1.08 L

final

pressure P₂= 90 atm

volume V2 =?

since the temperature is constant so the process is isothermal

so according to Boyles law

at constant temperatue ,pressure of a gas is inversely proportional to volume of ideal gas

P$\propto$1/V

or PV= constant

so at two different pressure and volume we can write

P₁V₁ =P₂V₂

substituting the values

1atm*1.08L=90 atm* V₂

(1atm*1.08L)/90 atm=V₂

0.012 L =V₂

The final answer is valid  because the volume and pressure has an inverse relationship as stated by Boyle's law. Since P₂>P₁, then V₂<V₁

_{Also given the gas was compressed so volume has to decrease}