A sample of an ideal gas at 1.00 atm and a volume of 1.82 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 15.0 atm , what was the volume of the sample? Assume that the temperature was held constant.

The answer to your question is Volume = 0.12 L

Explanation:

- Data

Pressure 1 = 1 atm

Volume 1 = 1.82 L

Pressure 2 = 15 atm

Volume 2 = ?

Temperature = constant

- Formula

- To solve this problem use the Boyle's law

            P1V1 = P2V2

- Solve for V2

            V2 = P1V1 / P2

- Substitution

             V2 = (1)(1.82) / 15

- Simplification

             V2 = 1.82 / 15

- Result

             V2 = 0.12 L

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

the answer is definitely c.