How much volume is needed, if given 92.0 L of Hydrogen gas in a container of 612 kPa and reduces to 228kPa

246.94 litres is the volume is needed, if given 92.0 L of Hydrogen gas in a container of 612 kPa and reduces to 228kPa.

Explanation:

Data given:

Initial volume of hydrogen gas in the container P1 = 92 litres

initial pressure of hydrogen gas in the container  V1 = 612 kPa

final pressure of hydrogen gas on the container P2= 228kPa

final volume of hydrogen gas on the container V2 =?

from the data we will require equation of Boyle's law to calculate the unknown volume:

P1V1 = P2V2

V2 =

putting the values in the equation:

V2 =

V2 = 246.94 Litres.

246.94 litres will be the volume of the hydrogen gas when pressure geta reduced to 228 kPa.

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.