What will have to happen to the temperature of a sample of methane if 1000 mL at 98.6 kPa and 25°c is given a pressure of 108.5 kPa and a volume of 900 mL?

The temperature of the methane gas will drop to 22.127 °C.

Explanation:

To solve the question, we note tht the combined gas equation is given by

Where:

P₁ = Initial pressure = 98.6 kPa

V₁ = Initial volume = 1000 mL

T₁ = Initial temperature = 25°c = 298.15 K

P₂ = Final pressure = 108.5 kPa

V₂ = Final volume = 900 mL

T₂ = Final temperature = Required

Therefore we have

= =  295.277 K

The temperature T₂ < T₁

Therefore what will happen is that the temperature of the methane gas will drop by 2.873 °C from 25 °C to 22.127 °C.

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.