Aballoon filled with 39.1 moles of helium has a volume of 876 l at 0.0'c and 1.00 atm pressure. the temperature of the balloon is increased to 38.0'c as it expands to a volume of 998 l, the pressure remaining constant. calculate q, w, and #e for the helium in the balloon. (the molar heat capacity for helium gas is 20.8 j/°c # mol.)

Q = 30.90 kJ, W = 12.36 kJ, ΔE = 18.54 kJ.

Explanation:

The expansion is happening at constant pressure without a phase change, thus, the heat can be calculated by:

Q = n*c*ΔT

Where n is the number of moles, c is the molar heat capacity, and ΔT is the temperature variation (final - initial), thus:

Q = 39.1*20.8*(38.0 - 0.0)

Q = 30904.64 J

Q = 30.90 kJ

The work done by a expansion at constant pressure is:

W = P*ΔV

Where P is the pressure (1 atm = 101325 Pa), and ΔV the volume variation (final - initial). Vfinal = 998 L = 0.998m³, Vinitial = 876 L = 0.876 m³.

W = 101325*(0.998 - 0.876)

W = 12361.65 J

W = 12.36 kJ

By the first law of the thermodynamics, the variation of the internal energy ΔE is:

ΔE = Q - W

ΔE = 30.90 - 12.36

ΔE = 18.54 kJ

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.