Question

Find the volume of methane measured at 298K and 1.19 atm required to convert 1.41L of water vapor at 373K.

The heat combustion of CH4 is 890.4 kJ/mole and the heat capacity of H2O is 75.2J/mol x K.

Answer 1

let the moles of methane required =x

Heat this is supplied= 890.4*x

It is assumed that the water vapor has to be evaporated at 373K

hence latent heat of vaporization of water= 2277 j/g

Density of water vapor = 18/22.4 ( at STP) =18/22.4 g/L (at STP)

density of water vapor at 373K= 0.803571 g/L*273.15/373= 0.588 g/L

mass of 1.41 L of water= 1.41*0.588 =0.83 gm

Latetn heat of water= 2277 (j/g)*0.83=1890 joules

890.4*1000x= 1890 ( heatt of combustion= heat required to convert water into vapor)

x= 1890/(890.4*1000)=0.002123 moles

from PV= nRT    V=volume of methane= nRT/P = 0.002123*0.08206*298/1.19=0.043619 L