Question

What quantity of energy, in joules, must be removed from a 164-g sample of CH3Cl(g) that is initially at 24.95°C to its condensation point, -24.09°C and then change the gas to a liquid at that temperature?

Answer 1

specific heat of CH3Cl = 0.04206kJ/(mol.K) molecular weight of CH3Cl= 12+3+35.5= 50.5

Latent heat of vaporization of CH3Cl= 431.44 Kj/kg=431.44 j/g

moles of CH3Cl in 164 gm= 164/50.5=3.25 moles

since CH3Cl specific heat is given in kj/mol.K, mass of CH3Cl needs to be converted into moles

Assumig specific heat remains constant during change in temperature from 24.95 deg.c to -24.09 deg.c, change in enthalpy = Moles of CH3Cl* specifc heat in Kj/mol.K * temperature difference= 3.25*0.04206*(-24.09-24.95)=6.703 Kj

At -24.09 deg.c,heat to be removed= -164 g*431.44 j/g=-70756.16 Joules= -70.76 Kj

since heat needs to be removed, it is shown with -ve sign

total heat needs to be removed= -6.703-70.76=77.463 Kj

Enthalpy of condensation =431.44 Kj/kg