Question

The system consists of 3.60 g of H2O in a diathermic cylinder sealed from the outside by a freely movable massless piston. In the initial state, the system is completely in the vapor phase, H2O (g), at 100. o C and 1.00 atm and is in equilibrium with the surroundings. In the final state, as a result of transferring heat away from the cylinder, the system is completely in the liquid phase, H2O (l), at 100.o C and 1.00 atm. Calculate the compression work, w, (in Joules with three significant figures) associated with the complete condensation process of 3.60 g of water vapor at 100.o C and 1.00 atm. In this question, you will assume the final volume of 3.60 g of water liquid at 100.o C and 1.00 atm to be zero in your calculations. To calculate the volume of vapor, you can again assume that water vapor behaves as an ideal gas. How do the results obtained in Questions 1 and 2 compare? What can you conclude from this comparison?

Answer 1

moles of H2O= mass/molar mass=3.60g/18g/mol=0.2 moles

initial state, T=273+100=373K

And p=1atm

final state=T=373K

And p=1 atm

final volume=V2=density*mass=0.9584g/cm3 *3.60g=3.45 cm3

initial volume can be calculated from ideal gas equation considering vapours behave as an ideal gas,

PV=nRT

where P=1atm

R=universal gas constant=0.0821L atm/K.mol

n=0.2 moles

T=373K

V=nRT/P=0.2*(0.0821 Latm/K.mol)*373K/1 atm=6.12 cm3

compression work=pV=p(v2-v1)=1 atm(3.45 cm3-6.12 cm3)=(-2.67 atm *cm3)

Applying unit conversion to get work done in joules W=(-2.67 atm *cm3)* (10^-3 L/cm3) *(8.314 J/K.mol)/0.0821 L atm/K.mol)

calculate

=-0.27 J