Work for the Condensation of Water WITH Approximation The system consists of 3.6

Question

Work for the Condensation of Water WITH Approximation 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 3 significant figures) associated with the complete condensation process of 3.60 g of water vapor at 100.o C and 1.00 atm. To calculate the final volume of 3.60 g of water liquid at 100.o C and 1.00 atm, use the value, d = 0.9584 g/cm3 , for the density of water. To calculate the volume of vapor, you can assume that water vapor behaves as an ideal gas. Hint: You need to pay attention to the actual amount of material that is involved in the process considered in THIS problem.

Explanation / Answer

GIVEN m = 3.6 g H20 = 0.0036 kg H2O
now state 1 : Water in steam form at 100 C
State 2 : Water in liquid form at 100 C

for state 1,
Pressure P = 1 atm = 1.01*10^5 Pa
Volume V = ?
temperature T = 100 C = 373.16 K
number of moles, n = m/18 = 3.6/18 = 0.2 moles
so using ideal gas equation for steam
PV = nRT
1.01*10^5*V = 0.2*8.31*373.16
so initial volume V = 6.14051*10^-3 m^3

state 2,
water is in liquid state with density rho = 0.9854 g/cm^3 = 985.4 kg/m^3
now, m = 0.0036 kg
so Volume Vf = m/rho = 0.0036/985.4 = 3.65*10^-6 m^3

so change in volume dV = Vf - V = -0.00613686 m^3
so, pressure volume work, ( compression work) done on the water in this process
W = PdV = -619.82286 J
-ve sign iundicates work is done on the water

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