Question

It takes 2260 J to vaporize a gram of liquid water to steam at its normal boiling point of 100^oC. What is ΔH for this process? What is the work, given that the water vapor expands against a pressure of 0.988 atm? What is ΔU for this process?

Answer 1

a)

What is ΔH for this process?

assume this ws performed at constan pressure

so

Q = ΔH

so

Q = 2260 J

ΔH = 2260 J (fo 1 gram of water)

What is the work, given that the water vapor expands against a pressure of 0.988 atm?

T = 100°C = 373 K

1 gram of water --> 1 mL

PV = nRT

1 g of water --> mol = 1/18 = 0.05555 mol

V = nRT/P = (0.05555)(0.082)(373)/0.98 = 1.7337 L = 1733.7 mL

so..

dV = 1733.7 - 1 = 1732.7 mL = 1.7327 L

then

W = -P*(dV) = -0.988*1.7327 = 1.71190 atm*L

1 atm*L = 101.35 Pa

so;

W = 1.71190*101.3 = 173.41547 J

What is ΔU for this process?

ΔU = ΔH - Q

ΔU =2260 - 173.41547 = 2086.58453 J