Question

Calculate the values of LlU, LlH, and LlS for the following process: 1 mole of liquid water 1 mole of steam -+ at 25°C and 1 atm at 100°C and 1 atm The molar heat of vaporization of water at 373 K is 40. 79 kJ mo1-1, and the molar heat capacity of water is 75.3 J K-1 mo1-1. Assume the molar heat capacity to be temperature independent and ideal-gas behavior.

Answer 1

1. 46.44 Kj

2. 126.3 J|K

3. 43.34 Kj

Step-by-step explanation:

change in temperature

25+274 = 298K

100+273 = 373

1.

∆H = nC∆T + n∆H

1x75.5x75+1x40.79

= 5647.5+40.9

we convert 5647.5 to Kj

= 5.6475+40.79

= 46.44Kj

2.

∆S = 1x0.0753x(ln373/298) + 1molx40.79/373

= 0.01690 + 0.1094

= 0.1263kj/k

∆S = 126.3J|K

3.

∆U = ∆H - RTn

= 46.44-0.0831x373x1mol

= 46.44-3.09963

= 43.34

these resulting values are the correct answer to this question.