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.

1. 46.44 Kj

2. 126.3 J|K

3. 43.34 Kj

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.

geosphere and hydrosphere.

this is because the geosphere has to do with geological events, while the hydrosphere controls all water on earth.

warm front forms when a moist, warm air mass slides up and over a cold air mass. as the warm air mass rises, it condenses into a broad area of clouds. a warm front brings gentle rain or light snow, followed by warmer, milder weather.

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