A 25-liter sample of steam at 100°c and 1.0 atm is cooled to 25°c and expanded until the pressure is 19.71 mmhg. if no water con denses, calculate the final volume of the water vapor. 770 l 240 l 1.3 x 10-3 l 1.0 l
2 answers:
I will use [pV/T] in the state 1 = [pV/T] in the state 2. State 1: p = 1.0 atm V = 25 liter T = 100 + 273.15 = 373.15 K State 2: p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm V= ? T = 25 + 273.15 = 298.15 K Application of the formula 1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K => V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter
the correct answer is 770 L. I had this question this morning.
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